Question

2. In the figure, PQRS is a parallelogram and
ZQPR = 90°, QP = 9 cm, QR = 15 cm and
PT I QR. Find (a) area of A QPR (b) area of
parallelogram PQRS.​

Answer 1

Answer:

(a) Since, AQPR is a right angled triangle because Z QPR = 90°

and, Hypotenuse = QR = 15cm (given)

height = PR = ? =

base = QP = 9cm (given)

So, By Pythagoras theorem,

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

=> (QR)2 = (PR)2 + (QP)2

=> (15)? = (PR)? + (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, a? - b² = (a+b)(a-b)=> (15 + 9)(15 9) = (PR)?

=> 24 x 6 = (PR)?

=> 144 = (PR)?

=> PR = 144

=> PR = V12 × 12

=> PR = 12 cm = height.

thus, area of APQR,

x base x height

=} x 12 x 9

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

= 6 x 9

54cm? (ANSWER)

(b) Since, PQRS is a parallelogram,so, opposite sides are equal,

.. QR = PS = 15cm

and,

PQ = RS = 9cm

so, area of AQPR = area of APRS (equation 1)

hence, area of PQRS,

= area of AQPR + area of APRS

= area of AQPR + area of AQPR (from equation 1)

= 2 x area of AQPR

= 2 x 54 cm

= 108cm2

Answer 2

Answer:

hope it helps u

Step-by-step explanation:

Since, ∆QPR is a right angled triangle because 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 =

=> PR =

=> PR = 12 cm = height.

thus, area of ∆PQR,

=  × base × height

=  ×  × 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 ,

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)

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