Question

In the given figure, ABC ~ PQR, PM is
median of PQR. If ar(ABC) = 289 cm²,
BC = 17 cm, MR= 6.5 cm, then the area of
PQM is :
A) 169 cm^2
B)13 cm^2
C)84.5cm^2
D)144.5cm^2​

Answer 1

Answer:

the correct answer of the question

(In the given figure, ABC ~ PQR, PM is

median of PQR. If ar(ABC) = 289 cm²,

BC = 17 cm, MR= 6.5 cm, then the area of

PQM is :

A) 169 cm^2

B)13 cm^2

C)84.5cm^2

D)144.5cm^2 )

Step-by-step explanation:

84.5

Answer 2

Given:

ABC is similar to PQR.

PM is the median of PQR, ar(ABC) = 289 cm², BC = 17 cm, MR= 6.5 cm.

To find:

Area of triangle PQM.

Solution:

As given, ABC is similar to PQR

So according to a theorem which states that- if two triangles are similar then the square of the ratio of their corresponding sides is equal to the ratio of their areas.

So, we have

$\frac{ar(ABC)}{ar(PQR)} = \frac{ {BC}^{2} }{ {QR}^{2} } \: \: \: (i)$

Now,

ar(ABC) = 289 cm², BC= 17cm, QR=?

So, first of all, we need to find out the length of the QR.

As PM is the median of triangle PQR, so it divides the base QR into equal halves.

So, we have

QM = MR

and

MR= 6.5cm

so,

QM= 6.5cm

Also,

QR= QM + MR

QR= 6.5 + 6.5

QR= 13cm

Now,

Putting the values in (i), we have

$\frac{ar(ABC)}{ar(PQR)} = \frac{ {17}^{2} }{ {13}^{2} }$

$\frac{289}{ar(PQR)} = \frac{289}{169}$

$ar(PQR) = \frac{289 \times 169}{289}$

$ar(PQR) = 169 {cm}^{2}$

Hence, ar(PQR) = 169 cm².

The correct option is (a) 169 cm².