Question

In a rhombus $PQRS,\angle SQR={40}^{\circ }$ and $PQ=3cm$. Find $\angle SPQ,\angle QSR$ and the perimeter of the rhombus

Answer 1

Given that,

$PQRS$ is a rhombus with $\angle SQR={40}^o$ and $PQ=3cm$.

To find out,

$\angle SPQ,\angle QSR$ and the perimeter of the rhombus.

On the basis of given information, we can draw a rhombus $PQRS$ as shown in the above figure.

We know that, all sides of a rhombus are equal.

Hence, in $\Delta QSQ$,

$SR=QR$

We know that, angles opposite to equal sides of a triangle are equal.

Hence, $\angle SQR=\angle QSR$

$\therefore \angle QSR={40}^o$

Also, the sum of all the interior angles of a triangle is ${180}^o$.

Hence, $\angle SQR+\angle QSR+\angle SRQ={180}^o$

$\Rightarrow {40}^o+{40}^o+\angle SRQ={180}^o$

$\therefore \angle SRQ={180}^o-{80}^o$

$={100}^o$

We also know that, opposite angles of a rhombus are equal.

Hence, $\angle SPQ=\angle SRQ$

$\therefore \angle SPQ={100}^o$

Now, perimeter of a rhombus $=4\times side$

Here, side of the rhombus $=3cm$

Hence, perimeter of the rhombus $=4\times 3$

$=12cm$.

Hence, $\angle SPQ={100}^o,\angle QSR={40}^o$ and perimeter of the rhombus $=12cm$.