Question

In the given figure, RS is parallel to PQ. If RS=3cm, PQ= 6cm and ar (angleTRS) = 15cm2, then ar ( triangle TPQ) =?​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Given:

In the given figure, RS is parallel to PQ. If RS = 3 cm, PQ = 6 cm and ar (angleTRS) = 15 cm², then ar ( triangle TPQ) =?​

To find:

Ar ( triangle TPQ) =?​

Solution:

The area of Δ TRS = 15 cm²  

The length of RS = 3 cm

The length of PQ = 6 cm

Since RS // PQ, in ΔTRS and Δ TPQ, we have

∠RTS = ∠QTP  . . . [common angle]

∠TRS = ∠TQP . . . [corresponding angles]

∴ ΔTRS ~ Δ TPQ . . . [By AA Similarity]

We know that,  

The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

So, based on the above theorem, we get  

$\frac{Area (\triangle\:TRS)}{Area (\triangle\:TPQ)} = \bigg(\frac{RS}{PQ} \bigg)^2$

 On substituting the given values, we get  

$\implies \frac{15}{Area (\triangle\:TPQ)} = \bigg(\frac{3}{6} \bigg)^2$  

$\implies \frac{15}{Area (\triangle\:TPQ)}=\frac{9}{36}$

$\implies Area (\triangle\:TPQ) = \frac{15 \:\times\: 36}{9}$

$\implies Area (\triangle\:TPQ) = 5 \times 12$

$\implies \bold{Area (\triangle\:TPQ) = 60\:cm^2}$  

 

Thus, the ar ( triangle TPQ) is → 60 cm².

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Also View:

The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

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Find the ratio of the areas of two similar triangles if two of their corresponding sides are of length 3 cm and 5 cm.

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