Question

ABCD is a rhombus in which Altitude from D to side AB bisects AB. Find the angles of the rhombus? Altitude from D to side AB bisects AB.
A. 110, 70, 110, 70
B. 120, 60, 120, 60
C. 125, 55, 125, 55
D. 135, 45, 135, 45

Answer 1

Hint: To solve this question, we need to know the basic theory related to the quadrilateral. As we know rhombus is a type of quadrilateral and also It is a special case of a parallelogram, whose diagonals intersect each other at 90 degrees. Here, by using various theorems or properties we will Find the angles of the rhombus.

Complete step-by-step answer:
Given that ABCD is a Rhombus and DE is the altitude on AB then AE= EB

In a $\vartriangle $AED and$\vartriangle $BED,
DE = DE (common line)
$\angle $AED = $\angle $BED (right angle)
AE = EB (DE is an altitude)
∴ $\vartriangle $AED ≅$\vartriangle $ BED (SAS property)
∴ AD = BD (by C.P.C.T)
But AD = AB ( Sides of rhombus are equal)
$ \Rightarrow $ AD = AB = BD
$\therefore $ ABD is an equilateral triangle.
$\therefore $ $\angle $A = ${60^0}$
\[ \Rightarrow \angle A{\text{ }} = \angle C{\text{ }} = {\text{ }}60^\circ \]( opposite angles of a rhombus are equal )
Always, when we add adjacent angles of a rhombus, it is supplementary in nature.
$\angle $ABC + $\angle $BCD = ${180^0}$
$ \Rightarrow $ $\angle $ABC + ${60^0} = {180^0}$
$ \Rightarrow $ $\angle $ABC = ${180^0} - {60^0} = {120^0}$
∴ $\angle $ABC = $\angle $ADC = ${120^0}$. (opposite angles of rhombus are equal)
∴ Angles of rhombus are \[\angle A{\text{ }} = {\text{ }}60^\circ and\angle C{\text{ }} = {\text{ }}60^\circ ,\angle B{\text{ }} = \angle D{\text{ }} = {\text{ }}120^\circ .\]
Therefore, option (B) is the correct answer.

Note:Rhombus has all its sides equal and so does a square. Also, the diagonals of any square are perpendicular (means ${90^ \circ }$) to each other and bisect the opposite angles. Therefore, a square is a type of rhombus. In rhombus the opposite angles are equal to each other. Also, in rhombus the diagonals bisect these angles.