MadeEasy Test Series: General Aptitude - Geometry

Question

Q.5

In the given figure, triangles ABC and PQR are right angled triangles with angle C and Q being right angles. QR is parallel to AC and AB = 300 cm, PQ = 20 cm and QR = 100 cm. The length of side BC to the nearest integer is ___________ cm.

Answer 1

Answer : $59 \ cm$

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In the right angled $\Delta PQR$, using Pythagoras theorem,

$PR^{2} = PQ^{2} + QR^{2}$.

$\Rightarrow PR = \sqrt{100^{2}+ 20^{2}}$,

$\Rightarrow PR = 102 \ cm$ ( after rounding off).

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Now in the right angled triangles $\Delta PQR$ and  $\Delta BCA$,

$\angle PQR = \angle BCA = 90^{\circ}$,

also $QR$ is parallel to $AC$, 

so $\angle QRP = \angle CAB$ ( as they are alternate interior angles).

Now since the above two pairs of angles in $\Delta PQR$ and  $\Delta BCA$ are equal, the third pair of angles  i.e. $\angle QPR$ and $\angle CBA$ will also be equal to each other.

Hence $\Delta PQR$ and  $\Delta BCA$ are similar triangles.

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Using a property of similar triangles, ratio of corresponding sides of both the triangles will be same,

that is $\frac{BC}{AB} = \frac{PQ}{PR}$,

$\Rightarrow BC = \frac{PQ \times AB}{PR}$,

$\Rightarrow BC = \frac{20 \times 300}{102} = 58.82 \ cm $.

Comments

In place of RQ you should take PQ.

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Thanks for noticing.

Edited.