NCERT Solutions: Class 10 Maths Chapter 6 - Triangles

All circles have the same shape, so they are similar. All squares have equal angles and proportional sides, so they are similar. All equilateral triangles have angles $60^\circ,60^\circ,60^\circ$, so they are similar. Similar polygons must have equal corresponding angles and proportional corresponding sides.

Similar figures have the same shape, though their sizes may differ. Circles have one common shape, and squares have one common shape. A circle and a square, or a triangle and a square, do not have the same shape.

Although their corresponding sides can be taken in the ratio $1.5:3 = 1:2$, their corresponding angles are not equal. ABCD is a square with each angle $90^\circ$, while PQRS is not a square. Therefore the quadrilaterals are not similar.

Since $DE \parallel BC$, by the Basic Proportionality Theorem, $\dfrac{AD}{DB}=\dfrac{AE}{EC}$. In (i), $\dfrac{1.5}{3}=\dfrac{1}{EC}$, so $EC=2$ cm. In (ii), $\dfrac{AD}{7.2}=\dfrac{1.8}{5.4}=\dfrac{1}{3}$, so $AD=2.4$ cm.

By the converse of the Basic Proportionality Theorem, $EF \parallel QR$ if $\dfrac{PE}{EQ}=\dfrac{PF}{FR}$. In (i), $\dfrac{3.9}{3}=1.3$ but $\dfrac{3.6}{2.4}=1.5$, so not parallel. In (ii), $\dfrac{4}{4.5}=\dfrac{8}{9}$ and $\dfrac{8}{9}=\dfrac{8}{9}$, so parallel. In (iii), $EQ=1.28-0.18=1.10$ and $FR=2.56-0.36=2.20$, so $\dfrac{PE}{EQ}=\dfrac{0.18}{1.10}=\dfrac{9}{55}$ and $\dfrac{PF}{FR}=\dfrac{0.36}{2.20}=\dfrac{9}{55}$; hence parallel.

In $\triangle ABC$, $LM \parallel CB$, so by BPT, $\dfrac{AM}{AB}=\dfrac{AL}{AC}$. In $\triangle ACD$, $LN \parallel CD$, so $\dfrac{AN}{AD}=\dfrac{AL}{AC}$. Therefore $\dfrac{AM}{AB}=\dfrac{AN}{AD}$.

In $\triangle BAE$, $DF \parallel AE$, so $\dfrac{BD}{DA}=\dfrac{BF}{FE}$. In $\triangle BAC$, $DE \parallel AC$, so $\dfrac{BD}{DA}=\dfrac{BE}{EC}$. Hence $\dfrac{BF}{FE}=\dfrac{BE}{EC}$.

Since $DE \parallel OQ$, BPT gives $\dfrac{PD}{DO}=\dfrac{PE}{EQ}$. Since $DF \parallel OR$, BPT gives $\dfrac{PD}{DO}=\dfrac{PF}{FR}$. Therefore $\dfrac{PE}{EQ}=\dfrac{PF}{FR}$. By the converse of BPT in $\triangle PQR$, $EF \parallel QR$.

In $\triangle OPQ$, $AB \parallel PQ$, so $\dfrac{OA}{AP}=\dfrac{OB}{BQ}$. In $\triangle OPR$, $AC \parallel PR$, so $\dfrac{OA}{AP}=\dfrac{OC}{CR}$. Hence $\dfrac{OB}{BQ}=\dfrac{OC}{CR}$. By the converse of BPT in $\triangle OQR$, $BC \parallel QR$.

Let $D$ be the mid-point of $AB$ in $\triangle ABC$, and let the line through $D$ parallel to $BC$ meet $AC$ at $E$. By Theorem 6.1, $\dfrac{AD}{DB}=\dfrac{AE}{EC}$. Since $D$ is the mid-point, $AD=DB$, so $\dfrac{AE}{EC}=1$, giving $AE=EC$. Thus the third side is bisected.

Let $D$ and $E$ be the mid-points of $AB$ and $AC$ of $\triangle ABC$. Then $AD=DB$ and $AE=EC$, so $\dfrac{AD}{DB}=\dfrac{AE}{EC}=1$. By Theorem 6.2, the converse of BPT, $DE \parallel BC$.

Since $AB \parallel DC$, $\angle ABO = \angle CDO$ and $\angle BAO = \angle DCO$ by alternate interior angles. Also, $\angle AOB = \angle COD$ vertically opposite. Thus $\triangle AOB \sim \triangle COD$, so corresponding sides are proportional: $\dfrac{AO}{CO}=\dfrac{BO}{DO}$. Rearranging gives $\dfrac{AO}{BO}=\dfrac{CO}{DO}$.

Given $\dfrac{AO}{BO}=\dfrac{CO}{DO}$, so $\dfrac{AO}{CO}=\dfrac{BO}{DO}$. Also, $\angle AOB=\angle COD$ vertically opposite. Therefore $\triangle AOB \sim \triangle COD$ by SAS similarity. Hence $\angle ABO=\angle CDO$, which are alternate interior angles, so $AB \parallel DC$. Thus ABCD has one pair of opposite sides parallel and is a trapezium.

In (i) and (vi), corresponding angles are equal, so AAA applies. In (ii), corresponding sides are proportional: $2:4 = 2.5:5 = 3:6 = 1:2$ with the correct correspondence. In (iv), two corresponding sides are proportional and the included angle is equal. In (iii) and (v), the available sides/angles do not satisfy a similarity criterion.

$\angle DOC$ and $\angle BOC$ are a linear pair, so $\angle DOC = 180^\circ - 125^\circ = 55^\circ$. In $\triangle ODC$, $\angle DCO = 180^\circ - 70^\circ - 55^\circ = 55^\circ$. Since $\triangle ODC \sim \triangle OBA$, $\angle DCO$ corresponds to $\angle OAB$, so $\angle OAB=55^\circ$.

In $\triangle AOB$ and $\triangle COD$, $\angle AOB=\angle COD$ vertically opposite. Since $AB \parallel DC$, $\angle OAB=\angle OCD$ and $\angle OBA=\angle ODC$ by alternate interior angles. Thus $\triangle AOB \sim \triangle COD$ by AAA, so corresponding sides give $\dfrac{OA}{OC}=\dfrac{OB}{OD}$.

Since $angle 1 = angle 2$, in $ riangle PQR$ the sides opposite these equal angles are equal, so $PQ = PR$. Substituting $PR = PQ$ in the given relation $dfrac{QR}{QS} = dfrac{QT}{PR}$ gives $dfrac{QR}{QS} = dfrac{QT}{PQ}$, which cross-multiplies to $QRcdot PQ = QScdot QT$, i.e. $dfrac{PQ}{QT} = dfrac{QS}{QR}$. Now compare $ riangle PQS$ and $ riangle TQR$: they share the angle at $Q$ ((angle PQS = angle TQR)), and the sides about this angle are proportional, $dfrac{PQ}{TQ} = dfrac{QS}{QR}$. Therefore, by the SAS similarity criterion, $ riangle PQS sim riangle TQR$.

Since $S$ lies on $PR$ and $T$ lies on $QR$, $\angle PRQ = \angle TRS$. Also, $\angle P = \angle RTS$ is given. Therefore two pairs of corresponding angles are equal, so $\triangle RPQ \sim \triangle RTS$ by AA similarity.

From $\triangle ABE \cong \triangle ACD$, corresponding sides give $AB=AC$ and $AE=AD$. Therefore $\dfrac{AD}{AE}=\dfrac{AC}{AB}$. Also, $\angle DAE=\angle CAB$ is common. Hence $\triangle ADE \sim \triangle ABC$ by SAS similarity.

Because $AD \perp BC$ and $CE \perp AB$, the relevant triangles contain right angles. Also, angles formed by the same intersecting lines are equal where used. Thus (i) $\angle AEP=\angle CDP=90^\circ$ and $\angle APE=\angle CPD$, so AA. (ii) $\angle ADB=\angle CEB=90^\circ$ and $\angle ABD=\angle CBE$, so AA. (iii) $\angle AEP=\angle ADB=90^\circ$ and $\angle EAP=\angle DAB$, so AA. (iv) $\angle PDC=\angle BEC=90^\circ$ and $\angle PCD=\angle EBC$, so AA.

In parallelogram ABCD, $AB \parallel CD$ and $AD \parallel BC$. Since E lies on AD produced and F lies on CD, $AE \parallel BC$ and $AB \parallel CF$. Hence $\angle ABE=\angle CFB$ and $\angle AEB=\angle CBF$. Therefore $\triangle ABE \sim \triangle CFB$ by AA similarity.

$\angle ABC=\angle AMP=90^\circ$, and $\angle BAC=\angle MAP$ is common. Hence $\triangle ABC \sim \triangle AMP$ by AA similarity. Corresponding sides of similar triangles are proportional, so $\dfrac{CA}{PA}=\dfrac{BC}{MP}$.

Since $\triangle ABC \sim \triangle FEG$, corresponding angles are equal: $\angle ACB=\angle EGF$, $\angle ABC=\angle FEG$, and $\angle BAC=\angle EFG$. Their bisectors give equal half-angles, so $\angle DCB=\angle HGE$ and $\angle DCA=\angle HGF$. Thus $\triangle DCB \sim \triangle HGE$ and $\triangle DCA \sim \triangle HGF$ by AA. From these similarities, corresponding sides are proportional, giving $\dfrac{CD}{GH}=\dfrac{AC}{FG}$.

Since $AD \perp BC$ and $EF \perp AC$, $\angle ADB=\angle EFC=90^\circ$. In isosceles $\triangle ABC$, $AB=AC$, so base angles are equal: $\angle ABC=\angle BCA$. Since E lies on CB produced and F lies on AC, $\angle ABD=\angle ECF$. Hence $\triangle ABD \sim \triangle ECF$ by AA similarity.

Given $\dfrac{AB}{PQ}=\dfrac{BC}{QR}=\dfrac{AD}{PM}$. Since AD and PM are medians, $BD=\dfrac{BC}{2}$ and $QM=\dfrac{QR}{2}$, so $\dfrac{BD}{QM}=\dfrac{BC}{QR}$. Thus in $\triangle ABD$ and $\triangle PQM$, $\dfrac{AB}{PQ}=\dfrac{BD}{QM}=\dfrac{AD}{PM}$, so $\triangle ABD \sim \triangle PQM$ by SSS. Hence $\angle ABD=\angle PQM$. Because D lies on BC and M lies on QR, $\angle ABC=\angle PQR$. With $\dfrac{AB}{PQ}=\dfrac{BC}{QR}$ and the included angle equal, $\triangle ABC \sim \triangle PQR$ by SAS.

Since D lies on BC, $\angle ACD=\angle BCA$. Also $\angle ADC=\angle BAC$ is given. Hence $\triangle ADC \sim \triangle BAC$ by AA similarity. Therefore $\dfrac{CD}{CA}=\dfrac{CA}{CB}$, so $CA^2=CB\cdot CD$.

Given $\dfrac{AB}{PQ}=\dfrac{AC}{PR}=\dfrac{AD}{PM}$. Extend AD to E so that $AD=DE$, and extend PM to N so that $PM=MN$. Since AD and PM are medians, quadrilaterals ABEC and PQNR have diagonals bisecting each other, so they are parallelograms. Thus $BE=AC$ and $QN=PR$. Now $\dfrac{AB}{PQ}=\dfrac{BE}{QN}=\dfrac{AE}{PN}$, so $\triangle ABE \sim \triangle PQN$ by SSS. Hence $\angle ABE=\angle PQN$. Since $BE \parallel AC$ and $QN \parallel PR$, this gives $\angle BAC=\angle QPR$. With $\dfrac{AB}{PQ}=\dfrac{AC}{PR}$ and the included angle equal, $\triangle ABC \sim \triangle PQR$ by SAS.

At the same time, the sun's elevation is the same, so the pole-shadow and tower-shadow triangles are similar. Therefore $\dfrac{\text{height of tower}}{28}=\dfrac{6}{4}$. Hence height of tower $=28\times\dfrac{6}{4}=42$ m.

Since $\triangle ABC \sim \triangle PQR$, $\dfrac{AB}{PQ}=\dfrac{BC}{QR}$ and $\angle ABC=\angle PQR$. Because AD and PM are medians, $BD=\dfrac{BC}{2}$ and $QM=\dfrac{QR}{2}$, so $\dfrac{BD}{QM}=\dfrac{BC}{QR}=\dfrac{AB}{PQ}$. Thus $\triangle ABD \sim \triangle PQM$ by SAS similarity. Hence corresponding sides give $\dfrac{AB}{PQ}=\dfrac{AD}{PM}$.