Triangles — NCERT Solutions, AI Tutor & Practice

(i) All circles are similar — any two circles have the same shape and differ only in radius.

(ii) All squares are similar — all angles are $90^\circ$ and sides are always in equal ratio.

(iii) Two polygons of the same number of sides are similar if their corresponding angles are equal and their corresponding sides are in the same ratio (proportional).

Both conditions must hold together for polygons.

By the Basic Proportionality Theorem (Thales' Theorem), since $DE \parallel BC$:

$\frac{AD}{DB} = \frac{AE}{EC}$

Substitute the values:

$\frac{3}{4} = \frac{4.5}{EC}$

Cross-multiply:

$3 \times EC = 4 \times 4.5 = 18$

$EC = \frac{18}{3} = 6 \text{ cm}$

First find the segment ratios.

$DB = AB - AD = 6 - 2 = 4 \text{ cm}$ $EC = AC - AE = 9 - 3 = 6 \text{ cm}$

Now compare:

$\frac{AD}{DB} = \frac{2}{4} = \frac{1}{2}, \qquad \frac{AE}{EC} = \frac{3}{6} = \frac{1}{2}$

Since $\frac{AD}{DB} = \frac{AE}{EC}$, by the converse of the Basic Proportionality Theorem, $DE \parallel BC$.

Find the missing angles using angle sum $= 180^\circ$.

In $\triangle ABC$: $\angle C = 180^\circ - 70^\circ - 60^\circ = 50^\circ$.

In $\triangle PQR$: $\angle P = 180^\circ - 60^\circ - 50^\circ = 70^\circ$.

Now compare: $\angle A = \angle P = 70^\circ, \quad \angle B = \angle Q = 60^\circ, \quad \angle C = \angle R = 50^\circ$

Since two (in fact all three) angles are equal, by the AA similarity criterion:

$\triangle ABC \sim \triangle PQR$

All three pairs of corresponding sides are in the same ratio:

$\frac{AB}{DE} = \frac{BC}{EF} = \frac{CA}{FD} = \frac{2}{3}$

By the SSS similarity criterion, the triangles are similar:

$\triangle ABC \sim \triangle DEF$

The scale factor from $\triangle ABC$ to $\triangle DEF$ is $\frac{2}{3}$, meaning $\triangle DEF$ is the larger triangle.

Consider triangles $\triangle OAB$ and $\triangle OCD$.

Since $AB \parallel DC$ and $AC$ is a transversal: $\angle OAB = \angle OCD \quad (\text{alternate angles})$

Since $AB \parallel DC$ and $BD$ is a transversal: $\angle OBA = \angle ODC \quad (\text{alternate angles})$

By the AA similarity criterion: $\triangle OAB \sim \triangle OCD$

Therefore corresponding sides are proportional: $\frac{OA}{OC} = \frac{OB}{OD}$

Hence proved.

In $\triangle RSQ$ and $\triangle RQP$:

$\angle QRS = \angle QRP \quad (\text{common angle } R)$ $\angle RSQ = \angle RQP \quad (\text{given } \angle P = \angle RSQ \text{ leads to this correspondence})$

By AA similarity, $\triangle RSQ \sim \triangle RQP$.

Corresponding sides are proportional: $\frac{RS}{RQ} = \frac{SQ}{QP} = \frac{RQ}{RP}$

Use $\frac{RS}{RQ} = \frac{RQ}{RP}$: $\frac{4}{6} = \frac{6}{RP} \;\Rightarrow\; RP = \frac{36}{4} = 9 \text{ cm} \;(\text{consistent})$

Now $QS$: use $\frac{RS}{RQ} = \frac{SQ}{QP}$. Since $\triangle RSQ \sim \triangle RQP$, the ratio $= \frac{RS}{RQ} = \frac{4}{6} = \frac{2}{3}$.

So $QS = \frac{2}{3} \times QP$. Because the correspondence gives $QS \leftrightarrow QP$ scaled by $\tfrac{2}{3}$, and using consistent similarity, $QS = \frac{2}{3}\times QP$. Taking $QP$ derived from the figure data, $QS = \frac{2}{3}\times 6 = 4$ cm.

In $\triangle ADC$ and $\triangle BAC$:

$\angle ADC = \angle BAC \quad (\text{given})$ $\angle ACD = \angle BCA \quad (\text{common angle } C)$

By the AA similarity criterion: $\triangle ADC \sim \triangle BAC$

Therefore corresponding sides are proportional: $\frac{CA}{CB} = \frac{CD}{CA}$

Cross-multiplying: $CA \times CA = CB \times CD$ $\boxed{CA^2 = CB \times CD}$

Hence proved.

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.