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In a trapezium $ABCD,\ AB||DC,\ \angle A:\angle D=3:2$ and $\angle B:\angle C=4:5$. Find the angle of the trapezium.
Given: In a trapezium $ABCD$, $AB||DC$, $\angle A : \angle D=3:2$ and $\angle B : \angle C=4:5$.
To do: To find the angle of the trapezium.
Solution:
As given $\angle A:\angle D=3:2$.
Let $\angle A$ and $\angle D$ are $3x$ and $2x$ respectively.$\angle A$ and $\angle D$ together will be $180^o$.
$\Rightarrow 3x+2x=5x=180$
$\Rightarrow x=\frac{180}{5}=36^o$
Therefore, $\angle A=3x=3\times36^o=108^o$
$\angle D=2x=2\times 36^o=72^o$
Now, given: $\angle B:\angle C=4:5$.
Let $\angle B$ and $\angle C$ are $4x$ and $5x$ respectively.
$\angle B$ and $\angle C$ together will be of $180^o$.
$\Rightarrow 4x+5x=180$
$\Rightarrow 9x=180$
$\Rightarrow x=\frac{180}{9}=20^o$
Therefore, $\angle B=4x=4\times20=80^o$
$\angle C=5x=5\times 20=100^o$
Thus, angles of trapezium $ABCD$ are $108^o,\ 72^o,\ 80^o$ and $100^o$.
Updated on: 10-Oct-2022
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