Question

In figure 11.36,ABCD is a trapezium .If x=4/3y,y=3/8z, then find the value of x.

Answer 1

Here is the required answer

Answer 2

Answer:

x = 48°, y = 36°, z = 96°

Step-by-step explanation:

Given information: ABCD is a trapezium, AB║CD $x=\frac{4}{3}y$ and $y=\frac{3}{8}z$.

If a transversal line intersect two parallel lines, then alternate interior angles are congruent.

$\angle ABD\cong \angle BDC$

$m\angle ABD=m\angle BDC$

$x=m\angle BDC$

According to angle sum property of triangle, the sum of interior angles of a triangle is 180°.

In triangle BCD,

$\angle BCD+\angle BDC+\angle CBD=180$

$z+x+y=180$

Substitute the given values.

$z+\frac{4}{3}y+\frac{3}{8}z=180$

$z+\frac{4}{3}\times \frac{3}{8}z+\frac{3}{8}z=180$

$z+\frac{1}{2}z+\frac{3}{8}z=180$

$\frac{8+4+3}{8}z=180$

$\frac{15}{8}z=180$

Isolate variable term.

$z=180\times \frac{8}{15}$

$z=96$

The value of z is 96°.

$y=\frac{3}{8}z=\frac{3}{8}(96)=36$

$x=\frac{4}{3}y=\frac{4}{3}(36)=48$

Therefore the value of y is 36° and value of x is 48°.