Question

In the diagram, ABCD is a rhombus. AEC and BED are straight lines. Find the value of $p+q+r+s+t$

• A. ${540}^{\circ }$
• B. ${360}^{\circ }$
• C. ${200}^{\circ }$
• D. ${270}^{\circ }$

Answer 1

The correct option is D ${270}^{\circ }$

In the rhombus ABCD,
$\angle A+\angle B+\angle C+\angle D={360}^{\circ }$ [Angle sum property]

The diagonals of a rhombus bisect the angles at the vertex.
$\therefore \angle DAC=\angle CAB=\frac{1}{2}\angle DAB$
$\Rightarrow p=\frac{1}{2}\angle A$
$\Rightarrow \angle A=2p$

Similarly, $\angle B=2q$, $\angle C=2t$ and $\angle D=2s$
$\Rightarrow 2p+2q+2t+2s={360}^{\circ }$
$\Rightarrow p+q+t+s={180}^{\circ }$

The diagonals of a rhombus intersect at ${90}^{\circ }$
$\therefore \angle AEB=r={90}^{\circ }$

Thus, $p+q+r+s+t={180}^{\circ }+{90}^{\circ }$
$\Rightarrow p+q+r+s+t={270}^{\circ }$