Construction of similar triangles - 3

Added: 2026-05-05

[00:00:00]this is the third type of questions based on similar triangles now we will do the second part triangle abc is similar to triangle pbr in triangle abc abs 4 bc is 8 and ac is 6. construct triangle abc and triangle pbr such that ac upon pr is equals to 2 upon 3 previously we did this as 3 upon 2 now we will do it the other way around that is 2 upon 3 so the rough work for that is this we know triangle abc is similar to triangle pbr therefore ratio of corresponding sides is equals to 2 upon 3 which is given in the equation that is corresponding sides of similar triangles next now in the ratio 2 is less than 3 we know this the smaller digit corresponds to the smaller triangle the bigger digit corresponds to the bigger triangle so sides of triangle abc will be smaller than the corresponding sides of pbr also vertex b is common in both of them so if we draw a rough figure suppose this is a b c again b is the common vertex b is common in both of them right abc and pbr so this has to be on the left hand side okay now sides of triangle abc is smaller than the corresponding sides of pbr that means the side br of triangle pbr will be larger or longer than the side bc of triangle abc this is the side bc of triangle abc so point r will be lying somewhere outside point r will be lying somewhere outside after bc so that when you join you will extend this and you join you have to get a triangle like this so this will be p b r the other triangle now let us see how to draw this question understand the concept behind it first so we know how to draw triangle abc now we will see step by step how to draw triangle pbr i understand the concept behind it the ratio is 3 upon 2 and 3 is greater than 2 that means we have to divide the given segment into three equal parts right so using bc bc will be two equal part reason let us take these two ratios bc upon br is equals to 2 upon 3 using bc bc will be two equal parts bc will be two equal parts br will be three equal parts but we don't have br we have only bc right so using bc we will construct point r how to do that first you will make and cut three equal arcs we know the second arc will be joined to point c why because two parts is bc right two equal parts using bc which will be two equal parts this is two equal parts so you will join correct and then plot point r such that cr forms the third equal part that means you have to draw an angle from this point suppose this is p 1 p 2 p 3 this angle should be congruent to the angle which i draw here so i'll have me make a cut here make a cut here how much this angle is opening this also how much it is opening and then you join both these so you will get your point r here and then again this angle should be congruent to the angle here so measure the angle and make a cut and then you join when you join you get your third point p this is your rough figure this is how you will draw okay so let us start drawing first we will draw triangle abc in triangle abc we have ab4 bc8 the base is bc which is 8 centimeters this is 8 centimeter this is 4 centimeter and ac is 6 centimeter so let us draw bc which is 8 centimeter so let us draw bc which is 8 centimeter this is b c 8 centimeter okay then you have a b which is 4 centimeter so let us take 4 centimeter on the compass 4 centimeter from b make a cut of 4 centimeter and ac is six centimeter this is six centimeter so take from point c make a cut of 6 centimeter it is the extra arcs you don't need them this is your point a join a to b this is your 4 centimeter point a to c this is your 6 centimeter so you've drawn triangle abc now to draw triangle pbr step one draw a array from point b and then you will cut it into three equal parts why three equal parts because 3 is greater than 2 out of the ratio 3 is greater so you will divide it into 3 equal parts so using the compass any measurement would do so from point b a cut from the point next point another cut from the next point another cut so you have got three equal parts now understand this the second point corresponds to point c this is your point c so second point so this is p 1 p 2 p 3 so this point corresponds to point c so we will join the second arc to point c now you have to draw an angle which is congruent to this from point p three let us see let us draw first procedure is you make a big arc using the same measurement from point p three also you make a big arc using the same measurement now you have to see how much this angle has opened this angle it has opened this much okay correct let us just check once more it is open this much so using the same measurement you have to make a cut on this arc also okay so now let us join p3 to this point ok so when we join you get a straight line like this now this will go beyond let us extend bc to get the point r so when you extend you get this as your point r okay now you have to draw an angle which is congruent to this angle bca at point r procedure is the same you have to first make a cut at point c any measurement would do but once you take the measurement do not change and make it from point r also now see how much this angle has opened take it on the compass like this okay this much it has opened so using this measurement from this point also you make a cut so you get the third this is side ac so now let us draw it from point r also so this is from point r this is your thing okay so this is the point of intersection right this is the point of intersection and then this goes up you just extend it wherever they meet this is your point p so this is p b r and both the triangles are sharing the same vertex b so the two types of questions which you have seen just now are these two okay in the first one the ratio was 3 upon 2 so triangle abc was bigger and pbr was inside the triangle abc and the other one the ratio was 2 upon 3 so triangle abc was smaller so it was inside the triangle pbr abc was inside the triangle pbr so these are the three types of questions which you get this is the first question where there is no vertex which is common and second question where you have a ratio where the numerator is bigger and the third type where the denominator is bigger thank you