Question

In the figure, AB = AC and AD || BC find the value of y + z - x.

• A. 40

Answer 1

The correct option is A 40
Given: AB || BC and AB = AC
$△$ABC is an isosceles triangle as its two sides are equal in length.
$\therefore$ y = z {Angles opposite to equal sides of an isosceles triangle are equal}
In $△$ABC,
x + y + z = 180${}^{\circ }$
x + 2y = 180${}^{\circ }$ ...(i)
$\angle EAD$ + $\angle DAB$ + $\angle BAC$ = 180${}^{\circ }$
{Angles on a straight line}
55${}^{\circ }$ + x + $\angle DAB$ = 180${}^{\circ }$
$\because$ AD || BC, $\angle DAB$ = y
$\therefore$ 55${}^{\circ }$ + x + y = 180${}^{\circ }$
x +y =125 ...(ii)
Solving (i) and (ii)
x + 2y = 180${}^{\circ }$
2x + 2y = 250
- - -
--------------------------
-x = -70
--------------------------
$\Rightarrow x={70}^{\circ }$
$\Rightarrow y=125-x={55}^{\circ }$
$\Rightarrow z=y={55}^{\circ }$
$\therefore$ x, y and z are 70${}^{\circ }$, 55${}^{\circ }$ and 55${}^{\circ }$ respectively.
Then, y+z-x = 55${}^{\circ }$ + 55${}^{\circ }$ - 70${}^{\circ }$ = 110${}^{\circ }$ - 70${}^{\circ }$ = 40${}^{\circ }$