Question

ABCD is a parallelogram.P is a point on CD .if area of ∆APB= 8cm sq. find the area of parallelogram ABCD.

Answer 1

• area of triangle APB = area of triangle ADP
• [two triangles having same base and lie between same parallels]
• = 8 cm sq.
• Similarly, ar(triangle APB) = at(triangle BPC)
• Therefore,ar(ABCD) = ar(ABP) + ar(ADP) +

ar(BPC)

• = 8 + 8 + 8
• = 24 cm sq.

Answer 2

Answer:

(Diagram given in attachment)

AC = 6 cm

⇒ AO + CO = 6

⇒ AO + AO = 6             [∵ AO = CO]

⇒ 2AO = 6

⇒ AO = 6 ÷ 2

⇒ AO = 3 cm

⇒ AO = CO = 3 cm

BD = 8 cm

⇒ BO + OD = 8

⇒ BO + BO = 8                [∵ BO = OD]

⇒ 2BO = 8

⇒ BO = 8 ÷ 2

⇒ BO = 4 cm

⇒ BO = OD = 4 cm

In ΔAOB,

By Pythagoras Theorem,

(AB)² = (AO)² + (BO)²

⇒ (AB)² = 3² + 4²

⇒ (AB)² = 9 + 16

⇒ (AB)² = 25

⇒ (AB) = √25

⇒ AB = ±5

Side cannot be negative

∴ Side of Rhombus = 5 cm

Now let's find the perimeter

Perimeter = 4(Side)

Perimeter = 4(5)

★ Perimeter = 20 cm ★