Question

in ∆ pqr M and N are the midpoints of pq and pr respectively if the area of ∆PMN is 24 cm² find the area of ∆ pqr​​

Answer 1

Answer:

AREA OF ∆PQR=96cm²

Step-by-step explanation:

Semi Perimeter of ∆PQR=

$\frac{pq + qr + pr}{2}$

PM=PQ/2,PN=PR/2

BY MID POINT THEOREM:

MN//QR & MN=QR/2

Semiperimer of ∆ PMN =

$\frac{1}{2} ( \frac{pq + qr + pr}{2} ) = \frac{s}{2}$

AREA OF ∆PMN=24cm²

$= \sqrt{ \frac{s}{2}( \frac{s}{2} - \frac{pq}{2} )( \frac{s}{2} - \frac{qr}{2})( \frac{s}{2} - \frac{pr}{2} ) } \\ = \frac{1}{4} \times x$

Here x represents the area of ∆PQR i.e

$\sqrt{s(s - pq)(s - qr)(s - pr)}$

A.T.Q

$\frac{1}{4} \times x = 24 \\ therefore \: \: x = 96 {cm}^{2}$

AREA OF ∆ PQR= 96 cm²

Answer 2

Answer:

ANS IS 96 CM square.

DO IT BY YOURSELF!

ALL THE BEST FOR BOARD EXAM!