CAT Areas of Similar Triangles MCQ - Practice Questions & Answers

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4 Questions around this concept.

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The areas of two similar triangles ABC and PQR are 25 cm² and 49 cm² respectively. If QR = 8.4 cm, find BC in cm.

5 cm

6 cm

7 cm

8 cm

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Concepts Covered - 1

Areas of Similar Triangles

You have learnt that in two similar triangles, the ratio of their corresponding sides is the same. Do you think there is any relationship between the ratio of their areas and the ratio of the corresponding sides?

You know that area is measured in square units. So, you may expect that this ratio is the square of the ratio of their corresponding sides. This is indeed true and we shall prove it in the next theorem.

Thoerem : The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

Given: $\triangle A B C \sim \triangle D E F$

To prove: $\frac{\operatorname{ar}(\Delta A B C)}{\operatorname{ar}(\Delta D E F)}=\frac{A B^{2}}{D E^{2}}=\frac{A C^{2}}{D F^{2}}=\frac{B C^{2}}{E F^{2}}$

Construction: $\text{Draw } A L \perp B C\text{ and }D M \perp E F$

Prrof: $\triangle A B C \sim \triangle D E F$

$\therefore \quad \angle A=\angle D,\;\; \angle B=\angle E,\;\; \angle C=\angle F$

$\text { and } \frac{A B}{D E}=\frac{B C}{E F}=\frac{A C}{D F}\qquad\ldots(i)$

$\text { Now, in }\Delta ABL\text{ and }\Delta DEM$

$\angle B =\angle E \qquad \text{(As } \Delta ABC \sim \Delta DEF )$

$\\ \angle L =\angle M \qquad \text { (Each is of }\left.90^{\circ}\right)$

$\therefore\quad\Delta ABL \sim \Delta DEM \qquad \text{(AA similarity criterion)}$

$\therefore\quad\frac{ AL }{ DM }=\frac{ AB }{ DE }\qquad\ldots(ii)$

Form (i) and (ii)

$\frac{AL}{DM}=\frac{B C}{E F}\qquad\ldots(iii)$

$\\\text { Now, } \operatorname{ar}(\triangle A B C)=\frac{1}{2} \times B C \times A L \\$

$\text { and }\;\; \operatorname{ar}(\triangle D E F)=\frac{1}{2} \times E F \times D M$

$\begin{aligned} \frac{\operatorname{ar}( ABC )}{\operatorname{ar}( DEF )} &=\frac{ AB }{ DE } \times \frac{ AL }{ DM }\\ &=\frac{ AB }{ DE } \times \frac{ AB }{ DE } \qquad\qquad\text{[From (iii)]} \\ &=\left(\frac{ AB }{ DE }\right)^{2} \end{aligned}$

$\text { Similarly, } \frac{\operatorname{ar}(\Delta A B C)}{\operatorname{ar}(\Delta D E F)}=\frac{A B^{2}}{D E^{2}} \text { and } \frac{\operatorname{ar}(\Delta A B C)}{\operatorname{ar}(\Delta D E F)}=\frac{A C^{2}}{D F^{2}} \\$

$\text { Hence, } \frac{\operatorname{ar}(\triangle A B C)}{\operatorname{ar}(\Delta D E F)}=\frac{A B^{2}}{D E^{2}}=\frac{A C^{2}}{D F^{2}}=\frac{B C^{2}}{E F^{2}}$