Exercise 6.3: Triangles — NCERT Solutions

For similar triangles, the ratio of areas equals the square of the ratio of corresponding sides:

$\frac{\text{Area}_1}{\text{Area}_2} = \left(\frac{3}{5}\right)^2 = \frac{9}{25}$

Let the larger area be $A$. The smaller area is $54\ \text{cm}^2$:

$\frac{54}{A} = \frac{9}{25}$

$A = \frac{54 \times 25}{9} = 6 \times 25 = 150 \text{ cm}^2$

The area of the larger triangle is $150\ \text{cm}^2$.

The pole with its shadow and the tower with its shadow form two similar triangles (same angle of elevation of the sun, both vertical objects give AA similarity).

Let $h$ be the height of the tower.

$\frac{\text{Height of pole}}{\text{Shadow of pole}} = \frac{\text{Height of tower}}{\text{Shadow of tower}}$

$\frac{6}{4} = \frac{h}{28}$

$h = \frac{6 \times 28}{4} = \frac{168}{4} = 42 \text{ m}$

The height of the tower is $42$ m.

Ratio of areas: $\frac{121}{64}$

Ratio of corresponding sides is the square root: $\frac{\text{side}_{\text{large}}}{\text{side}_{\text{small}}} = \sqrt{\frac{121}{64}} = \frac{11}{8}$

Let the longest side of the smaller triangle be $x$. The corresponding side of the larger triangle is $22$ cm:

$\frac{22}{x} = \frac{11}{8}$

$x = \frac{22 \times 8}{11} = \frac{176}{11} = 16 \text{ cm}$

The longest side of the smaller triangle is $16$ cm.

In right triangle $ABC$ with the right angle at $A$, $AD \perp BC$.

Consider $\triangle ADB$ and $\triangle CDA$.

$\angle ADB = \angle CDA = 90^\circ$

Also, $\angle DAB = \angle DCA$ (both equal $90^\circ - \angle B$; since $\angle BAC=90^\circ$, $\angle DAB + \angle DAC = 90^\circ$, and in $\triangle ADC$, $\angle DCA + \angle DAC = 90^\circ$).

By AA similarity: $\triangle ADB \sim \triangle CDA$

Therefore: $\frac{AD}{CD} = \frac{BD}{AD}$

Cross-multiplying: $AD^2 = BD \times DC$

Hence proved.

For similar triangles, the ratio of perimeters equals the ratio of corresponding sides:

$\frac{\text{Perimeter}_1}{\text{Perimeter}_2} = \frac{30}{20} = \frac{3}{2}$

Let the corresponding side of the second triangle be $x$. The side of the first triangle is $12$ cm:

$\frac{12}{x} = \frac{3}{2}$

$x = \frac{12 \times 2}{3} = \frac{24}{3} = 8 \text{ cm}$

The corresponding side of the second triangle is $8$ cm.

Let each side of the equilateral triangle be $AB = BC = CA = a$.

Draw $AE \perp BC$. In an equilateral triangle the foot $E$ is the midpoint, so: $BE = \frac{a}{2}$

Given $BD = \frac{1}{3}BC = \frac{a}{3}$.

Then: $DE = BE - BD = \frac{a}{2} - \frac{a}{3} = \frac{3a - 2a}{6} = \frac{a}{6}$

In right triangle $AEB$: $AE^2 = AB^2 - BE^2 = a^2 - \frac{a^2}{4} = \frac{3a^2}{4}$

In right triangle $AED$: $AD^2 = AE^2 + DE^2 = \frac{3a^2}{4} + \frac{a^2}{36}$

Take LCM $36$: $AD^2 = \frac{27a^2}{36} + \frac{a^2}{36} = \frac{28a^2}{36} = \frac{7a^2}{9}$

Therefore: $9\,AD^2 = 7a^2 = 7\,AB^2$

Hence proved.