Understanding Quadrilaterals Test - 3

In the given figure, ABCD is a rhombus and DP_AB such that AP=PB.
Join BD. Let AB = BC = CD = AD = 2a
So, AP = a = PB In ΔAPD, DP² = AD² − AP²
(Pythagoras theorem) = (2a)² − a² = 4a² − a² = 3a²
In ΔDPB
we have BD² = DP² + PB² = 3a² + a² = 4a²
∴ BD = 2a So, AB = BO = AD = 2a
∴ ΔABD is an equilateral triangle.
∴ ∠A = 60^∘ and ∠ABD = 60^∘
Similarly,
ΔBDCis an equilateral triangle and ∠DBC=60^∘
Now, ∠ABC =∠ABD +∠DBC = 60^∘ + 60^∘ = 120^∘
So, ∠A = 60^∘ and ∠B= 120^∘.

The diagonals of a rhombus bisect each other at right angles,

Let ABCD be the parallelogram in which AB = 3x cm and BC = 5x cm.
Since opposite sides of a parallelogram are equal,
∴ AB = DC = 3x cm and BC = AD = 5x cm
Perimeter of the parallelogram = 48 cm
⇒ AB + BC + CD + DA = 48cm
⇒ (3x + 5x + 3x + 5x) = 48cm
⇒ 16x = 48
⇒ x = 3
∴ Length of the sides = (3×3)cm = 9cm
and
(5×3)cm = 15cm (3x + 5x + 3x + 5x) = 48cm
⇒ 16x = 48
⇒ x = 3
∴ Length of the sides = (3×3)cm = 9cm
and
(5×3)cm = 15cm.