Question

In two triangles ABC an DEF, angle A=angleD . The sum of the angles A and B is equal to the sum of the angles D and E. If BC=6 cm and EF=8cm , find the ratio of the areas of the triangles, ABC and DEF.​

Answer 1

Given: In two triangles ABC an DEF, angle A=angleD . The sum of the angles A and B is equal to the sum of the angles D and E. If BC=6 cm and EF=8cm.

To find: find the ratio of the areas of the triangles, ABC and DEF.

Solution:

Tip: Find similarly of triangles

ATQ

$\angle A=\angle D$ ...eq1

$\angle A+\angle B=\angle D+\angle E$...eq2

From eq1 and eq2

It is clear that

$\angle B=\angle E$ ...eq3

From AA criteria of similarity,∆ABC ≈ ∆DEF

Thus

$\frac{AB}{DE}=\frac{BC}{EF}=\frac{AC}{DF}$

Thus,

Ratio of area of ∆ABC to ∆DEF is

$\frac{ar(\triangle ABC)}{ar(\triangle DEF)}=\frac{(BC)^2}{(EF)^2}$

$\frac{ar(\triangle ABC)}{ar(\triangle DEF)}=\frac{(6)^2}{(8)^2}$

$\frac{ar(\triangle ABC)}{ar(\triangle DEF)}=\frac{36}{64}$

$\frac{ar(\triangle ABC)}{ar(\triangle DEF)}=\frac{9}{16}$

Final answer:

$\bold{\pink{\frac{ar(\triangle ABC)}{ar(\triangle DEF)}=\frac{9}{16}}}$

Hope it helps you.

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