Question

. If the angles of a quadrilateral are x, (2x+13), (3x+10) and (x-6). Find all the four angles.

Answer 1

Answer:

Given :-

Angles of Quadrilateral are x, (2x +13),(3x+10),(x-6)

To Find :-

All four angles

Solution :-

As we know that Sum of all angles in a Quadrilateral is 360⁰.

$\sf \: x + (2x + 13) + (3x + 10) + (x - 6) = 360$

Revoming Bracket

$\sf \: x + 2x + 13 + 3x + 10 + x - 6 = 360$

Adding variable

$\sf \: 7x + 13 + 10 - 6 = 360$

Adding 13 and 10

$\sf \: 7x + 23 - 6 = 360$

Subtracting 6 from 23

$\sf \: 7x + 17 = 360$

Subtracting 17 from 360

$\sf \: 7x = 360 - 17$

$\sf \: 7x = 343$

Dividing 343 and 7 to find x

$\sf \: x = \dfrac{343}{7}$

$\sf \: x = 49$

Finding Angles

$\bf \: x = 49$

$\bf \: 2x + 13 = 2(49) + 13 = 98 + 13 = 111$

$\bf \: 3x + 10 = 3(49) + 10 = 147 + 10 = 157$

$\bf \: x - 6 = 49 - 6 = 43$

Let's verify :-

49 + 111 + 157 + 43 = 360

160 + 200 = 360

360 = 360

And, Done.