Question

ABCD is a rhombus in which altitude from point D to the side AB bisects AB. The angles of the rhombus are:

(a) 150°, 30°, 150°, 30°

(b) 135°, 45°, 135°, 45°

(c) 116\(\frac 13\)°, 63\(\frac 23\)°, 116\(\frac 13\)°, 62\(\frac 23\)°

(d) 120°, 60°, 120°, 60°

Answer 1

Correct option is (d) 120°, 60°, 120°, 60°

Given that ABCD is a rhombus and the altitude on AB such that AE = EB.

In a △AED and △BED, we have

DE = DE (common line)

∠AED = ∠BED (right angle)

AE = EB (DE bisects AB)

∴ △AED ≅ △BED ( by SAS property)

∴ AD = BD (by C.P.C.T)

But AD = AB (sides of rhombus are equal)

⇒ AD = AB = BD

∴ ABD is an equilateral triangle.

∴∠A = 60^∘

⇒ ∠A = ∠C = 60^∘ (opposite angles of rhombus are equal)

But Sum of adjacent angles of a rhombus is supplementary.

∠ABC + ∠BCD = 180^∘

⇒ ∠ABC + 60^∘ = 180^∘ 

⇒ ∠ABC = 180^∘ − 60^∘ = 120^∘

∴ ∠ABC = ∠ADC = 120^∘ (opposite angles of rhombus are equal)

Hence, Angles of rhombus are ∠A = 60^∘ and ∠C = 60^∘, ∠B = ∠D = 120^∘.