How to Solve Geometry Problems: A Step-by-Step Guide

To solve geometry problems, it's essential to break down the problem into smaller parts and use basic geometric concepts. Let's take a look at how we can solve two different types of geometry problems.

First Problem: Breaking Down Triangles

Let's start with a problem involving breaking down triangles into smaller triangles. We have a triangle OAB with sides OA, OB, and AB. Our goal is to find the length of side AB using basic geometric concepts.

We can break down the triangle into two smaller triangles, BOC and AOC. By finding the area of both triangles, we can then equate them to solve for the value of x. The area of a triangle can be found using the formula: Area = (base * height) / 2. Since both triangles share the same base (OA), we can set up an equation based on their areas.

Let's assume the length of side AB is x. We know that OA + OB = x, so we can substitute this into our equation for the area of triangle OAB. By equating the areas of both triangles and solving for x, we get x = 7. This value is then used to find the lengths of sides AB and AC.

Second Problem: Tangents to a Circle

Now, let's move on to a problem involving tangents to a circle. We have two parallel lines, PQ and XY, that intersect at points P and Q. A third tangent line, AB, intersects both lines at points C and D. Our goal is to find the measure of angle AOB.

To solve this problem, we can construct a line joining point O (the center of the circle) and point C. By doing so, we create two congruent triangles, OCP and OPA. These triangles are congruent because they share a common side (OA), have equal angles due to being external points tangents from an external point, and meet at a common angle.

Since these triangles are congruent, their corresponding angles are also equal. We can then equate the measures of these angles to solve for the value of angle AOB.

Using the fact that PQ is parallel to XY, we know that angle P and angle Q are supplementary (their sum equals 180 degrees). By adding this information to our previous equation, we get: Angle 1 + Angle 2 + Angle 3 + Angle 4 = 180 degrees. Substituting the values of Angle 1 and Angle 2 from the congruent triangles, we can solve for Angle 3.

Simplifying further, we find that 2 * Angle 2 = 90 degrees. Therefore, Angle AOB is equal to 90 degrees, which implies that angle AOB is a right angle.

"WEBVTTKind: captionsLanguage: enwelcome to pramar academy today we will discuss about the 10th chapter of great time that is circles up in ninth grade may circles could discuss kiawah that what is circle what is the radius of circle what is the center of circle what is the diameter relationship between diameter and radius arc of circle chord of circle segment of circle and the sector of circles is means we will discuss the properties of straight line with the circle and we will do the applications of that also so let's start first of all there are three possible conditions as combinedly we will represent a circle and a straight line the first condition is jumper upgrade so straight line there is nothing common between the circle and the straight between straight line and now we will discuss the second point first of all here the straight line test the circle at so that's why this is this line is called a tangent line and the common point this common point is called point of contact point of this line is called second and these two points are called points of intersection points of intersection tangent is a line that does the circle boundary of the circle at just one point now a circle has how many tangents is x circle now we know that a circle has infinite number of points at end at every point it has one tangent so a circle has infinite number of tangents so look at here this is circle and at this point there is a tangent at another this point there is another tangent at this point there is another tangent like this so at every point each and every point of circle a circle has a tangent and we know that a circle has infinite number of points so it has infinite number of tangents so first is a circle has dash number of tangents that is infinite this one the second is a circle has dash number of parallel tangents so the answer is look at here first of all this is circle and we will draw a diameter that passes through center now at this point we will draw tangent at the other end we will draw another tangent now these two are the parallel tangents these two are the parent is that means a circle has only two parallel tangents so answer is a circle has how many parallel tangents that is two only the next one this first shape here this a is the point of contact this is the tangent at point a i'm saying this is tangent at point a so it is clear now what is tangent at point means that point is on the boundary of the circle in the second one case when it is tangent from point from from this point circle boundary that is external point to the circle so from that external point look at this that this is external point so from that external point three tangents draw construction to the circle so very easy look at here from that point i can draw two tangents to the circle the first tangent is p a circle p a tangent and the second one is pb tangent so this is also one of the very important question from an external point we can draw easily draw two tangents only the next one is if we are talking about the tangent at a point there is only one single tangent and if we're talking about the tangent from a point there are exactly two changes we can draw the next one is normal is perpendicular to tangent and passes through the circle so a line that passes through a center and perpendicular to tangent is the normal look at here a line this line that is passes through the center and point of contact and it is perpendicular to a tangent line that line is called normal normal or tangent that are perpendicular to each other this is the first property that normal and tangent are perpendicular to the each other and normal joe all that always passes through the center and the point of contact so this one is normal is always perpendicular this is the to the tangent this is the first property is always perpendicular to the tangent and it always passes through the center of circle and the point of contact next one is the length of tangent drawn from an external point to a circle are equal means care this one means we are drawing two tangents that first is p a tangent and second is p b tangents so in length if p a is equal to 10 centimeter then p b is also equal to 10 now there are two triangles triangle p a o and triangle p b o s we need three things that are equal so first is that first thing is po is equal to po that is the common one po is equal to po that is common side second thing is oa is equal to ob that is radius of circle radii of circle and radius is always equal to a circle and the third thing is that is angle o a p is equal to equal to angle o b p that is 90 degree each as we know that that this is a tangent line and this is the normal line that passes through the center and point of context so this angle is always 90 degree this angle is always 90 degree now this this satisfies the rule that side and that are that is the property of rhs rhs means we have having the one right angle one side and one is the hypotenuse so one hypothetic is equal right angle we are also having and this is one of the side so by r h s rule triangle o a p o a p is congruent to triangle o b p by rhs rule these two triangles are congruent now if these two triangles are congruent their corresponding parts are equal as we know that that cpct that is corresponding parts of congruent triangles are equal so look at here corresponding to pa there is a bp so pa is equal to pb by cp ctc that is corresponding parts of congruent triangles are equal here c p a is equal to p b and what is p and p b that are the lengths of triangles slants of tangents so we will say that this is a result that from if through tangents are drawn from external point to a circle their lengths are equal so this is the proof for that property look at here and note it down the next one is perpendicular from the center of circle to any code of the circle bisects the code means circle centers a circle means i have to prove that ap is equal to bp look at here first of all what is given to us that given is op is perpendicular to a b that op is perpendicular to a b and what we have to prove we have to prove that that line is bisected code is bisected at point p that means ap is equal to bb so we will start the solution here before that we will construct oa and ob means i have to join oa and ob both are the radius of circle now is and triangle o b p the first one common thing is that o p is equal to o p that is common side common side of both the triangles second thing is o a is equal to ob that is the radius of circle circle and the third thing is angle o p a is equal to angle o p b both are 90 degree angle o p a is equal to angle o p b both are each 90 degree so now by again by the property of rhs by rule of rhs both triangles are congruent triangle o p a is congruent to triangle o p b and we know that if two triangles are congruent their corresponding parts are equal so the corresponding part corresponding to p a there is p b so p a is equal to p b by c p c t c that is corresponding parts of congruent triangle are equal so the result is if we draw a perpendicular from center to any code of the circle that perpendicular bisects the code the next one is prove that the angle between the two tangents drawn from an external point to the center is supplementary to the angle subtended by the line segment joining the points of contact to the center what this question means this is first of all this is very interesting and easy one question so according to this question if from an external point p we will draw two tangents that is p a and p b and we will join a and b with the center of circle then these two angles are supplementary so look at the statement proved at the angle between the tangents that means this one angle angle between the tangents drawn from an external from from an external point to a circle is supplementary that means their sum is 180 degree and the angle subtended by the segments joining the point to the contact to the center you owning the point of context to the center means these two angles are supplementary so first of all we know that p a is tangent to the circle and this one o a is the line that is passes through the point of contact a and the center also so this clearly o a line is the normal and we know that normal is perpendicular to the circle so angle o a p is 90 degree similarly angle o b p is also 90 degree because ob is normal to tangent p b now look at here carefully p a o b is a correlator p a o b is a coordinator and we know that the angle some property of a quadrature that sum of all the four angles of quadrilateral is 360 degree so now we will use that angle some property of coordinate angle sum property of quadrilateral that sum of all the four angle is 360 degree so angle a p b means this angle plus angle p b o means this one angle plus angle b o a plus angle o a p that is 360 degree and angle p a o and p b are already 90 and 90 each so angle this one angle pbo and that is this one op a both are 90 degrees so angle a p b plus 90 plus that is angle b o a plus 90 is equal to 360 degree so angle a p b plus 90 plus 90 that is 180 plus angle b o a is 360 degree now 180 gets subtracted from right hand side so angle a p b plus angle b o a is equal to 360 minus 180 that is angle a p b plus angle b o a is equal to 180 degree so look at here angle apb means this one angle and angle boa bo means this one angle both sum of both angle is 180 degree means these two angles are supplementary angles this is what about the question that some of these two angles is always 180 degree means if this is 60 degree angle we clearly say that this is 120 degree because the sum is 180 degree if this is 50 this is 130 if this is 80 this is 100 okay so now one more thing this property holds only when these two lines meet at ori's center of circle if these two lines meet out of center this property can't hold so the next one is find the value of x this thing is given that pb is tangent and pa is also tangent that means two tangent from an external point p and here this o is the center of the circle and this angle is 110 given and we have to find the value of x as we are as we use the last rule that if we draw two tangent from an external point and join the point of context with the center these two angles are supplementary so clearly angle b p a plus angle a o b is equal to 180 degree and you will use the statement of last question here and now we will solve angle p that is already given x plus this is given 110 that is equal to 180 degrees so x is equal to 180 minus 110 so x is equal to 70 degrees so this is the angle with the 70 degree so it's very easy to find the opposite angle because this relationship is known that both the angles are supplementary the next one is the length of tangents from a point a at distance five centimeter from the center of the circle is four find the radius of circle so look at the statement carefully what is given to us and what we have to find so first of all we will draw the diagram this is center of circle first of what is given the length of tangent from a point a from a point a look at carry very carefully from at from from point a we have to draw a tangent from point a that look at let this is a point so b point so a b is the tangent at a distance five center from the center your point at that point a is five centimeter from the center so your point a that is five centimeter from the center that is given five of the circle is four the length of tangent of a four f length of tangent is 4 this is 4 given and we have to find the radius of find the radius of circle means we have to find the radius so this is the diagram for this question so clearly a b is the tangent again and this is the ob is the line that passes through the point of contact and the center and we know that this line is the normal to the tangent a b and we know that angle between tangent and normally is always 96 degree so clearly triangle oba is a right angle triangle so this implies triangle o b a is right angle triangle at point at point b means this is right angle triangle at point b b is the 90 degree angle so o is the hypotenuse a b is the perpendicular and ob is the base now we will use pythagoras theorem here by pythagoras theorem here we say that perpendicular square plus b square is equal to the hypotenuse square so that means o square is equal to ob square plus a b square so what is o a that is already given 5 pi square is equal to ob that is our radius we have to find it plus a b square that is 4 4 square 5 square is 25 that is equal to r square plus 4 square is 16 16 gets subtracted in left hand side that is 25 minus 16 is equal to r square and this is 9 is equal to r square so clearly r is what that is 3 centimeter so radius of the circle is three centimeter this is one of the very interesting question related to the straight line and the circle that if a line a b that is changing to the circle and the distance of the from point that is external point to the certain center is always also given we will easily find the radius of the circle the next one is prove that the parallelogram circumscribe a circle is a rhombus in this question we have to prove that there is a parallelogram in which a circle is circumscribed and we have to prove that parallelogram is also rhombus so for that we will draw a diagram that this is a some parallelogram which some circumscribe a circle and we have to prove that this parallelogram is a rhombus so name it as a b c d these are the four corners of the parallelogram p q r s these are the four point of contacts or the common points to the circle and the parallelogram now look at here very carefully solution look at the a point a is an external 0.0 circle that is out of the boundary of the circle and clearly a p and a s are the two tangents to the circle from point a and we know that from an external point if we draw two tangents their lengths are equal so a is external point from which we are drawing the two tangents so a is external point external point and ap is equal to a s we will use the statement here that that the two tangents are drawn from an external point they are equal in length length of tangents from an external point are equal so that's why ap is equal to a s similarly b is also the external point to the circle and from b we will draw two tangents that is b p and b q so their lines are also equal so i will write uh this is equation number one now bp is equal to bq for the same reason bp is equal to bq this is your equation number two again c is also the external point we are drawing two tangents that is c q and c are their lines are also equal so cq is equal to cr4 the same reason cq is equal to cr this is equation number three we are using the same reasons for all for the last d is also external point from which two tangents d s and d are and their lines are also equal so d d r d r is equal to d s this is your equation number four and the same reason so we are having the four equations now what we have to do next step add all the equations so here add one two question number two equation number three and equation number four what we will get we will get ap plus bp plus cr plus d r is equal to a s plus b q plus c cube plus d s now look at here what is a p plus b p a p plus b p that is combining a b so this is equal to a b c r plus d r t r plus d r that is combined with c d that is equal to a s plus d s a s plus d s that is combinedly a d a d plus b q plus c cube b q plus c q that is combining bc so this is equation number y please note up to here that we have done that that first of all a is an external point from that ap and a s are two tangents to the circle and their lines are equal so ap is equal to a s this is the first equation again b is external point from there two tangents b p and b q are to the circle and their lines are also equal similarly for the point c and the point for the point d and we are having the four equations and the next step we will add all the four equations that is a p plus b p plus c i plus d r is equal to a s plus b q plus c q plus d s now a p plus b p combinedly represents a b c r plus d r that is c r plus d i combinedly represents c d similarly a s plus d s completely represents a d and c q plus that is c q plus b q combini represents b c this is our fifth equation now after this so what is given to us this is a parallelogram as abcd is a parallelogram and we know that in a parallelogram opposite sides are equal so a b is equal to cd and bc is equal to 80 and you have to write it as because abcd is a parallelogram abcd is a parallel graph in a parallelogram opposite sides are equal now we will use that two things equations here look at here a b plus what is c d what is the value of c d that is also a b so i will use replace c d by a b here that is equal to 80 plus what is the value of bc that is also equal to 80 so i will replace bc by a d this is 2 times a b is equal to 2 times a d now 2 get cancelled out so a b is equal to a d this is your last equation now look at here carefully because of the parallelogram opposite sides are already equal and now the two axis and sides are also equal because of the parallelogram opposite sides are equal a b is equal to c d and a d is equal to b c and from this equation two as the sides a b and a d are also equal so from that we will conclude that all the four sides are equal and you know that in a quadrilateral if all the four sides are equal that coordinate returns as a rhombus so from here we will conclude that a b is equal to bc equal to cd is equal to da and this completes that abcd is a rhombus a b c d is a chromosome this is the total proof of this theorem so i will repeat prove that the parallelogram circumstance a circle is rhombus and this is the complete solution this is a very important equation in ncit book this question repeats two time in first question you have to prove it up to here only that a b plus c d is equal to a d plus b c and in the next question you have to also prove that parallelogram is a rhombus if satisfy the same conditions so for that you have to prove up to here at the last point the next one is prove that the perpendicular at the point of contact to the tangent to a circle passes through the center we have to prove for this question we have to prove that if we draw a perpendicular from the point of contact that always passes through the center so let's start for that we will draw first of all diagram what is given to us this is circle and this is one of the tangent to a circle and this is a is a point of contact that is common point now we have to prove that that the perpendicular from point a passes through the center so to prove that we will let that that that point never passes through the center that point passes through some another point this is the center of circle o and i will let that that the perpendicular draw from some another point or dash so what i have to prove that to prove to prove that perpendicular on tangent perpendicular to tangent at point of contact at point of contact passes through center of circle we have to prove this so i let let if possible that if possible that perpendicular passes through some another point that is o dash that is not the center of circle that if possible perpendicular passes through another point another point that is o dash that is not the center of circle because o is the center of circle so let this is x and y now look at it carefully we'll let that perpendicular y o y a o dash that is passing through the odh clearly so by this one by this one it is clear that angle y a o by a o dash is 90 degree this is your first equation equation number one and we also know that that normal that normal line passes through the point of contact and the center of circle that has also perpendicular to the tangent so join o and a this is normal this is normal line and we know that normal is perpendicular to the tangent as oa is normal to tangent normal to tangent x y and normal is always perpendicular tangent this implies angle y a o is also 90 degree this is your second equation now look at these two questions very carefully angle y a o dash that is also 90 and angle y a o that is center is also 90 degree this is possible only if the case if o and o dash are the same points means they coincide with each other so this is contradiction to our supposition this is contradiction to our supposition this is contradiction to a supposition that point passes through another point and this will both the equation possible simultaneously if o and o dash are coincide means they are same points so this is possible only if o and o dash inside each other that means o and o dash are the same points so the next one prove that opposite sides of a quadrilateral circumcise a circle substance supplementary angle at the center of circle in this question we have to prove that that angle at the center of circle this is center of circle and angle made by opposite sides at the center that this angle is supplementary that this is a b c and d and these are the obtained common points of context that are p q r and s we have to prove that these these angles these two angles are supplementary and these two angles are also supply mandatory for that i have to do i have to do construction that is i will join o r o s o p and o a so clearly there are eight angles at the center let this is angle 1 angle 2 angle 3 angle 4 angle 5 angle 6 angle 7 and angle 8 there are eight angles at the center of the circle now what we have to do next so look at here look at these two triangles these this one a blue triangle and this one green triangle look at the these two triangles these two triangles are clearly congruent how this is already done in the previous question that ap is equal to a s that is a length of tangent from an external point is equal op is equal to os that is the radius of circle and o a is equal to o a that is the common side of the triangles so by ss rule these two triangles are congruent and if these two triangles are congruent they are these one angle one and angle five are equal by cpct by corresponding parts of congruent triangles are equal so let's do it in triangle o s a and triangle o p a means i will say in triangle in blue color triangle and in green color triangle so three reasons are reason one o a is equal to o a common side of the triangles o a is equal to a common side of both the triangles second reason is a s is equal to ap that is if we can draw a two tangent from a same external point their length is equal so this is a rule tangent from our tangent from an same external point tangent from same external point so that's why their length is equal and the third reason is os is equal to ob that is radius of circle so by by side side psi rule triangle o s a is congruent to triangle o p e now you know that in congruent triangle corresponding parts are equal so angle 1 is equal to angle 5 angle 1 is equal to angle pi by cp ct this is the first question similarly in another for these combinations for this these two triangles angle 6 and 7 for the same reason angle 6 is equal to angle 7 similarly similarly angle 6 is equal to angle 7 angle 1 is equal to angle 8 and angle 3 is equal to angle 2. these are four equations equation number two question number three question number four so i think you understand how we can prove that angle 1 is equal to angle 5 just similar we will prove that these two triangles are congruent and the corresponding parts are equal now the last one step is so we know that sum of all these angle is 360 degree because this completes a circle so sum of all the eight angles is 360 degree angle one angle 2 angle 3 angle 4 angle 5 angle 6 angle 7 at angle 8 is equal to 360 degree because this completes a circle now angle 1 angle 2 angle 3 angle 3 is equal to angle 2 so i will replace this by angle 2. angle 4 that angle 4 is equal to angle 5 so i will replace it by angle pi this is angle 5 angle 6 angle 7 is equal to angle 6 also so simon angle is replaced by 6 also and angle 8 is equal to angle 1 so i can replace angle 8 by angle 1. now look at here angle 1 repeats twice angle 2 repeats twice angle 5 repeats twice and angle 6 repeats twice also so all the angles are repeated twice so i will get 2 get common that is 2 angle 1 angle 2 angle 5 and angle 6 that is equal to 360 degree now the two get divided at the right hand side so angle 1 plus angle 2 plus angle 5 plus angle 6 is equal to 180 degree now combining angle 1 and 2 angle 1 and 2 represents angle c o d that is angle c o d and 5 and 6 combining represents angle ao b angle ao b that is 180 degree and that is what we have to prove that that at the end these two angles represent the supplementary angles the next is find the value of a b and ac in this diagram so look at here carefully abc is a triangle in which a circle is circumscribed of radius 4 the radius of the circle is 4 and the length of a at this point is p so length of bp is 6 and length of cp is a so we have to find the length of a b and ac so that this point is q and this point is r now simply if from this is this b is external point and you know that from an external point if we draw two tangents their length is equal so if length of tangent bp is 6 the length of tangent br is also so 6 so this length is 6. similarly from external point c if the length of tangent c p is eight the length of this tangent c q is also eight now for the last a a is also the external point from that point we will draw two tangents that is c r and c q and their lines are also equal so let the length this length is x and this length is also x now join ob oc and o a and from draw o or join or at oq clearly or is normal to a b that is perpendicular oq is normal to ac and op is normal to bc now look at here so next what we have to do we have to find the area of triangle abc in two different ways firstly we will find the area using heron's method of grade 9 and secondly we will find the area of triangle this abc triangle using half into base into height so first of all i am finding the area of this triangle using heron's formula now for that i need length of all the three sides means a b b c and c a so what is the side of a b what is the side of a b that is 6 plus x that is 6 plus x what is the side of bc that is 6 plus 8 14 and last what is the side of ca that is 8 plus x this is a side of or this is the length of all three sides now using errors formula we need semi parameter that is s so first step s is equal to sum of all the sides that is 6 plus x plus second side 14 plus third side x 8 plus x whole divided by 2 so that is 6 plus 8 14 plus 14 28 plus x plus x 2 x whole divided by 2 that is equal to 14 plus so this is the sami parameter now the second step to find the area of triangle that is equal to s s minus first side s minus length of second side as minus length of third side that is equal to s that is 14 plus x 14 plus x second is 14 plus x minus first side that is 6 plus x and that is 6 minus 6 minus x again next is 14 plus x minus second side that is only 14 and the third is s minus that is 14 plus x minus third side that is 8 plus x here this is minus 8 minus x now this is 14 plus x this is 14 minus 6 8 plus x and minus x guys cancel out 14 and 14 get cancel out only x and this is x minus and x plus cancel out 14 plus eight that is a six so look at here this fourteen into x fourteen x plus x into x x square here this is eight is written as four into 2 and 6 is written as 2 into 3 so here 4 get common as this get 2 because of the under root here this is a pair of 2 this is also get common as 2 the remaining portion is 14 plus 14 x plus x square into 3 that is 4 under root 3 times 14 x plus x square so this is the area of triangle abc using heroes formula i will write it here that is 4 times under root 3 that is 14 x plus x square this is the area of triangle abc now i will find the area of same triangle using the formula half into base into height using the half into base into so for that i will break the triangle abc in three similar equal congruent triangles that are triangle b oc triangle aob and triangle aoc that means the area of triangle aob is equal to sum of all the three triangles that is look at here area of triangle aob is equal to sum of these three triangles that is sum of aob triangle area of aob triangle area of bo a triangle and area of co 8 angle so look at here area of triangle a b c is equal to area of three triangles the first one is boc plus area of triangle aob plus triangle area of triangle ao c so for that look at here what is the area of triangle b oc that is half into base into height that is half what is the base the base is 6 plus 18 that is 14 and what is the height that is the radius of circle that is 4 plus area of triangle aob again that is half into base what is the base that is 6 plus x and what is the height that is radius again so this is a 4 plus third is half base is 8 plus x and the height is 4 also so calculate it 2 cancel out with 4 that is 14 to the 28 2 cancel out with 4 again 2 2 that is 12 plus 2 x plus 2 cancels out with 2 that is 2 that is 16 plus 2x combinedly 8 28 plus 12 that is 40 plus that is 56 plus 4 x so this is the area of triangle abc now look at here we are having the area of triangle in two different forms the one is 56 plus 4x and second is 4 under root 3 times 14 x plus x 4. now both are the areas of same triangle so they are equal so from here this is equation number one and equation number two because both those represent the area of same triangle so 56 plus 4x is equal to 4 times under root 3 into 14 x plus x square now 4 get divided this side we will left with 456 4 divided by 4 that is 14 plus 4 divided by 4 that is x is equal to 3 under root that is under root 3 14 x plus x square now because of under root in the right hand side so squaring both sides we will remove the under root that is squaring both sides that square is 14 is 196 plus x square plus 28 x that is a square plus b square plus 2 a b that is equal to 3 into 14 x plus x square so this is 3 get divided that is 196 plus x square plus 28 x is equal to 14 theta 42 x plus 3 x square now get all the terms at one side that 3 x square minus x square that is 2 x square 42 x minus 28 x that is 14 x minus 190 6 that is equal to 0 now to get common from all the terms we will left with x square plus 7x minus 98. so this is simply a quadratic in variable x now we know how to solve the quadratic we have to split the middle term in such a way that whose sum is plus minus is 7 and their product is minus 98 so we have to break the middle term in two portions that the sum is seven and their product is ninety eight so what are the two numbers clearly the numbers are first of all x square plus seven x minus 90 8 is equal to 0 we have to break these two in two parts these two parts are x square plus 14x minus 7x minus 90 8 is equal to 0 14 x minus 7 x that is again 7 x and 40 x minus 14 x minus into 7 x is equal to minus 98 x square so what is common from first two parts the only x we left with x plus 14 and what is common from here that is minus 7 and left with again x plus 14. that is equal to 0 and the factor is x plus 14 and x minus 7 that is 0 so values for x are x is equal to minus 14 and x is equal to plus sum so values of x are x is equal to minus 14 from the first from first factor and x is equal to plus 7 from the second factor and you know look at here x is clearly the length of side length of the that portion a p a k a r and a q so it can't be negative so minus 14 gets rejected we will be left with x is equal to 7 so this value of x is 7 now what is the main part main question what we have to find actually we have to find the length of size a b and a b and ac now look at here what is the length of side a b that is 7 plus 6 that is 13 so length of a b is 13 and what we find is find second that is length of ac that is 7 plus a that is 50 so length of ac is 15 so this is how to solve this question i will repeat how to find the value of x or how to find the length of these two portions first of all break the triangles in or break the triangle oab into these triangles small triangles b o c c o a and o a b and find the area of whole triangle using half into base into height and again find that area of triangle a b c using the hero's formula because of these are the areas of the same triangle so both are equal and after equating we will at the end we will get a quadratic and after solving we will get the value of x the one value is negative and other is positive because x is the length that can't be negative so x is equal to minus 14 get rejected so actual value of x is that is 7 on put the value of x7 here we will get the length of a b that is 7 plus 6 that is 13 and the length of ac is 15 that is all about this question come to the next last question the last question is tangents x y and x y dash x dash y dash at point p and q are parallel to each other and the third tangent a b that is at point c we have to find the value of angle aob so this is also very interesting question for that we will construct join oc so now look at here there are two triangles a and o p a we have to prove these two triangles as congruent so in triangle o c a and triangle o p a we need three things to equal to prove that it is congruent triangles so first thing is oc is equal to ob because it is the radius of circle so oc is equal to op that is radii of circle second thing that clearly a is an external point and we know that from external point the length of tangent is same so ap is equal to ac that is ap is equal to ac that is tangent from an external point tangents from same external point same external point that that's why their lines are equal and oa is the common side of both the triangles so o a is equal to o a that is common side that is common side of both the triangles so by clearly by side side side rule both the triangles are congruent triangle o c a o c is congruent to triangle opa now these two triangles are congruent so let these angles are angle 1 angle 2 angle 3 and angle 4. now if as these two triangles are congruent so angle 1 and angle 2 are equal because of the cpct does the corresponding parts of congruent triangles are equal so clearly angle 1 is equal to angle 2 this is our first equation now using the same method the we get these two triangles are also congruent and angle 3 and 4 are equal similarly similarly angle 3 is equal to angle 4 this is equation number 2. now clearly p o q is a straight line so by the angle atmo property that the angles at the one side of a line is sum is 180 degree so angle 1 plus angle 2 plus angle 3 plus angle 4 is equal to 180 degree that is other linear angles at the same side of a straight line so look like look at here angle angle 1 is replaced by angle 2. this is angle 2 angle 3 and angle 4 is also equal to angle 2 so it is replaced by angle 3 that is equal to 180 degree now this is 2 times angle 2 and 2 times angle 3 so 2 get common that is angle 2 plus angle 3 is equal to 180 degree now 2 get divided that side that is angle 2 plus angle 3 is 90 degree so what is what is angle 2 and angle 3 angle 2 plus angle 3 angle 2 plus angle 3 that is a o b angle that implies angle a o b is equal to 90 degrees so hence proved that the angle aob is 90 degree so that's all about the today's video thanks for watching the video please like share and subscribe the channel to watch more videos thank youwelcome to pramar academy today we will discuss about the 10th chapter of great time that is circles up in ninth grade may circles could discuss kiawah that what is circle what is the radius of circle what is the center of circle what is the diameter relationship between diameter and radius arc of circle chord of circle segment of circle and the sector of circles is means we will discuss the properties of straight line with the circle and we will do the applications of that also so let's start first of all there are three possible conditions as combinedly we will represent a circle and a straight line the first condition is jumper upgrade so straight line there is nothing common between the circle and the straight between straight line and now we will discuss the second point first of all here the straight line test the circle at so that's why this is this line is called a tangent line and the common point this common point is called point of contact point of this line is called second and these two points are called points of intersection points of intersection tangent is a line that does the circle boundary of the circle at just one point now a circle has how many tangents is x circle now we know that a circle has infinite number of points at end at every point it has one tangent so a circle has infinite number of tangents so look at here this is circle and at this point there is a tangent at another this point there is another tangent at this point there is another tangent like this so at every point each and every point of circle a circle has a tangent and we know that a circle has infinite number of points so it has infinite number of tangents so first is a circle has dash number of tangents that is infinite this one the second is a circle has dash number of parallel tangents so the answer is look at here first of all this is circle and we will draw a diameter that passes through center now at this point we will draw tangent at the other end we will draw another tangent now these two are the parallel tangents these two are the parent is that means a circle has only two parallel tangents so answer is a circle has how many parallel tangents that is two only the next one this first shape here this a is the point of contact this is the tangent at point a i'm saying this is tangent at point a so it is clear now what is tangent at point means that point is on the boundary of the circle in the second one case when it is tangent from point from from this point circle boundary that is external point to the circle so from that external point look at this that this is external point so from that external point three tangents draw construction to the circle so very easy look at here from that point i can draw two tangents to the circle the first tangent is p a circle p a tangent and the second one is pb tangent so this is also one of the very important question from an external point we can draw easily draw two tangents only the next one is if we are talking about the tangent at a point there is only one single tangent and if we're talking about the tangent from a point there are exactly two changes we can draw the next one is normal is perpendicular to tangent and passes through the circle so a line that passes through a center and perpendicular to tangent is the normal look at here a line this line that is passes through the center and point of contact and it is perpendicular to a tangent line that line is called normal normal or tangent that are perpendicular to each other this is the first property that normal and tangent are perpendicular to the each other and normal joe all that always passes through the center and the point of contact so this one is normal is always perpendicular this is the to the tangent this is the first property is always perpendicular to the tangent and it always passes through the center of circle and the point of contact next one is the length of tangent drawn from an external point to a circle are equal means care this one means we are drawing two tangents that first is p a tangent and second is p b tangents so in length if p a is equal to 10 centimeter then p b is also equal to 10 now there are two triangles triangle p a o and triangle p b o s we need three things that are equal so first is that first thing is po is equal to po that is the common one po is equal to po that is common side second thing is oa is equal to ob that is radius of circle radii of circle and radius is always equal to a circle and the third thing is that is angle o a p is equal to equal to angle o b p that is 90 degree each as we know that that this is a tangent line and this is the normal line that passes through the center and point of context so this angle is always 90 degree this angle is always 90 degree now this this satisfies the rule that side and that are that is the property of rhs rhs means we have having the one right angle one side and one is the hypotenuse so one hypothetic is equal right angle we are also having and this is one of the side so by r h s rule triangle o a p o a p is congruent to triangle o b p by rhs rule these two triangles are congruent now if these two triangles are congruent their corresponding parts are equal as we know that that cpct that is corresponding parts of congruent triangles are equal so look at here corresponding to pa there is a bp so pa is equal to pb by cp ctc that is corresponding parts of congruent triangles are equal here c p a is equal to p b and what is p and p b that are the lengths of triangles slants of tangents so we will say that this is a result that from if through tangents are drawn from external point to a circle their lengths are equal so this is the proof for that property look at here and note it down the next one is perpendicular from the center of circle to any code of the circle bisects the code means circle centers a circle means i have to prove that ap is equal to bp look at here first of all what is given to us that given is op is perpendicular to a b that op is perpendicular to a b and what we have to prove we have to prove that that line is bisected code is bisected at point p that means ap is equal to bb so we will start the solution here before that we will construct oa and ob means i have to join oa and ob both are the radius of circle now is and triangle o b p the first one common thing is that o p is equal to o p that is common side common side of both the triangles second thing is o a is equal to ob that is the radius of circle circle and the third thing is angle o p a is equal to angle o p b both are 90 degree angle o p a is equal to angle o p b both are each 90 degree so now by again by the property of rhs by rule of rhs both triangles are congruent triangle o p a is congruent to triangle o p b and we know that if two triangles are congruent their corresponding parts are equal so the corresponding part corresponding to p a there is p b so p a is equal to p b by c p c t c that is corresponding parts of congruent triangle are equal so the result is if we draw a perpendicular from center to any code of the circle that perpendicular bisects the code the next one is prove that the angle between the two tangents drawn from an external point to the center is supplementary to the angle subtended by the line segment joining the points of contact to the center what this question means this is first of all this is very interesting and easy one question so according to this question if from an external point p we will draw two tangents that is p a and p b and we will join a and b with the center of circle then these two angles are supplementary so look at the statement proved at the angle between the tangents that means this one angle angle between the tangents drawn from an external from from an external point to a circle is supplementary that means their sum is 180 degree and the angle subtended by the segments joining the point to the contact to the center you owning the point of context to the center means these two angles are supplementary so first of all we know that p a is tangent to the circle and this one o a is the line that is passes through the point of contact a and the center also so this clearly o a line is the normal and we know that normal is perpendicular to the circle so angle o a p is 90 degree similarly angle o b p is also 90 degree because ob is normal to tangent p b now look at here carefully p a o b is a correlator p a o b is a coordinator and we know that the angle some property of a quadrature that sum of all the four angles of quadrilateral is 360 degree so now we will use that angle some property of coordinate angle sum property of quadrilateral that sum of all the four angle is 360 degree so angle a p b means this angle plus angle p b o means this one angle plus angle b o a plus angle o a p that is 360 degree and angle p a o and p b are already 90 and 90 each so angle this one angle pbo and that is this one op a both are 90 degrees so angle a p b plus 90 plus that is angle b o a plus 90 is equal to 360 degree so angle a p b plus 90 plus 90 that is 180 plus angle b o a is 360 degree now 180 gets subtracted from right hand side so angle a p b plus angle b o a is equal to 360 minus 180 that is angle a p b plus angle b o a is equal to 180 degree so look at here angle apb means this one angle and angle boa bo means this one angle both sum of both angle is 180 degree means these two angles are supplementary angles this is what about the question that some of these two angles is always 180 degree means if this is 60 degree angle we clearly say that this is 120 degree because the sum is 180 degree if this is 50 this is 130 if this is 80 this is 100 okay so now one more thing this property holds only when these two lines meet at ori's center of circle if these two lines meet out of center this property can't hold so the next one is find the value of x this thing is given that pb is tangent and pa is also tangent that means two tangent from an external point p and here this o is the center of the circle and this angle is 110 given and we have to find the value of x as we are as we use the last rule that if we draw two tangent from an external point and join the point of context with the center these two angles are supplementary so clearly angle b p a plus angle a o b is equal to 180 degree and you will use the statement of last question here and now we will solve angle p that is already given x plus this is given 110 that is equal to 180 degrees so x is equal to 180 minus 110 so x is equal to 70 degrees so this is the angle with the 70 degree so it's very easy to find the opposite angle because this relationship is known that both the angles are supplementary the next one is the length of tangents from a point a at distance five centimeter from the center of the circle is four find the radius of circle so look at the statement carefully what is given to us and what we have to find so first of all we will draw the diagram this is center of circle first of what is given the length of tangent from a point a from a point a look at carry very carefully from at from from point a we have to draw a tangent from point a that look at let this is a point so b point so a b is the tangent at a distance five center from the center your point at that point a is five centimeter from the center so your point a that is five centimeter from the center that is given five of the circle is four the length of tangent of a four f length of tangent is 4 this is 4 given and we have to find the radius of find the radius of circle means we have to find the radius so this is the diagram for this question so clearly a b is the tangent again and this is the ob is the line that passes through the point of contact and the center and we know that this line is the normal to the tangent a b and we know that angle between tangent and normally is always 96 degree so clearly triangle oba is a right angle triangle so this implies triangle o b a is right angle triangle at point at point b means this is right angle triangle at point b b is the 90 degree angle so o is the hypotenuse a b is the perpendicular and ob is the base now we will use pythagoras theorem here by pythagoras theorem here we say that perpendicular square plus b square is equal to the hypotenuse square so that means o square is equal to ob square plus a b square so what is o a that is already given 5 pi square is equal to ob that is our radius we have to find it plus a b square that is 4 4 square 5 square is 25 that is equal to r square plus 4 square is 16 16 gets subtracted in left hand side that is 25 minus 16 is equal to r square and this is 9 is equal to r square so clearly r is what that is 3 centimeter so radius of the circle is three centimeter this is one of the very interesting question related to the straight line and the circle that if a line a b that is changing to the circle and the distance of the from point that is external point to the certain center is always also given we will easily find the radius of the circle the next one is prove that the parallelogram circumscribe a circle is a rhombus in this question we have to prove that there is a parallelogram in which a circle is circumscribed and we have to prove that parallelogram is also rhombus so for that we will draw a diagram that this is a some parallelogram which some circumscribe a circle and we have to prove that this parallelogram is a rhombus so name it as a b c d these are the four corners of the parallelogram p q r s these are the four point of contacts or the common points to the circle and the parallelogram now look at here very carefully solution look at the a point a is an external 0.0 circle that is out of the boundary of the circle and clearly a p and a s are the two tangents to the circle from point a and we know that from an external point if we draw two tangents their lengths are equal so a is external point from which we are drawing the two tangents so a is external point external point and ap is equal to a s we will use the statement here that that the two tangents are drawn from an external point they are equal in length length of tangents from an external point are equal so that's why ap is equal to a s similarly b is also the external point to the circle and from b we will draw two tangents that is b p and b q so their lines are also equal so i will write uh this is equation number one now bp is equal to bq for the same reason bp is equal to bq this is your equation number two again c is also the external point we are drawing two tangents that is c q and c are their lines are also equal so cq is equal to cr4 the same reason cq is equal to cr this is equation number three we are using the same reasons for all for the last d is also external point from which two tangents d s and d are and their lines are also equal so d d r d r is equal to d s this is your equation number four and the same reason so we are having the four equations now what we have to do next step add all the equations so here add one two question number two equation number three and equation number four what we will get we will get ap plus bp plus cr plus d r is equal to a s plus b q plus c cube plus d s now look at here what is a p plus b p a p plus b p that is combining a b so this is equal to a b c r plus d r t r plus d r that is combined with c d that is equal to a s plus d s a s plus d s that is combinedly a d a d plus b q plus c cube b q plus c q that is combining bc so this is equation number y please note up to here that we have done that that first of all a is an external point from that ap and a s are two tangents to the circle and their lines are equal so ap is equal to a s this is the first equation again b is external point from there two tangents b p and b q are to the circle and their lines are also equal similarly for the point c and the point for the point d and we are having the four equations and the next step we will add all the four equations that is a p plus b p plus c i plus d r is equal to a s plus b q plus c q plus d s now a p plus b p combinedly represents a b c r plus d r that is c r plus d i combinedly represents c d similarly a s plus d s completely represents a d and c q plus that is c q plus b q combini represents b c this is our fifth equation now after this so what is given to us this is a parallelogram as abcd is a parallelogram and we know that in a parallelogram opposite sides are equal so a b is equal to cd and bc is equal to 80 and you have to write it as because abcd is a parallelogram abcd is a parallel graph in a parallelogram opposite sides are equal now we will use that two things equations here look at here a b plus what is c d what is the value of c d that is also a b so i will use replace c d by a b here that is equal to 80 plus what is the value of bc that is also equal to 80 so i will replace bc by a d this is 2 times a b is equal to 2 times a d now 2 get cancelled out so a b is equal to a d this is your last equation now look at here carefully because of the parallelogram opposite sides are already equal and now the two axis and sides are also equal because of the parallelogram opposite sides are equal a b is equal to c d and a d is equal to b c and from this equation two as the sides a b and a d are also equal so from that we will conclude that all the four sides are equal and you know that in a quadrilateral if all the four sides are equal that coordinate returns as a rhombus so from here we will conclude that a b is equal to bc equal to cd is equal to da and this completes that abcd is a rhombus a b c d is a chromosome this is the total proof of this theorem so i will repeat prove that the parallelogram circumstance a circle is rhombus and this is the complete solution this is a very important equation in ncit book this question repeats two time in first question you have to prove it up to here only that a b plus c d is equal to a d plus b c and in the next question you have to also prove that parallelogram is a rhombus if satisfy the same conditions so for that you have to prove up to here at the last point the next one is prove that the perpendicular at the point of contact to the tangent to a circle passes through the center we have to prove for this question we have to prove that if we draw a perpendicular from the point of contact that always passes through the center so let's start for that we will draw first of all diagram what is given to us this is circle and this is one of the tangent to a circle and this is a is a point of contact that is common point now we have to prove that that the perpendicular from point a passes through the center so to prove that we will let that that that point never passes through the center that point passes through some another point this is the center of circle o and i will let that that the perpendicular draw from some another point or dash so what i have to prove that to prove to prove that perpendicular on tangent perpendicular to tangent at point of contact at point of contact passes through center of circle we have to prove this so i let let if possible that if possible that perpendicular passes through some another point that is o dash that is not the center of circle that if possible perpendicular passes through another point another point that is o dash that is not the center of circle because o is the center of circle so let this is x and y now look at it carefully we'll let that perpendicular y o y a o dash that is passing through the odh clearly so by this one by this one it is clear that angle y a o by a o dash is 90 degree this is your first equation equation number one and we also know that that normal that normal line passes through the point of contact and the center of circle that has also perpendicular to the tangent so join o and a this is normal this is normal line and we know that normal is perpendicular to the tangent as oa is normal to tangent normal to tangent x y and normal is always perpendicular tangent this implies angle y a o is also 90 degree this is your second equation now look at these two questions very carefully angle y a o dash that is also 90 and angle y a o that is center is also 90 degree this is possible only if the case if o and o dash are the same points means they coincide with each other so this is contradiction to our supposition this is contradiction to our supposition this is contradiction to a supposition that point passes through another point and this will both the equation possible simultaneously if o and o dash are coincide means they are same points so this is possible only if o and o dash inside each other that means o and o dash are the same points so the next one prove that opposite sides of a quadrilateral circumcise a circle substance supplementary angle at the center of circle in this question we have to prove that that angle at the center of circle this is center of circle and angle made by opposite sides at the center that this angle is supplementary that this is a b c and d and these are the obtained common points of context that are p q r and s we have to prove that these these angles these two angles are supplementary and these two angles are also supply mandatory for that i have to do i have to do construction that is i will join o r o s o p and o a so clearly there are eight angles at the center let this is angle 1 angle 2 angle 3 angle 4 angle 5 angle 6 angle 7 and angle 8 there are eight angles at the center of the circle now what we have to do next so look at here look at these two triangles these this one a blue triangle and this one green triangle look at the these two triangles these two triangles are clearly congruent how this is already done in the previous question that ap is equal to a s that is a length of tangent from an external point is equal op is equal to os that is the radius of circle and o a is equal to o a that is the common side of the triangles so by ss rule these two triangles are congruent and if these two triangles are congruent they are these one angle one and angle five are equal by cpct by corresponding parts of congruent triangles are equal so let's do it in triangle o s a and triangle o p a means i will say in triangle in blue color triangle and in green color triangle so three reasons are reason one o a is equal to o a common side of the triangles o a is equal to a common side of both the triangles second reason is a s is equal to ap that is if we can draw a two tangent from a same external point their length is equal so this is a rule tangent from our tangent from an same external point tangent from same external point so that's why their length is equal and the third reason is os is equal to ob that is radius of circle so by by side side psi rule triangle o s a is congruent to triangle o p e now you know that in congruent triangle corresponding parts are equal so angle 1 is equal to angle 5 angle 1 is equal to angle pi by cp ct this is the first question similarly in another for these combinations for this these two triangles angle 6 and 7 for the same reason angle 6 is equal to angle 7 similarly similarly angle 6 is equal to angle 7 angle 1 is equal to angle 8 and angle 3 is equal to angle 2. these are four equations equation number two question number three question number four so i think you understand how we can prove that angle 1 is equal to angle 5 just similar we will prove that these two triangles are congruent and the corresponding parts are equal now the last one step is so we know that sum of all these angle is 360 degree because this completes a circle so sum of all the eight angles is 360 degree angle one angle 2 angle 3 angle 4 angle 5 angle 6 angle 7 at angle 8 is equal to 360 degree because this completes a circle now angle 1 angle 2 angle 3 angle 3 is equal to angle 2 so i will replace this by angle 2. angle 4 that angle 4 is equal to angle 5 so i will replace it by angle pi this is angle 5 angle 6 angle 7 is equal to angle 6 also so simon angle is replaced by 6 also and angle 8 is equal to angle 1 so i can replace angle 8 by angle 1. now look at here angle 1 repeats twice angle 2 repeats twice angle 5 repeats twice and angle 6 repeats twice also so all the angles are repeated twice so i will get 2 get common that is 2 angle 1 angle 2 angle 5 and angle 6 that is equal to 360 degree now the two get divided at the right hand side so angle 1 plus angle 2 plus angle 5 plus angle 6 is equal to 180 degree now combining angle 1 and 2 angle 1 and 2 represents angle c o d that is angle c o d and 5 and 6 combining represents angle ao b angle ao b that is 180 degree and that is what we have to prove that that at the end these two angles represent the supplementary angles the next is find the value of a b and ac in this diagram so look at here carefully abc is a triangle in which a circle is circumscribed of radius 4 the radius of the circle is 4 and the length of a at this point is p so length of bp is 6 and length of cp is a so we have to find the length of a b and ac so that this point is q and this point is r now simply if from this is this b is external point and you know that from an external point if we draw two tangents their length is equal so if length of tangent bp is 6 the length of tangent br is also so 6 so this length is 6. similarly from external point c if the length of tangent c p is eight the length of this tangent c q is also eight now for the last a a is also the external point from that point we will draw two tangents that is c r and c q and their lines are also equal so let the length this length is x and this length is also x now join ob oc and o a and from draw o or join or at oq clearly or is normal to a b that is perpendicular oq is normal to ac and op is normal to bc now look at here so next what we have to do we have to find the area of triangle abc in two different ways firstly we will find the area using heron's method of grade 9 and secondly we will find the area of triangle this abc triangle using half into base into height so first of all i am finding the area of this triangle using heron's formula now for that i need length of all the three sides means a b b c and c a so what is the side of a b what is the side of a b that is 6 plus x that is 6 plus x what is the side of bc that is 6 plus 8 14 and last what is the side of ca that is 8 plus x this is a side of or this is the length of all three sides now using errors formula we need semi parameter that is s so first step s is equal to sum of all the sides that is 6 plus x plus second side 14 plus third side x 8 plus x whole divided by 2 so that is 6 plus 8 14 plus 14 28 plus x plus x 2 x whole divided by 2 that is equal to 14 plus so this is the sami parameter now the second step to find the area of triangle that is equal to s s minus first side s minus length of second side as minus length of third side that is equal to s that is 14 plus x 14 plus x second is 14 plus x minus first side that is 6 plus x and that is 6 minus 6 minus x again next is 14 plus x minus second side that is only 14 and the third is s minus that is 14 plus x minus third side that is 8 plus x here this is minus 8 minus x now this is 14 plus x this is 14 minus 6 8 plus x and minus x guys cancel out 14 and 14 get cancel out only x and this is x minus and x plus cancel out 14 plus eight that is a six so look at here this fourteen into x fourteen x plus x into x x square here this is eight is written as four into 2 and 6 is written as 2 into 3 so here 4 get common as this get 2 because of the under root here this is a pair of 2 this is also get common as 2 the remaining portion is 14 plus 14 x plus x square into 3 that is 4 under root 3 times 14 x plus x square so this is the area of triangle abc using heroes formula i will write it here that is 4 times under root 3 that is 14 x plus x square this is the area of triangle abc now i will find the area of same triangle using the formula half into base into height using the half into base into so for that i will break the triangle abc in three similar equal congruent triangles that are triangle b oc triangle aob and triangle aoc that means the area of triangle aob is equal to sum of all the three triangles that is look at here area of triangle aob is equal to sum of these three triangles that is sum of aob triangle area of aob triangle area of bo a triangle and area of co 8 angle so look at here area of triangle a b c is equal to area of three triangles the first one is boc plus area of triangle aob plus triangle area of triangle ao c so for that look at here what is the area of triangle b oc that is half into base into height that is half what is the base the base is 6 plus 18 that is 14 and what is the height that is the radius of circle that is 4 plus area of triangle aob again that is half into base what is the base that is 6 plus x and what is the height that is radius again so this is a 4 plus third is half base is 8 plus x and the height is 4 also so calculate it 2 cancel out with 4 that is 14 to the 28 2 cancel out with 4 again 2 2 that is 12 plus 2 x plus 2 cancels out with 2 that is 2 that is 16 plus 2x combinedly 8 28 plus 12 that is 40 plus that is 56 plus 4 x so this is the area of triangle abc now look at here we are having the area of triangle in two different forms the one is 56 plus 4x and second is 4 under root 3 times 14 x plus x 4. now both are the areas of same triangle so they are equal so from here this is equation number one and equation number two because both those represent the area of same triangle so 56 plus 4x is equal to 4 times under root 3 into 14 x plus x square now 4 get divided this side we will left with 456 4 divided by 4 that is 14 plus 4 divided by 4 that is x is equal to 3 under root that is under root 3 14 x plus x square now because of under root in the right hand side so squaring both sides we will remove the under root that is squaring both sides that square is 14 is 196 plus x square plus 28 x that is a square plus b square plus 2 a b that is equal to 3 into 14 x plus x square so this is 3 get divided that is 196 plus x square plus 28 x is equal to 14 theta 42 x plus 3 x square now get all the terms at one side that 3 x square minus x square that is 2 x square 42 x minus 28 x that is 14 x minus 190 6 that is equal to 0 now to get common from all the terms we will left with x square plus 7x minus 98. so this is simply a quadratic in variable x now we know how to solve the quadratic we have to split the middle term in such a way that whose sum is plus minus is 7 and their product is minus 98 so we have to break the middle term in two portions that the sum is seven and their product is ninety eight so what are the two numbers clearly the numbers are first of all x square plus seven x minus 90 8 is equal to 0 we have to break these two in two parts these two parts are x square plus 14x minus 7x minus 90 8 is equal to 0 14 x minus 7 x that is again 7 x and 40 x minus 14 x minus into 7 x is equal to minus 98 x square so what is common from first two parts the only x we left with x plus 14 and what is common from here that is minus 7 and left with again x plus 14. that is equal to 0 and the factor is x plus 14 and x minus 7 that is 0 so values for x are x is equal to minus 14 and x is equal to plus sum so values of x are x is equal to minus 14 from the first from first factor and x is equal to plus 7 from the second factor and you know look at here x is clearly the length of side length of the that portion a p a k a r and a q so it can't be negative so minus 14 gets rejected we will be left with x is equal to 7 so this value of x is 7 now what is the main part main question what we have to find actually we have to find the length of size a b and a b and ac now look at here what is the length of side a b that is 7 plus 6 that is 13 so length of a b is 13 and what we find is find second that is length of ac that is 7 plus a that is 50 so length of ac is 15 so this is how to solve this question i will repeat how to find the value of x or how to find the length of these two portions first of all break the triangles in or break the triangle oab into these triangles small triangles b o c c o a and o a b and find the area of whole triangle using half into base into height and again find that area of triangle a b c using the hero's formula because of these are the areas of the same triangle so both are equal and after equating we will at the end we will get a quadratic and after solving we will get the value of x the one value is negative and other is positive because x is the length that can't be negative so x is equal to minus 14 get rejected so actual value of x is that is 7 on put the value of x7 here we will get the length of a b that is 7 plus 6 that is 13 and the length of ac is 15 that is all about this question come to the next last question the last question is tangents x y and x y dash x dash y dash at point p and q are parallel to each other and the third tangent a b that is at point c we have to find the value of angle aob so this is also very interesting question for that we will construct join oc so now look at here there are two triangles a and o p a we have to prove these two triangles as congruent so in triangle o c a and triangle o p a we need three things to equal to prove that it is congruent triangles so first thing is oc is equal to ob because it is the radius of circle so oc is equal to op that is radii of circle second thing that clearly a is an external point and we know that from external point the length of tangent is same so ap is equal to ac that is ap is equal to ac that is tangent from an external point tangents from same external point same external point that that's why their lines are equal and oa is the common side of both the triangles so o a is equal to o a that is common side that is common side of both the triangles so by clearly by side side side rule both the triangles are congruent triangle o c a o c is congruent to triangle opa now these two triangles are congruent so let these angles are angle 1 angle 2 angle 3 and angle 4. now if as these two triangles are congruent so angle 1 and angle 2 are equal because of the cpct does the corresponding parts of congruent triangles are equal so clearly angle 1 is equal to angle 2 this is our first equation now using the same method the we get these two triangles are also congruent and angle 3 and 4 are equal similarly similarly angle 3 is equal to angle 4 this is equation number 2. now clearly p o q is a straight line so by the angle atmo property that the angles at the one side of a line is sum is 180 degree so angle 1 plus angle 2 plus angle 3 plus angle 4 is equal to 180 degree that is other linear angles at the same side of a straight line so look like look at here angle angle 1 is replaced by angle 2. this is angle 2 angle 3 and angle 4 is also equal to angle 2 so it is replaced by angle 3 that is equal to 180 degree now this is 2 times angle 2 and 2 times angle 3 so 2 get common that is angle 2 plus angle 3 is equal to 180 degree now 2 get divided that side that is angle 2 plus angle 3 is 90 degree so what is what is angle 2 and angle 3 angle 2 plus angle 3 angle 2 plus angle 3 that is a o b angle that implies angle a o b is equal to 90 degrees so hence proved that the angle aob is 90 degree so that's all about the today's video thanks for watching the video please like share and subscribe the channel to watch more videos thank you\n"