Question

Please help! I think I have the equation set up.

Answer 1

$\bold{\huge{\underline{ Solution }}}$

Given :-

• Here, we have given one quadrilateral that is quadrilateral ABCD
• We also have given the angles of quadrilateral that is ( 6x + 5)° , ( 9x - 10)° , 80° and a right angle

To Find :-

Here, we have to find the value of x

Let's Begin :-

We have given quadrilateral ABCD here , whose angles are as follows

• Angle A = ( 6x + 5)°
• Angle B = ( 9x - 10)°
• Angle C = 80°
• Angle D = A right angled triangle

[ The measure of right angled triangle is 90° ]

We know that,

• Sum of the angles of quadrilateral is equal to 360°

That is

$\bold{\angle{ A + }}{\bold{\angle{B +}}}{\bold{\angle{C + }}}{\bold{\angle{D + = 360{\degree}}}}$

Subsitute the required values

$\sf{ ( 6x + 5){\degree} + 80{\degree}+ 90{\degree} + (9x - 10){\degree} = 360{\degree}}$

$\sf{ 6x + 5 + 170{\degree} + 9x - 10 = 360{\degree}}$

$\sf{ 15x - 5 = 360{\degree} - 170{\degree}}$

$\sf{ 15x - 5 = 190 {\degree}}$

$\sf{ 15x = 190 {\degree} + 5 }$

$\sf{ 15x = 195 {\degree}}$

$\sf{ x = }{\sf{\dfrac{ 195}{15}}}$

$\sf{ x = }{\sf{\cancel{\dfrac{ 195}{15}}}}$

$\bold{ x = 13 }$

Hence, The value of x is 13 .

$\bold{\huge{\underline{\red{ Verification }}}}$

Measure of Angle A

$\sf{ = (6x + 5){\degree}}$

$\sf{ = (6(13) + 5 ){\degree}}$

$\sf{ = (78 + 5 ){\degree}}$

$\sf{ = 83{\degree}}$

Measure of Angle D

$\sf{ = (9x - 10){\degree}}$

$\sf{ = (9(13) - 10){\degree}}$

$\sf{ = (117 - 10 ){\degree}}$

$\sf{ = 107{\degree}}$

Now, we know that,

• Sum of angles of triangles is equal to 360°

That is,

$\sf{ 83{\degree} + 80{\degree}+ 90{\degree} + 107 {\degree} = 360{\degree}}$

$\sf{ 163{\degree}+ 90{\degree} + 107 {\degree} = 360{\degree}}$

$\sf{ 253{\degree}+ 107 {\degree} = 360{\degree}}$

$\sf{ 253{\degree}+ 107 {\degree} = 360{\degree}}$

$\bold{ 360{\degree} = 360{\degree}}$

$\bold{ LHS = RHS }$

Hence, Proved.