Question

ABCD is a parallelogram in which ∠DAB = (3x - 50)° and ∠BCD = (x + 40)°, find the value of x. *​

Answer 1

Answer :-

The value of x is 45.

Step-by-step explanation :-

Given that,

$\setlength{\unitlength}{1 cm}\begin{picture}(0,0)\thicklines\qbezier(1,1)(1,1)(6,1)\put(0.4,0.5){\bf B}\qbezier(1,1)(1,1)(1.6,4)\put(6.2,0.5){\bf C}\qbezier(1.6,4)(1.6,4)(6.6,4)\put(1,4){\bf A}\qbezier(6,1)(6,1)(6.6,4)\put(6.9,3.8){\bf D}\qbezier(5.4,1)(5.2,1.48)(6.1,1.5)\qbezier(1.5,3.5)(1.7,3.2)(2.2,4)\put(2.2,3.5){\sf (3x - 50)^ \circ }\put(3.9,1.5){\sf (x + 40)^\circ }\end{picture}$

Solution :-

We know that,

If in a parallelogram ABCD,

∠DAB = ∠BCD ........ opposite angles are equal

Therefore,

$\sf{\implies \: 3x - 50 = x + 40}$

Transposing x to L.H.S and - 50 to R.H.S

$\sf{\implies \: 3x - x = 40 + 50}$

$\sf{\implies \: 2x = 90}$

$\sf{\implies \: x = \dfrac{90}{4}}$

Dividing 90 and 2 with 2

$\sf{\implies \: x = \dfrac{45}{1}}$

$\sf{\leadsto \: x = 45}$

$\underline{\boxed{ \sf{Therefore, \: \: the \: \: value \: \: of \: \: x \: \: is \: \: 45}}}$