NCERT Solutions for Class 10 Maths Chapter 6 Triangles Exercise 6.3#

From the figure, in △ABC and △PQR:
$\angle A = \angle P = 60°$
$\angle B = \angle Q = 80°$
$\angle C = \angle R = 40°$

∴ By AAA similarity criterion, $\triangle ABC \sim \triangle PQR$.

From the figure, AB = 2, BC = 2.5, CA = 3 and PQ = 6, QR = 4, RP = 5.

In △ABC and △QRP:

$$\frac{AB}{QR} = \frac{2}{4} = \frac{1}{2}$$

$$\frac{BC}{RP} = \frac{2.5}{5} = \frac{1}{2}$$

$$\frac{CA}{PQ} = \frac{3}{6} = \frac{1}{2}$$$$\Rightarrow \frac{AB}{QR} = \frac{BC}{RP} = \frac{CA}{PQ}$$

∴ All corresponding sides are proportional, so by SSS similarity criterion, $\triangle ABC \sim \triangle QRP$.

From the figure, LM = 2.7, MP = 2, LP = 3 and DE = 4, EF = 5, DF = 6.

Comparing the sides in order of size:

$$\frac{MP}{DE} = \frac{2}{4} = \frac{1}{2}$$

$$\frac{LP}{DF} = \frac{3}{6} = \frac{1}{2}$$

$$\frac{LM}{EF} = \frac{2.7}{5} = \frac{27}{50}$$$$\Rightarrow \frac{1}{2} \neq \frac{27}{50}$$

∴ The sides are not all proportional, so the triangles are not similar.

From the figure, in △MNL, MN = 2.5, ML = 5, ∠M = 70°, and in △QPR, QP = 5, QR = 10, ∠Q = 70°.

In △MNL and △QPR:

$$\frac{MN}{QP} = \frac{2.5}{5} = \frac{1}{2}$$

$$\frac{ML}{QR} = \frac{5}{10} = \frac{1}{2}$$$$\Rightarrow \frac{MN}{QP} = \frac{ML}{QR}$$

and the included angles are equal:
$\angle M = \angle Q = 70°$

∴ By SAS similarity criterion, $\triangle MNL \sim \triangle QPR$.

From the figure, in △ABC, AB = 2.5, BC = 3, ∠A = 80°, and in △DEF, DF = 5, EF = 6, ∠F = 80°.

$$\frac{AB}{DF} = \frac{2.5}{5} = \frac{1}{2}$$

$$\frac{BC}{EF} = \frac{3}{6} = \frac{1}{2}$$

The sides are proportional, but the angle included between AB and BC is ∠B, not ∠A. The value of ∠B is not known.

∴ SAS cannot be applied, so the triangles are not similar from the given information.

From the figure, in △DEF, ∠D = 70°, ∠E = 80°, and in △PQR, ∠Q = 80°, ∠R = 30°.

Using the angle sum property:
$\angle F = 180° - 70° - 80° = 30°$
$\angle P = 180° - 80° - 30° = 70°$

In △DEF and △PQR:
$\angle D = \angle P = 70°$
$\angle E = \angle Q = 80°$
$\angle F = \angle R = 30°$

∴ By AA similarity criterion, $\triangle DEF \sim \triangle PQR$.

DOB is a straight line, so:

$$\angle DOC + \angle BOC = 180°$$

$$\Rightarrow \angle DOC + 125° = 180°$$

$$\Rightarrow \angle DOC = 55°$$

In △DOC, by the angle sum property:

$$\angle DOC + \angle CDO + \angle DCO = 180°$$

$$\Rightarrow 55° + 70° + \angle DCO = 180°$$

$$\Rightarrow \angle DCO = 55°$$

△ODC ~ △OBA, so the corresponding angles are equal:

$$\angle OAB = \angle OCD$$

$$\Rightarrow \angle OAB = 55°$$

∴ $\angle DOC = \mathbf{55°}$, $\angle DCO = \mathbf{55°}$ and $\angle OAB = \mathbf{55°}$

In △OAB and △OCD:
$\angle AOB = \angle COD$ (vertically opposite angles)
$\angle OAB = \angle OCD$ (alternate angles, AB ∥ DC with transversal AC)
$\angle OBA = \angle ODC$ (alternate angles, AB ∥ DC with transversal BD)

So by AAA similarity:

$$\triangle OAB \sim \triangle OCD$$

The corresponding sides of similar triangles are proportional, so:

$$\frac{OA}{OC} = \frac{OB}{OD}$$

Hence proved. $\blacksquare$

In the figure, $\angle 1 = \angle PQR$ and $\angle 2 = \angle PRQ$.

In △PQR, $\angle 1 = \angle 2$, so the sides opposite these angles are equal:

$$PQ = PR \quad \cdots (1)$$

Given:

$$\frac{QR}{QS} = \frac{QT}{PR}$$

Substituting $PR = PQ$ from equation (1):

$$\frac{QR}{QS} = \frac{QT}{PQ}$$

$$\Rightarrow \frac{QS}{QR} = \frac{PQ}{QT} \quad \cdots (2)$$

In △PQS and △TQR:
$\dfrac{PQ}{TQ} = \dfrac{QS}{QR}$ (from equation (2))
$\angle PQS = \angle TQR$ (common angle $\angle 1$)

So by SAS similarity:

$$\triangle PQS \sim \triangle TQR$$

Hence proved. $\blacksquare$

In △RPQ and △RTS:
$\angle RPQ = \angle RTS$ (given)
$\angle PRQ = \angle TRS$ (common angle $\angle R$)

So by AA similarity:

$$\triangle RPQ \sim \triangle RTS$$

Hence proved. $\blacksquare$

△ABE ≅ △ACD, so the corresponding parts are equal (CPCT):

$$AB = AC \quad \cdots (1)$$

$$AE = AD \quad \cdots (2)$$

Dividing equation (2) by equation (1):

$$\frac{AD}{AB} = \frac{AE}{AC}$$

In △ADE and △ABC:
$\dfrac{AD}{AB} = \dfrac{AE}{AC}$ (shown above)
$\angle DAE = \angle BAC$ (common angle $\angle A$)

So by SAS similarity:

$$\triangle ADE \sim \triangle ABC$$

Hence proved. $\blacksquare$

In △AEP and △CDP:
$\angle AEP = \angle CDP = 90°$ (CE ⊥ AB and AD ⊥ BC)
$\angle APE = \angle CPD$ (vertically opposite angles)

So by AA similarity:

$$\triangle AEP \sim \triangle CDP$$

Hence proved. $\blacksquare$

In △ABD and △CBE:
$\angle ADB = \angle CEB = 90°$ (AD ⊥ BC and CE ⊥ AB)
$\angle ABD = \angle CBE$ (common angle $\angle B$)

So by AA similarity:

$$\triangle ABD \sim \triangle CBE$$

Hence proved. $\blacksquare$

In △AEP and △ADB:
$\angle AEP = \angle ADB = 90°$ (CE ⊥ AB and AD ⊥ BC)
$\angle PAE = \angle BAD$ (common angle $\angle A$)

So by AA similarity:

$$\triangle AEP \sim \triangle ADB$$

Hence proved. $\blacksquare$

In △PDC and △BEC:
$\angle PDC = \angle BEC = 90°$ (AD ⊥ BC and CE ⊥ AB)
$\angle PCD = \angle BCE$ (common angle $\angle C$)

So by AA similarity:

$$\triangle PDC \sim \triangle BEC$$

Hence proved. $\blacksquare$

In △ABE and △CFB:
$\angle BAE = \angle FCB$ (opposite angles of parallelogram ABCD)
$\angle AEB = \angle CBF$ (alternate angles, AE ∥ BC with transversal BE)

So by AA similarity:

$$\triangle ABE \sim \triangle CFB$$

Hence proved. $\blacksquare$

In △ABC and △AMP:
$\angle ABC = \angle AMP = 90°$ (given)
$\angle BAC = \angle MAP$ (common angle $\angle A$)

So by AA similarity:

$$\triangle ABC \sim \triangle AMP$$

Hence proved. $\blacksquare$

From part (i), △ABC ~ △AMP.

The corresponding sides of similar triangles are proportional, so:

$$\frac{CA}{PA} = \frac{BC}{MP}$$

Hence proved. $\blacksquare$

△ABC ~ △FEG, so the corresponding angles are equal:
$\angle A = \angle F$
$\angle B = \angle E$
$\angle ACB = \angle FGE$

CD and GH bisect the equal angles ∠ACB and ∠FGE, so:
$\angle ACD = \dfrac{1}{2}\angle ACB$
$\angle FGH = \dfrac{1}{2}\angle FGE$

$$\Rightarrow \angle ACD = \angle FGH$$

In △ACD and △FGH:
$\angle A = \angle F$ (shown above)
$\angle ACD = \angle FGH$ (shown above)

So by AA similarity:

$$\triangle ACD \sim \triangle FGH$$

The corresponding sides of similar triangles are proportional, so:

$$\frac{CD}{GH} = \frac{AC}{FG}$$

Hence proved. $\blacksquare$

CD and GH bisect the equal angles ∠ACB and ∠FGE, so:
$\angle DCB = \dfrac{1}{2}\angle ACB$
$\angle HGE = \dfrac{1}{2}\angle FGE$

$$\Rightarrow \angle DCB = \angle HGE$$

In △DCB and △HGE:
$\angle B = \angle E$ (△ABC ~ △FEG)
$\angle DCB = \angle HGE$ (shown above)

So by AA similarity:

$$\triangle DCB \sim \triangle HGE$$

Hence proved. $\blacksquare$

From part (i):

$$\angle DCA = \angle HGF$$

In △DCA and △HGF:
$\angle A = \angle F$ (△ABC ~ △FEG)
$\angle DCA = \angle HGF$ (shown above)

So by AA similarity:

$$\triangle DCA \sim \triangle HGF$$

Hence proved. $\blacksquare$

In △ABC, AB = AC, so the angles opposite them are equal:

$$\angle ABC = \angle ACB$$

$$\Rightarrow \angle ABD = \angle ECF \quad \cdots (1)$$

In △ABD and △ECF:
$\angle ABD = \angle ECF$ (from equation (1))
$\angle ADB = \angle EFC = 90°$ (AD ⊥ BC and EF ⊥ AC)

So by AA similarity:

$$\triangle ABD \sim \triangle ECF$$

Hence proved. $\blacksquare$

Given:

$$\frac{AB}{PQ} = \frac{BC}{QR} = \frac{AD}{PM} \quad \cdots (1)$$

AD and PM are medians, so $BC = 2BD$ and $QR = 2QM$. Then:

$$\frac{BC}{QR} = \frac{2BD}{2QM} = \frac{BD}{QM}$$

Putting this in equation (1):

$$\frac{AB}{PQ} = \frac{BD}{QM} = \frac{AD}{PM}$$

So all three sides of △ABD and △PQM are proportional. By SSS similarity:

$$\triangle ABD \sim \triangle PQM$$

$$\Rightarrow \angle ABD = \angle PQM$$

$$\Rightarrow \angle ABC = \angle PQR \quad \cdots (2)$$

In △ABC and △PQR:
$\dfrac{AB}{PQ} = \dfrac{BC}{QR}$ (from equation (1))
$\angle ABC = \angle PQR$ (from equation (2))

So by SAS similarity:

$$\triangle ABC \sim \triangle PQR$$

Hence proved. $\blacksquare$

In △BAC and △ADC:
$\angle BAC = \angle ADC$ (given)
$\angle BCA = \angle ACD$ (common angle $\angle C$)

So by AA similarity:

$$\triangle BAC \sim \triangle ADC$$

The corresponding sides of similar triangles are proportional, so:

$$\frac{CA}{CD} = \frac{CB}{CA}$$

$$\Rightarrow CA \times CA = CB \times CD$$

$$\Rightarrow CA^2 = CB \cdot CD$$

Hence proved. $\blacksquare$

Given:

$$\frac{AB}{PQ} = \frac{AC}{PR} = \frac{AD}{PM} \quad \cdots (1)$$

Construction: Produce AD to E so that AD = DE, and produce PM to N so that PM = MN. Join BE, CE, QN and RN.

In quadrilateral ABEC, the diagonals AE and BC bisect each other at D, so ABEC is a parallelogram. Therefore:

$$BE = AC$$

$$CE = AB$$

Similarly, PQNR is a parallelogram. Therefore:

$$QN = PR$$

$$RN = PQ$$

Also $AE = 2AD$ and $PN = 2PM$, so:

$$\frac{AE}{PN} = \frac{2AD}{2PM} = \frac{AD}{PM}$$

Putting these in equation (1):

$$\frac{AB}{PQ} = \frac{BE}{QN} = \frac{AE}{PN}$$

So by SSS similarity:

$$\triangle ABE \sim \triangle PQN$$

$$\Rightarrow \angle BAE = \angle QPN \quad \cdots (2)$$

In the same way, using $CE = AB$ and $RN = PQ$:

$$\frac{AC}{PR} = \frac{CE}{RN} = \frac{AE}{PN}$$

So by SSS similarity:

$$\triangle ACE \sim \triangle PRN$$

$$\Rightarrow \angle CAE = \angle RPN \quad \cdots (3)$$

Adding equation (2) and equation (3):

$$\angle BAE + \angle CAE = \angle QPN + \angle RPN$$

$$\Rightarrow \angle BAC = \angle QPR \quad \cdots (4)$$

In △ABC and △PQR:
$\dfrac{AB}{PQ} = \dfrac{AC}{PR}$ (from equation (1))
$\angle BAC = \angle QPR$ (from equation (4))

So by SAS similarity:

$$\triangle ABC \sim \triangle PQR$$

Hence proved. $\blacksquare$

Let AB be the pole of length 6 m with shadow BC = 4 m, and PQ be the tower of height $h$ m with shadow QR = 28 m.

In △ABC and △PQR:
$\angle ABC = \angle PQR = 90°$ (pole and tower are vertical)
$\angle ACB = \angle PRQ$ (the sun’s rays make the same angle with the ground at the same time)

So by AA similarity:

$$\triangle ABC \sim \triangle PQR$$

The corresponding sides of similar triangles are proportional, so:

$$\frac{AB}{PQ} = \frac{BC}{QR}$$

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

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

$$\Rightarrow h = 42$$

∴ The height of the tower is 42 m.

△ABC ~ △PQR, so the corresponding sides are proportional and the corresponding angles are equal:

$$\frac{AB}{PQ} = \frac{BC}{QR} \quad \cdots (1)$$

$$\angle B = \angle Q \quad \cdots (2)$$

AD and PM are medians, so $BC = 2BD$ and $QR = 2QM$. Then:

$$\frac{BC}{QR} = \frac{2BD}{2QM} = \frac{BD}{QM}$$

Putting this in equation (1):

$$\frac{AB}{PQ} = \frac{BD}{QM} \quad \cdots (3)$$

In △ABD and △PQM:
$\dfrac{AB}{PQ} = \dfrac{BD}{QM}$ (from equation (3))
$\angle ABD = \angle PQM$ (from equation (2))

So by SAS similarity:

$$\triangle ABD \sim \triangle PQM$$

The corresponding sides of similar triangles are proportional, so:

$$\frac{AB}{PQ} = \frac{AD}{PM}$$

Hence proved. $\blacksquare$