Understanding Quadrilaterals Test - 1

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, D² = AD² - AP² 

(Pythagoras theorem) = (2a)² - a² = 40a² - a² = 3a² In ΔDPB. We have BD² = DP² + PB² = 3a² + a² = 4a² 

∴ BD = 2a

so, AB = BO = AD = 2a

∴ Δ ABD is an equilateral triangle.

∴ ∠A = 60° and ∠ABD = 60° Similarly, ΔBDC is 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) = 48 cm ⇒ 16x = 48 ⇒ x = 3

∴ Length of the sides = (3 × 3) cm = 9cm and (5 × 3) cm = 15cm.