Question

solve this question find area of pqrs

Answer 1

(a) Since, ∆QPR is a right angled triangle because $\angle$QPR = 90°

and, Hypotenuse = QR = 15cm (given)

height = PR = ?

base = QP = 9cm (given)

So, By Pythagoras theorem,

( Hypotenuse )² = ( Base )² + ( height )²

=> (QR)² = (PR)² + (QP)²

=> (15)² = (PR)² + (9)²

Now, take (9)² to other side as it is positive, so taking it other side , it will become negative.

=> (15)² - (9)² = (PR)²

using the identity, a² - b² = (a+b)(a-b) :-

=> (15 + 9)(15 - 9) = (PR)²

=> 24 × 6 = (PR)²

=> 144 = (PR)²

=> PR = $\sqrt{144}$

=> PR = $\sqrt{12 × 12}$

=> PR = 12 cm = height.

thus, area of ∆PQR,

= $\frac{1}{2}$ × base × height

= $\frac{1}{\cancel{2}}$ × $\cancel{12}$ × 9

when we divide 12 by 2 we get 6 , i.e. 12 ÷ 2 = 6

= 6 × 9

= 54cm² (ANSWER)

(b) Since, PQRS is a parallelogram,

so, opposite sides are equal ,

$\therefore$ QR = PS = 15cm

and,

PQ = RS = 9cm

so, area of ∆QPR = area of ∆PRS ----(equation 1)

hence, area of PQRS,

= area of ∆QPR + area of ∆PRS

= area of ∆QPR + area of ∆QPR (from equation 1)

= 2 × area of ∆QPR

= 2 × 54 cm²

= 108cm² (Answer)

Answer 2

Answer:

Step-by-step explanation:

A=B*H

=> 15*8

=>120 CM PER 2