What is the ratio of areas of two similar triangles whose sides are in the ratio 3 is to 5?

Answer

What is the ratio of areas of two similar triangles whose sides are in the ratio 3 is to 5?
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Hint: If two triangles are similar, then the ratio of the areas of both the triangles is equal to the ratio of the squares of their corresponding sides. Here the ratio of the sides of the triangles is given as 4:9. Using this, we are finding the ratio of their areas.

Complete step-by-step answer:

In two triangles, if the corresponding angles are similar and corresponding sides are in proportion, then those triangles are said to be similar. When 3 angles are equal, 2 sides and one angle are equal in two triangles, then we can say that they both are similar. We are given to find the ratio of areas of two similar triangles when the ratio of their sides is given.Let the similar triangles be ABC and PQR.

What is the ratio of areas of two similar triangles whose sides are in the ratio 3 is to 5?

And the ratio of the sides of these two similar triangles be AB: PQ which is 4:9. If two triangles are similar, then the ratio of the areas of both the triangles is equal to the ratio of the squares of their corresponding sides. This means, the ratio of areas of triangles ABC and PQR will be $ \dfrac{{Ar\left( {\vartriangle ABC} \right)}}{{Ar\left( {\vartriangle PQR} \right)}} = \dfrac{{A{B^2}}}{{P{Q^2}}} = {\left( {\dfrac{{AB}}{{PQ}}} \right)^2} $ We already know that $ \dfrac{{AB}}{{PQ}} $ is equal to $ \dfrac{4}{9} $ Therefore, $ \dfrac{{Ar\left( {\vartriangle ABC} \right)}}{{Ar\left( {\vartriangle PQR} \right)}} = {\left( {\dfrac{4}{9}} \right)^2} = \dfrac{{16}}{{81}} = 16:81 $

So, the correct answer is “16:81”.

Note: Do not confuse similar triangles with congruent triangles, because congruent triangles have similar three sides and similar three angles which also mean similar areas, whereas the areas may or may not be the same in similar triangles. Congruent triangles are always similar whereas similar triangles may not be congruent always.