Question

Point S is on the side PR of ∆PMR such that 3SR=2SP.
Seg ST || seg PM . If A(∆PMR)=50 CM^2.
Find (a) A(∆RST) (b) A(□PMTS).​

Answer 1

Given:

Point S is on the side PR of ∆PMR such that 3SR=2SP.

Seg ST || seg PM

A(∆PMR)=50 cm²

To find:

(a) A(∆RST)

(b) A(□PMTS)

Solution:

We have,

3SR = 2SP  

⇒ $\frac{SR}{SP} = \frac{2}{3}$ ↔ Equation1

Finding the area of triangle RST:

Consider ∆RST and ∆PRM, we get

∠SRT = ∠PRM → [common angle]

∠RST = ∠RPM → [corresponding angles]

∴ By AA similarity → ∆RST ~ ∆PRM

We know that → the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides.

Therefore,

$\frac{Area(\triangle RST)}{Area(\triangle PRM} = \frac{(RS)^2}{(PR)^2}$

$\implies \frac{Area(\triangle RST)}{Area(\triangle PMR)} = \frac{(SR)^2}{(SR + SP)^2}$

On substituting the values from equation1 and A(∆PMR)=50 cm²

$\implies \frac{Area(\triangle RST)}{50} = \frac{(2)^2}{(2 + 3)^2}$

$\implies Area(\triangle RST) = \frac{(2)^2\times 50}{(5)^2}$

$\implies Area(\triangle RST) = \frac{4 \times 50}{25}$

$\implies \boxed{\bold{Area(\triangle RST) = 8\:cm^2}}$

Finding the area of quadrilateral PMTS

The area of quadrilateral PMTS is given by,

= [Area of Δ PMR] – [Area of Δ RST]  

= 50 cm² – 8 cm²

= $\boxed{\bold{42\:cm^2}}$

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Also View:

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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