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CAT Areas of Similar Triangles - Practice Questions & MCQ

Edited By admin | Updated on Oct 04, 2023 04:20 PM | #CAT

Syllabus

Quick Facts

2 Questions around this concept.

Solve by difficulty

DIFFICULT

If triangles ABC and DBC are on the same base BC and AD intersects BC at O

then $\frac{\operatorname{ar}(\Delta A B C)}{\operatorname{ar}(\Delta D B C)}$

$\frac{A O}{D O}$

$\left (\frac{A O}{D O} \right )^2$

$\frac{A L}{D M}$

Both (a) and (c)

View Solution

In the given figure, the line segment XY is parallel to the side AC of the triangle ABC and it divides the triangle into two parts of equal area. Then AX:AB is

1:2

1:4

$1:\sqrt2$

None of these

View Solution

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)

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

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

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

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

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

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