Test Index

NDA Math Polygons and Quadrilaterals

DIRECTIONS (Qs. 36-37) : Read the following information and answer the two items that follow:

\text{A B C D}

is a trapezium such that

\text{A B}

and

\text{C D}

are parallel and

\text{B C}

is perpendicular to them. Let

\angle ADB = \theta, \angle ABD = \alpha, BC = p

and

CD = q

and

CD = q

[NDA-I 2020]

Question : 12 of 12

Marks: +1, -0

What is

A B

Solution: 👈: Video Solution

(a) From

\Delta A B D

\frac{AB}{\sin \theta} = \frac{BD}{\sin (180^{\circ} - (\alpha + \theta))}

\Rightarrow AB = \frac{BD \cdot \sin \theta}{\sin (\alpha + \theta)} = \frac{BD \cdot \sin \theta}{\sin \alpha \cdot \cos \theta + \cos \alpha \cdot \sin \theta}

Again from

\Delta B C D

BD = \sqrt{(BC)^{2} + (CD)^{2}} = \sqrt{p^{2} + q^{2}}

\sin \alpha = \frac{BC}{BD} = \frac{p}{\sqrt{p^{2} + q^{2}}}

\cos \alpha = \frac{CD}{BD} = \frac{q}{\sqrt{p^{2} + q^{2}}}

\therefore AB = \frac{\sqrt{p^{2}+q^{2}} \cdot \sin \theta}{\frac{p \cdot \cos \theta}{\sqrt{p^{2}+q^{2}}} + \frac{q \cdot \sin \theta}{\sqrt{p^{2}+q^{2}}}}

= \frac{(p^{2}+q^{2}) \cdot \sin \theta}{p \cdot \cos \theta + q \cdot \sin \theta}

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