Question

If the angles of a quadrilateral takes in order are proportional to $1:2:3:4$ . Prove that it is trapezium.

Answer 1

Angles of a quadrilateral are in ratio $1:2:3:4$

Let the quadrilateral be ABCD

$A:B:C:D=1:2:3:4$

Let $A=x$

$x:2x:3x:4x=A:B:C:D$

We know that

$\angle A+\angle B+\angle C+\angle D={360}^o$

$x+2x+3x+4x={360}^o$

$10x={360}^o$

$x={36}^o$

$\angle A={36}^o$ …………$\left(1\right)$

$\angle B={72}^o$ …………..$\left(2\right)$

$\angle C={108}^o$ ………….$\left(3\right)$

$\angle D={144}^o$ …………$\left(4\right)$

By adding $\left(1\right)$ & $\left(4\right)$

We get

$\angle A+\angle D={180}^o$

By adding $\left(2\right)$ & $\left(3\right)$

We get

$\angle C+\angle B={180}^o$

So we can conclude that AB$||$CD.

So it is a trapezium.