Question

The diagonal of a quadrilateral ABCD bisect each other at O. If angle A is 70 degree,then angle B in degrees is
(1 Point)​

Answer 1

Answer:

$\sf{\angle B = {110}^{\circ}}$

Step-by-step explanation:

Given :

$\sf{\angle A = {70}^{\circ}}$

The diagonal of the quadrilateral ABCD bisect each other

To find :

$\sf{\angle B \:in \:degrees}$

Concept :

Use the parallelogram theorem .

The diagonal of the parallelogram bisect each other . So , it is a parallelogram .

So , sum of the adjacent angle is 180° .

So , to find angle B use this formula -:

$\boxed{\sf{\angle A + \angle B = {180}^{\circ}}}$

Solution :

$:\implies\sf{{70}^{\circ}+\angle B = {180}^{\circ}}$

$:\implies\sf{\angle B = {180}^{\circ}-{70}^{\circ}}$

$:\implies\sf{\angle B = {110}^{\circ}}$

So , the angle B is 110° .

Extra information :

▪︎To find area of the Rhombus -:

$\boxed{\sf{A=\dfrac{1}{2}\times d_{1}\times d_{2}}}$

Where,

A = Area of the Rhombus

$\sf{d_{1} = First \:Diagonal }$

$\sf{d_{2} = Second\: Diagonal }$

▪︎To find the perimeter of the Rhombus -:

$\boxed{\sf{P=4a}}$

Where,

P = Perimeter of the Rhombus

a = Side