Question

(b) P is a point of side AD of parallelogram
ABCD. Find the area of parallelogram
ABCD if PB = 8 cm, BC = 17 cm and
<BPC = 90°.
[3]​

Answer 1

Step-by-step explanation:

I hope....it will be helful for you

Answer 2

Given :

BP = 8cm

BC = 17cm

<BPC = 90°

To Find:

Area of parallelogram ABCD

Solution :

•since, Triangle BPC is Right angled triangle

So , by Pythagoras theorem

BP²+PC² = BC²

(8)²+ PC² =( 17)²

PC² = 289-64

PC² = 225

PC = 15 cm

•

Area of triangle = 1/2 × base × height

Area of triangle BPC = 1/2 ×8×15

Area of triangle BPC = 4×15 = 60 cm²

•Consider the theorm ,

If triangle and a parallelogram are on the same base and between the

same parallels, then prove that the area of the triangle is equal to half the area of the

parallelogram.

• So , In triangle BPC & parallelogram ABCD

BC is common base & both lies between same parallels i.e. BC & AD

=> Area of ∆BPC = 1/2 Area of parallelogram ABCD

=> 60 cm² = 1/2 Area of parallelogram ABCD

=>Area of parallelogram ABCD = 2×60

=>Area of parallelogram ABCD = 120 cm²