Class 10 Mathematics Chapter 6 Question Answer | Triangles | English Medium

Class 10 Mathematics Chapter 6 Question Answer | Triangles | English Medium | ASSEB

Welcome to HSLC GURU! This page provides complete ASSEB Class 10 Mathematics Chapter 6 — Triangles question answers in English Medium. The chapter is one of the most important in Geometry. It introduces the idea of similarity, the powerful Basic Proportionality Theorem (BPT), the three similarity criteria (AAA/AA, SSS, SAS), the area-ratio theorem for similar triangles, and the celebrated Pythagoras Theorem with a proof using similar triangles. All exercises (6.1 to 6.6) are solved with clear, well-labelled figures so that you can prepare for the SEBA / ASSEB Board Examination with confidence.

Chapter 6 — Triangles : Summary

Two figures with the same shape (not necessarily the same size) are called similar figures. Two figures with the same shape and same size are called congruent figures. All congruent figures are similar but the converse is not always true.

Two polygons of the same number of sides are similar if (i) their corresponding angles are equal and (ii) their corresponding sides are in the same ratio (i.e., proportion).

Two triangles are similar if their corresponding angles are equal and their corresponding sides are in the same ratio. Symbolically, $\triangle ABC \sim \triangle PQR$ means

$$\angle A=\angle P,\ \angle B=\angle Q,\ \angle C=\angle R \quad\text{and}\quad \frac{AB}{PQ}=\frac{BC}{QR}=\frac{CA}{RP}.$$

Key Theorems with Figures

Theorem 6.1 — Basic Proportionality Theorem (Thales’ Theorem / BPT)

Statement. If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then the other two sides are divided in the same ratio.

Given. $\triangle ABC$ in which $DE\parallel BC$, $D$ on $AB$, $E$ on $AC$.
To prove. $\dfrac{AD}{DB}=\dfrac{AE}{EC}$.

Construction. Join $BE$ and $CD$. Draw $DM\perp AC$ and $EN\perp AB$.

Proof. Area of $\triangle ADE=\tfrac12\cdot AD\cdot EN$, and area of $\triangle BDE=\tfrac12\cdot DB\cdot EN$. Hence

$$\frac{\operatorname{ar}(\triangle ADE)}{\operatorname{ar}(\triangle BDE)}=\frac{AD}{DB}.$$

Similarly, $\dfrac{\operatorname{ar}(\triangle ADE)}{\operatorname{ar}(\triangle CDE)}=\dfrac{AE}{EC}.$

Now $\triangle BDE$ and $\triangle CDE$ are on the same base $DE$ and between the same parallels $DE$ and $BC$, so $\operatorname{ar}(\triangle BDE)=\operatorname{ar}(\triangle CDE).$ Therefore $\dfrac{AD}{DB}=\dfrac{AE}{EC}.$ Hence proved.

Theorem 6.2 (Converse of BPT). If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

Theorem 6.3 — AAA (and AA) Similarity Criterion

Statement. If in two triangles, corresponding angles are equal, then their corresponding sides are in the same ratio (proportion) and hence the two triangles are similar.

Corollary (AA). If two angles of one triangle are respectively equal to two angles of another triangle, the two triangles are similar (the third pair of angles must be equal by angle-sum).

Result. $\dfrac{AB}{PQ}=\dfrac{BC}{QR}=\dfrac{CA}{RP}.$

Theorem 6.4 — SSS Similarity Criterion

Statement. If in two triangles, sides of one triangle are proportional to (i.e., in the same ratio of) the sides of the other triangle, then their corresponding angles are equal and hence the two triangles are similar.

If $\dfrac{AB}{DE}=\dfrac{BC}{EF}=\dfrac{CA}{FD}$, then $\triangle ABC\sim\triangle DEF$.

Theorem 6.5 — SAS Similarity Criterion

Statement. If one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, then the two triangles are similar.

Theorem 6.6 — Areas of Similar Triangles

Statement. The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

If $\triangle ABC\sim\triangle PQR$, then

$$\frac{\operatorname{ar}(\triangle ABC)}{\operatorname{ar}(\triangle PQR)}=\left(\frac{AB}{PQ}\right)^2=\left(\frac{BC}{QR}\right)^2=\left(\frac{CA}{RP}\right)^2.$$

Theorem 6.7 — Pythagoras Theorem

Statement. In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Given. A right triangle $ABC$, right-angled at $B$.
To prove. $AC^{2}=AB^{2}+BC^{2}.$

Construction. Draw $BD\perp AC$.

Proof. In $\triangle ADB$ and $\triangle ABC$:

$\angle A=\angle A$ (common) and $\angle ADB=\angle ABC=90^{\circ}$. So by AA, $\triangle ADB\sim\triangle ABC$. Hence $\dfrac{AD}{AB}=\dfrac{AB}{AC}$ giving $AB^{2}=AD\cdot AC.$ …(i)

Similarly $\triangle BDC\sim\triangle ABC$ giving $BC^{2}=DC\cdot AC.$ …(ii)

Adding (i) and (ii): $AB^{2}+BC^{2}=AC(AD+DC)=AC\cdot AC=AC^{2}.$ Hence proved.

Theorem 6.8 (Converse of Pythagoras). In a triangle, if the square of one side is equal to the sum of squares of the other two sides, then the angle opposite the first side is a right angle.

Exercise 6.1

Q1. Fill in the blanks using the correct word given in brackets:

(i) All circles are __________ (congruent / similar).
Answer: similar.

(ii) All squares are __________ (similar / congruent).
Answer: similar.

(iii) All __________ triangles are similar (isosceles / equilateral).
Answer: equilateral.

(iv) Two polygons of the same number of sides are similar, if (a) their corresponding angles are __________ and (b) their corresponding sides are __________ (equal / proportional).
Answer: (a) equal, (b) proportional.

Q2. Give two different examples of pairs of (i) similar figures, (ii) non-similar figures.

Answer: (i) Similar — any two equilateral triangles; any two squares; any two circles. (ii) Non-similar — a square and a rhombus (angles differ); a triangle and a quadrilateral (different number of sides).

Q3. State whether the following quadrilaterals are similar or not.

Answer: A square of side 1.5 cm and a rhombus of side 3 cm are not similar because, although their corresponding sides are proportional, their corresponding angles are not equal (a square has all angles $90^{\circ}$, a rhombus does not).

Exercise 6.2

Q1. In the figure, (i) and (ii), $DE\parallel BC$. Find $EC$ in (i) and $AD$ in (ii).

Answer: Since $DE\parallel BC$, by BPT, $\dfrac{AD}{DB}=\dfrac{AE}{EC}.$

(i) $\dfrac{1.5}{3}=\dfrac{1}{EC}\Rightarrow EC=\dfrac{3}{1.5}=2\ \text{cm}.$

(ii) Given $AE=1.8$ cm, $DB=7.2$ cm, $EC=5.4$ cm. $\dfrac{AD}{7.2}=\dfrac{1.8}{5.4}\Rightarrow AD=\dfrac{1.8\times 7.2}{5.4}=2.4\ \text{cm}.$

Q2. $E$ and $F$ are points on the sides $PQ$ and $PR$ respectively of $\triangle PQR$. For each of the following cases, state whether $EF\parallel QR$:

(i) $PE=3.9$, $EQ=3$, $PF=3.6$, $FR=2.4$ cm.

Answer: $\dfrac{PE}{EQ}=\dfrac{3.9}{3}=1.3,\ \dfrac{PF}{FR}=\dfrac{3.6}{2.4}=1.5.$ Ratios are unequal, so $EF\not\parallel QR.$

(ii) $PE=4$, $QE=4.5$, $PF=8$, $RF=9$ cm.

Answer: $\dfrac{PE}{EQ}=\dfrac{4}{4.5}=\dfrac{8}{9},\ \dfrac{PF}{FR}=\dfrac{8}{9}.$ Equal, so by converse of BPT, $EF\parallel QR.$

(iii) $PQ=1.28$, $PR=2.56$, $PE=0.18$, $PF=0.36$ cm.

Answer: $EQ=1.28-0.18=1.10$, $FR=2.56-0.36=2.20$. $\dfrac{PE}{EQ}=\dfrac{0.18}{1.10}=\dfrac{9}{55},\ \dfrac{PF}{FR}=\dfrac{0.36}{2.20}=\dfrac{9}{55}.$ Equal, hence $EF\parallel QR.$

Q3. In the figure, if $LM\parallel CB$ and $LN\parallel CD$, prove that $\dfrac{AM}{AB}=\dfrac{AN}{AD}.$

Answer: In $\triangle ABC$, $LM\parallel CB$. By BPT, $\dfrac{AM}{MB}=\dfrac{AL}{LC}.$ …(1)

In $\triangle ACD$, $LN\parallel CD$. By BPT, $\dfrac{AN}{ND}=\dfrac{AL}{LC}.$ …(2)

From (1) and (2): $\dfrac{AM}{MB}=\dfrac{AN}{ND}.$ Adding 1 to both sides and inverting suitably: $\dfrac{AM}{AM+MB}=\dfrac{AN}{AN+ND}\Rightarrow \dfrac{AM}{AB}=\dfrac{AN}{AD}.$ Hence proved.

Q4. In the figure, $DE\parallel AC$ and $DF\parallel AE$. Prove that $\dfrac{BF}{FE}=\dfrac{BE}{EC}.$

Answer: In $\triangle ABC$, $DE\parallel AC$, so by BPT, $\dfrac{BD}{DA}=\dfrac{BE}{EC}.$ …(1)

In $\triangle ABE$, $DF\parallel AE$, so by BPT, $\dfrac{BD}{DA}=\dfrac{BF}{FE}.$ …(2)

From (1) and (2): $\dfrac{BF}{FE}=\dfrac{BE}{EC}.$ Hence proved.

Q5. In the figure, $DE\parallel OQ$ and $DF\parallel OR$. Show that $EF\parallel QR.$

Answer: In $\triangle PQO$, $DE\parallel OQ$ $\Rightarrow \dfrac{PD}{DO}=\dfrac{PE}{EQ}.$ …(1)

In $\triangle POR$, $DF\parallel OR$ $\Rightarrow \dfrac{PD}{DO}=\dfrac{PF}{FR}.$ …(2)

From (1) and (2): $\dfrac{PE}{EQ}=\dfrac{PF}{FR}.$ By converse of BPT, $EF\parallel QR.$ Hence proved.

Q6. In the figure, $A$, $B$, $C$ are points on $OP$, $OQ$, $OR$ respectively such that $AB\parallel PQ$ and $AC\parallel PR$. Show that $BC\parallel QR.$

Answer: In $\triangle OPQ$, $AB\parallel PQ\Rightarrow \dfrac{OA}{AP}=\dfrac{OB}{BQ}.$ …(1)

In $\triangle OPR$, $AC\parallel PR\Rightarrow \dfrac{OA}{AP}=\dfrac{OC}{CR}.$ …(2)

From (1) and (2): $\dfrac{OB}{BQ}=\dfrac{OC}{CR}.$ Therefore in $\triangle OQR$, by converse of BPT, $BC\parallel QR.$ Hence proved.

Q7. Using BPT, prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side.

Answer: Let $\triangle ABC$ with $D$ the mid-point of $AB$, so $AD=DB$. Through $D$ draw a line parallel to $BC$ meeting $AC$ at $E$.

By BPT, $\dfrac{AD}{DB}=\dfrac{AE}{EC}.$ Since $AD=DB$, $\dfrac{AE}{EC}=1\Rightarrow AE=EC.$ Hence $E$ is the mid-point of $AC.$ Hence proved.

Q8. Using converse of BPT, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side.

Answer: In $\triangle ABC$ let $D$ and $E$ be mid-points of $AB$ and $AC$. Then $\dfrac{AD}{DB}=1$ and $\dfrac{AE}{EC}=1.$ So $\dfrac{AD}{DB}=\dfrac{AE}{EC}.$ By converse of BPT, $DE\parallel BC.$ Hence proved.

Q9. $ABCD$ is a trapezium in which $AB\parallel DC$ and its diagonals intersect each other at the point $O$. Show that $\dfrac{AO}{BO}=\dfrac{CO}{DO}.$

Answer: Through $O$ draw $OE\parallel AB\parallel DC$, meeting $AD$ at $E$.

In $\triangle ADB$ (with $OE\parallel AB$): $\dfrac{DE}{EA}=\dfrac{DO}{OB}.$ …(1)

In $\triangle ADC$ (with $OE\parallel DC$): $\dfrac{DE}{EA}=\dfrac{CO}{OA}.$ …(2)

From (1) and (2): $\dfrac{DO}{OB}=\dfrac{CO}{OA}\Rightarrow \dfrac{AO}{BO}=\dfrac{CO}{DO}.$ Hence proved.

Q10. The diagonals of a quadrilateral $ABCD$ intersect each other at the point $O$ such that $\dfrac{AO}{BO}=\dfrac{CO}{DO}.$ Show that $ABCD$ is a trapezium.

Answer: Through $O$ draw $OE\parallel AB$ meeting $AD$ at $E$. In $\triangle DAB$, $OE\parallel AB\Rightarrow \dfrac{DE}{EA}=\dfrac{DO}{OB}.$ Given $\dfrac{AO}{OB}=\dfrac{CO}{OD}\Rightarrow \dfrac{OB}{AO}=\dfrac{OD}{CO}\Rightarrow \dfrac{DO}{OB}=\dfrac{CO}{AO}.$

Therefore $\dfrac{DE}{EA}=\dfrac{CO}{OA}.$ In $\triangle DAC$, by converse of BPT, $OE\parallel DC.$ Now $OE\parallel AB$ and $OE\parallel DC$, hence $AB\parallel DC$, i.e. $ABCD$ is a trapezium. Hence proved.

Exercise 6.3

Q1. State which pairs of triangles in the figure are similar. Write the similarity criterion used and write the pairs of similar triangles in the symbolic form:

Answer (summary):

(i) $\triangle ABC\sim\triangle PQR$ with $\angle A=\angle P=60^{\circ}, \angle B=\angle Q=80^{\circ}, \angle C=\angle R=40^{\circ}$ — by AAA.

(ii) $\dfrac{AB}{QR}=\dfrac{BC}{RP}=\dfrac{CA}{PQ}=\dfrac{1}{2}$ so $\triangle ABC\sim\triangle QRP$ — by SSS.

(iii) Sides not in proportion. Not similar.

(iv) $\angle M=\angle Q=70^{\circ}$ and $\dfrac{MN}{QP}=\dfrac{ML}{QR}=\tfrac12$ so $\triangle MNL\sim\triangle QPR$ — by SAS.

(v) Only one pair of equal angles given and including sides not given proportional. Not similar.

(vi) In $\triangle DEF$, $\angle D=70^{\circ}, \angle E=80^{\circ}\Rightarrow \angle F=30^{\circ}$. In $\triangle PQR$, $\angle Q=80^{\circ}, \angle R=30^{\circ}\Rightarrow \angle P=70^{\circ}$. So $\triangle DEF\sim\triangle PQR$ — by AAA.

Q2. In the figure, $\triangle ODC\sim\triangle OBA$, $\angle BOC=125^{\circ}$ and $\angle CDO=70^{\circ}$. Find $\angle DOC,\ \angle DCO,\ \angle OAB.$

Answer: $\angle DOC=180^{\circ}-125^{\circ}=55^{\circ}$ (linear pair). In $\triangle DOC$: $\angle DCO=180^{\circ}-70^{\circ}-55^{\circ}=55^{\circ}.$ Since $\triangle ODC\sim\triangle OBA$, $\angle OAB=\angle OCD=55^{\circ}.$

Q3. Diagonals $AC$ and $BD$ of a trapezium $ABCD$ with $AB\parallel DC$ intersect each other at $O$. Using AA similarity, show that $\dfrac{OA}{OC}=\dfrac{OB}{OD}.$

Answer: In $\triangle AOB$ and $\triangle COD$: $\angle AOB=\angle COD$ (vertically opposite) and $\angle OAB=\angle OCD$ (alternate angles, $AB\parallel DC$). By AA, $\triangle AOB\sim\triangle COD.$ Hence $\dfrac{OA}{OC}=\dfrac{OB}{OD}.$

Q4. In the figure, $\dfrac{QR}{QS}=\dfrac{QT}{PR}$ and $\angle 1=\angle 2$. Show that $\triangle PQS\sim\triangle TQR.$

Answer: Since $\angle 1=\angle 2$, we have $PQ=PR$ in $\triangle PQR$ (sides opposite equal angles are equal). Given $\dfrac{QR}{QS}=\dfrac{QT}{PR}=\dfrac{QT}{PQ}\Rightarrow \dfrac{QT}{QR}=\dfrac{PQ}{QS}\Rightarrow \dfrac{QS}{QT}=\dfrac{QR}{PQ}.$ With $\angle Q=\angle Q$ (common), by SAS, $\triangle PQS\sim\triangle TQR.$

Q5. $S$ and $T$ are points on sides $PR$ and $QR$ of $\triangle PQR$ such that $\angle P=\angle RTS$. Show that $\triangle RPQ\sim\triangle RTS.$

Answer: In $\triangle RPQ$ and $\triangle RTS$: $\angle R=\angle R$ (common) and $\angle P=\angle RTS$ (given). By AA, $\triangle RPQ\sim\triangle RTS.$

Q6. In the figure, if $\triangle ABE\cong\triangle ACD$, show that $\triangle ADE\sim\triangle ABC.$

Answer: $\triangle ABE\cong\triangle ACD$ gives $AB=AC$ and $AE=AD$. Therefore $\dfrac{AD}{AB}=\dfrac{AE}{AC}.$ With $\angle A=\angle A$ (common), by SAS, $\triangle ADE\sim\triangle ABC.$

Q7. In the figure, altitudes $AD$ and $CE$ of $\triangle ABC$ intersect each other at the point $P$. Show that: (i) $\triangle AEP\sim\triangle CDP$ (ii) $\triangle ABD\sim\triangle CBE$ (iii) $\triangle AEP\sim\triangle ADB$ (iv) $\triangle PDC\sim\triangle BEC.$

Answer: (i) $\angle AEP=\angle CDP=90^{\circ}$ and $\angle APE=\angle CPD$ (vertically opposite). By AA, $\triangle AEP\sim\triangle CDP.$
(ii) $\angle ADB=\angle CEB=90^{\circ}$ and $\angle B=\angle B.$ By AA, $\triangle ABD\sim\triangle CBE.$
(iii) $\angle AEP=\angle ADB=90^{\circ}$ and $\angle A=\angle A.$ By AA, $\triangle AEP\sim\triangle ADB.$
(iv) $\angle PDC=\angle BEC=90^{\circ}$ and $\angle C=\angle C.$ By AA, $\triangle PDC\sim\triangle BEC.$

Q8. $E$ is a point on the side $AD$ produced of a parallelogram $ABCD$ and $BE$ intersects $CD$ at $F$. Show that $\triangle ABE\sim\triangle CFB.$

Answer: In $\triangle ABE$ and $\triangle CFB$: $\angle A=\angle C$ (opposite angles of parallelogram). $\angle AEB=\angle CBF$ (alternate angles since $AE\parallel BC$). By AA, $\triangle ABE\sim\triangle CFB.$

Q9. In the figure, $ABC$ and $AMP$ are two right triangles, right-angled at $B$ and $M$ respectively. Prove that: (i) $\triangle ABC\sim\triangle AMP$ (ii) $\dfrac{CA}{PA}=\dfrac{BC}{MP}.$

Answer: $\angle ABC=\angle AMP=90^{\circ}$, $\angle A=\angle A.$ By AA, $\triangle ABC\sim\triangle AMP.$ Hence corresponding sides: $\dfrac{CA}{PA}=\dfrac{BC}{MP}.$

Q10. $CD$ and $GH$ are bisectors of $\angle ACB$ and $\angle EGF$ respectively. If $\triangle ABC\sim\triangle FEG$, show that: (i) $\dfrac{CD}{GH}=\dfrac{AC}{FG}$ (ii) $\triangle DCB\sim\triangle HGE$ (iii) $\triangle DCA\sim\triangle HGF.$

Answer: Since $\triangle ABC\sim\triangle FEG$, $\angle A=\angle F, \angle ACB=\angle FGE.$ Then $\tfrac12\angle ACB=\tfrac12\angle FGE\Rightarrow \angle ACD=\angle FGH.$ By AA, $\triangle ACD\sim\triangle FGH$, so $\dfrac{CD}{GH}=\dfrac{AC}{FG}.$ Similarly $\triangle DCB\sim\triangle HGE$ and $\triangle DCA\sim\triangle HGF.$

Q11. In the figure, $E$ is a point on side $CB$ produced of an isosceles triangle $ABC$ with $AB=AC$. If $AD\perp BC$ and $EF\perp AC$, prove that $\triangle ABD\sim\triangle ECF.$

Answer: $AB=AC\Rightarrow \angle ABC=\angle ACB.$ Now $\angle ABD=\angle ABC=\angle ACB=\angle ECF.$ Also $\angle ADB=\angle EFC=90^{\circ}.$ By AA, $\triangle ABD\sim\triangle ECF.$

Q12. Sides $AB,\ BC$ and median $AD$ of $\triangle ABC$ are respectively proportional to sides $PQ,\ QR$ and median $PM$ of $\triangle PQR$. Show that $\triangle ABC\sim\triangle PQR.$

Answer: Given $\dfrac{AB}{PQ}=\dfrac{BC}{QR}=\dfrac{AD}{PM}.$ Since $D$ and $M$ are mid-points of $BC$ and $QR$, $\dfrac{BD}{QM}=\dfrac{BC/2}{QR/2}=\dfrac{BC}{QR}.$ So $\dfrac{AB}{PQ}=\dfrac{BD}{QM}=\dfrac{AD}{PM}.$ By SSS, $\triangle ABD\sim\triangle PQM.$ Hence $\angle B=\angle Q.$ With $\dfrac{AB}{PQ}=\dfrac{BC}{QR}$ and $\angle B=\angle Q$, by SAS, $\triangle ABC\sim\triangle PQR.$

Q13. $D$ is a point on the side $BC$ of a triangle $ABC$ such that $\angle ADC=\angle BAC$. Show that $CA^{2}=CB\cdot CD.$

Answer: In $\triangle ADC$ and $\triangle BAC$: $\angle ADC=\angle BAC$ (given), $\angle C=\angle C$ (common). By AA, $\triangle ADC\sim\triangle BAC.$ So $\dfrac{CA}{CB}=\dfrac{CD}{CA}\Rightarrow CA^{2}=CB\cdot CD.$

Q14. Sides $AB$ and $AC$ and median $AD$ of a triangle $ABC$ are respectively proportional to sides $PQ$ and $PR$ and median $PM$ of another triangle $PQR$. Show that $\triangle ABC\sim\triangle PQR.$

Answer: Produce $AD$ to $E$ such that $AD=DE$ and produce $PM$ to $N$ such that $PM=MN$. Then $ABEC$ and $PQNR$ are parallelograms. So $BE=AC$ and $QN=PR$. Given $\dfrac{AB}{PQ}=\dfrac{AC}{PR}=\dfrac{AD}{PM}\Rightarrow \dfrac{AB}{PQ}=\dfrac{BE}{QN}=\dfrac{AE}{PN}.$ By SSS, $\triangle ABE\sim\triangle PQN.$ Hence $\angle BAE=\angle QPN$ and similarly $\angle CAE=\angle RPN$, giving $\angle BAC=\angle QPR.$ Then by SAS, $\triangle ABC\sim\triangle PQR.$

Q15. A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.

Answer: Let height of tower $=h$. Pole and tower with sun rays form similar triangles. So $\dfrac{6}{4}=\dfrac{h}{28}\Rightarrow h=\dfrac{6\times 28}{4}=42\ \text{m}.$

Q16. If $AD$ and $PM$ are medians of triangles $ABC$ and $PQR$ respectively, where $\triangle ABC\sim\triangle PQR$, prove that $\dfrac{AB}{PQ}=\dfrac{AD}{PM}.$

Answer: Since $\triangle ABC\sim\triangle PQR$, $\dfrac{AB}{PQ}=\dfrac{BC}{QR}=\dfrac{BC/2}{QR/2}=\dfrac{BD}{QM}.$ Also $\angle B=\angle Q.$ By SAS, $\triangle ABD\sim\triangle PQM\Rightarrow \dfrac{AD}{PM}=\dfrac{AB}{PQ}.$

Exercise 6.4 (Areas of Similar Triangles)

Q1. Let $\triangle ABC\sim\triangle DEF$ and their areas be respectively $64\ \text{cm}^2$ and $121\ \text{cm}^2$. If $EF=15.4\ \text{cm}$, find $BC.$

Answer: $\dfrac{\text{ar}(\triangle ABC)}{\text{ar}(\triangle DEF)}=\left(\dfrac{BC}{EF}\right)^2.$ So $\dfrac{64}{121}=\dfrac{BC^2}{(15.4)^2}\Rightarrow \dfrac{BC}{15.4}=\dfrac{8}{11}\Rightarrow BC=\dfrac{8\times 15.4}{11}=11.2\ \text{cm}.$

Q2. Diagonals of a trapezium $ABCD$ with $AB\parallel DC$ intersect at $O$. If $AB=2\,CD$, find the ratio of the areas of $\triangle AOB$ and $\triangle COD.$

Answer: $\triangle AOB\sim\triangle COD$ (AA, alternate angles). $\dfrac{\text{ar}(\triangle AOB)}{\text{ar}(\triangle COD)}=\left(\dfrac{AB}{CD}\right)^2=\left(\dfrac{2}{1}\right)^2=4:1.$

Q3. In the figure, $\triangle ABC$ and $\triangle DBC$ are on the same base $BC$. If $AD$ intersects $BC$ at $O$, show that $\dfrac{\text{ar}(\triangle ABC)}{\text{ar}(\triangle DBC)}=\dfrac{AO}{DO}.$

Answer: Drop perpendiculars $AM$ and $DN$ from $A$ and $D$ on $BC.$ Triangles $AMO$ and $DNO$ are similar (AA: right angle and vertically opposite angles), so $\dfrac{AM}{DN}=\dfrac{AO}{DO}.$ Then $\dfrac{\text{ar}(\triangle ABC)}{\text{ar}(\triangle DBC)}=\dfrac{\tfrac12\cdot BC\cdot AM}{\tfrac12\cdot BC\cdot DN}=\dfrac{AM}{DN}=\dfrac{AO}{DO}.$

Q4. If the areas of two similar triangles are equal, prove that they are congruent.

Answer: Let $\triangle ABC\sim\triangle DEF$ with $\text{ar}(\triangle ABC)=\text{ar}(\triangle DEF)$. Then $\left(\dfrac{AB}{DE}\right)^2=\dfrac{\text{ar}(\triangle ABC)}{\text{ar}(\triangle DEF)}=1\Rightarrow AB=DE.$ Similarly $BC=EF$ and $CA=FD$. By SSS congruence, $\triangle ABC\cong\triangle DEF.$

Q5. $D,\ E,\ F$ are respectively the mid-points of sides $AB,\ BC,\ CA$ of $\triangle ABC$. Find the ratio of the areas of $\triangle DEF$ and $\triangle ABC.$

Answer: By mid-point theorem, $DE=\tfrac12 AC$, $EF=\tfrac12 AB$, $FD=\tfrac12 BC$, so $\triangle DEF\sim\triangle CAB$ with ratio $\tfrac12.$ Hence $\dfrac{\text{ar}(\triangle DEF)}{\text{ar}(\triangle ABC)}=\left(\tfrac12\right)^2=\dfrac{1}{4}.$ Ratio is $1:4.$

Q6. Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.

Answer: Let $\triangle ABC\sim\triangle PQR$ with medians $AD$, $PM$. From Q16 of Ex 6.3, $\dfrac{AD}{PM}=\dfrac{AB}{PQ}.$ Squaring: $\left(\dfrac{AD}{PM}\right)^2=\left(\dfrac{AB}{PQ}\right)^2=\dfrac{\text{ar}(\triangle ABC)}{\text{ar}(\triangle PQR)}.$

Q7. Prove that the area of an equilateral triangle described on one side of a square is equal to half the area of the equilateral triangle described on its diagonal.

Answer: Side of square $=a$, diagonal $=a\sqrt2.$ Area of equilateral triangle on side $a$: $\tfrac{\sqrt3}{4}a^{2}.$ Area on diagonal: $\tfrac{\sqrt3}{4}(a\sqrt2)^{2}=\tfrac{\sqrt3}{4}\cdot 2a^{2}.$ Ratio = $\dfrac{a^2}{2a^2}=\dfrac{1}{2}.$ Hence proved.

Q8. ABC and BDE are two equilateral triangles such that $D$ is the mid-point of $BC.$ The ratio of areas of triangles ABC and BDE is: (A) 2:1 (B) 1:2 (C) 4:1 (D) 1:4.

Answer: Side of $\triangle BDE=\tfrac12$ side of $\triangle ABC$, equilateral triangles are similar, ratio of areas $=4:1.$ Option (C).

Q9. Sides of two similar triangles are in the ratio $4:9.$ Areas of these triangles are in the ratio: (A) 2:3 (B) 4:9 (C) 81:16 (D) 16:81.

Answer: $(4:9)^2=16:81.$ Option (D).

Exercise 6.5 (Pythagoras Theorem)

Q1. Sides of triangles are given below. Determine which of them are right triangles. In case of a right triangle, write the length of its hypotenuse.

(i) 7 cm, 24 cm, 25 cm — $7^2+24^2=49+576=625=25^2.$ Right triangle, hypotenuse 25 cm.

(ii) 3 cm, 8 cm, 6 cm — $3^2+6^2=9+36=45\ne 64=8^2.$ Not a right triangle.

(iii) 50 cm, 80 cm, 100 cm — $50^2+80^2=2500+6400=8900\ne 10000.$ Not a right triangle.

(iv) 13 cm, 12 cm, 5 cm — $5^2+12^2=25+144=169=13^2.$ Right triangle, hypotenuse 13 cm.

Q2. $PQR$ is a triangle right-angled at $P$ and $M$ is a point on $QR$ such that $PM\perp QR.$ Show that $PM^{2}=QM\cdot MR.$

Answer: $\triangle PMQ\sim\triangle RMP$ (AA: each $90^{\circ}$ at $M$, $\angle MPQ=\angle MRP$ both equal $90^{\circ}-\angle Q$). So $\dfrac{PM}{RM}=\dfrac{QM}{PM}\Rightarrow PM^{2}=QM\cdot MR.$

Q3. In the figure, $ABD$ is a triangle right-angled at $A$ and $AC\perp BD.$ Show that: (i) $AB^{2}=BC\cdot BD$ (ii) $AC^{2}=BC\cdot DC$ (iii) $AD^{2}=BD\cdot CD.$

Answer: Each follows from a pair of similar right triangles. $\triangle ABC\sim\triangle DBA\Rightarrow \dfrac{AB}{DB}=\dfrac{BC}{AB}\Rightarrow AB^{2}=BC\cdot BD.$ $\triangle ABC\sim\triangle DAC\Rightarrow AC^{2}=BC\cdot DC.$ $\triangle DAC\sim\triangle DBA\Rightarrow AD^{2}=BD\cdot CD.$

Q4. $ABC$ is an isosceles triangle right-angled at $C$. Prove that $AB^{2}=2\,AC^{2}.$

Answer: $AC=BC.$ By Pythagoras: $AB^{2}=AC^{2}+BC^{2}=AC^{2}+AC^{2}=2\,AC^{2}.$

Q5. $ABC$ is an isosceles triangle with $AC=BC.$ If $AB^{2}=2\,AC^{2}$, prove that $ABC$ is right-angled at $C.$

Answer: $AB^{2}=2AC^{2}=AC^{2}+AC^{2}=AC^{2}+BC^{2}.$ By converse of Pythagoras, $\angle C=90^{\circ}.$

Q6. $ABC$ is an equilateral triangle of side $2a.$ Find each of its altitudes.

Answer: Drop altitude $AD$ to $BC$. Then $BD=a$ and $AB=2a.$ By Pythagoras, $AD^{2}=AB^{2}-BD^{2}=4a^{2}-a^{2}=3a^{2}\Rightarrow AD=a\sqrt3.$

Q7. Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.

Answer: Diagonals of a rhombus bisect each other at right angles. Let diagonals be $2p$ and $2q.$ In each right triangle formed: side $^2=p^2+q^2.$ With four equal sides, $4(\text{side}^2)=4(p^2+q^2)=(2p)^2+(2q)^2=$ sum of squares of diagonals.

Q8. In the figure, $O$ is a point in the interior of triangle $ABC$, $OD\perp BC$, $OE\perp AC$, $OF\perp AB.$ Show that: (i) $OA^{2}+OB^{2}+OC^{2}-OD^{2}-OE^{2}-OF^{2}=AF^{2}+BD^{2}+CE^{2}$ (ii) $AF^{2}+BD^{2}+CE^{2}=AE^{2}+CD^{2}+BF^{2}.$

Answer: By Pythagoras in right triangles formed: $OA^{2}=OF^{2}+AF^{2}=OE^{2}+AE^{2}$, etc. Adding gives both results.

Q9. A ladder 10 m long reaches a window 8 m above the ground. Find the distance of the foot of the ladder from base of the wall.

Answer: $d^{2}=10^{2}-8^{2}=100-64=36\Rightarrow d=6\ \text{m}.$

Q10. A guy wire attached to a vertical pole of height 18 m is 24 m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut?

Answer: $x^{2}=24^{2}-18^{2}=576-324=252\Rightarrow x=\sqrt{252}=6\sqrt7\ \text{m}.$

Q11. An aeroplane leaves an airport and flies due north at a speed of 1000 km/hr. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1200 km/hr. How far apart will be the two planes after $1\tfrac12$ hours?

Answer: Distances: $1000\times \tfrac32=1500$ km north and $1200\times \tfrac32=1800$ km west. Distance apart: $\sqrt{1500^{2}+1800^{2}}=\sqrt{2250000+3240000}=\sqrt{5490000}=300\sqrt{61}\ \text{km}.$

Q12. Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between the feet of the poles is 12 m, find the distance between their tops.

Answer: Vertical difference $=11-6=5$ m, horizontal $=12$ m. Top distance $=\sqrt{12^{2}+5^{2}}=\sqrt{169}=13\ \text{m}.$

Q13. $D$ and $E$ are points on the sides $CA$ and $CB$ respectively of a triangle $ABC$ right-angled at $C.$ Prove that $AE^{2}+BD^{2}=AB^{2}+DE^{2}.$

Answer: By Pythagoras: $AE^{2}=AC^{2}+CE^{2},\ BD^{2}=BC^{2}+CD^{2},\ AB^{2}=AC^{2}+BC^{2},\ DE^{2}=CD^{2}+CE^{2}.$ Adding: $AE^{2}+BD^{2}=AC^{2}+BC^{2}+CD^{2}+CE^{2}=AB^{2}+DE^{2}.$

Q14. The perpendicular from $A$ on side $BC$ of a $\triangle ABC$ intersects $BC$ at $D$ such that $DB=3\,CD.$ Prove that $2\,AB^{2}=2\,AC^{2}+BC^{2}.$

Answer: Let $CD=x$, then $DB=3x$, $BC=4x.$ By Pythagoras: $AB^{2}=AD^{2}+DB^{2}=AD^{2}+9x^{2},\ AC^{2}=AD^{2}+x^{2}.$ So $AB^{2}-AC^{2}=8x^{2}=\tfrac12 BC^{2}\Rightarrow 2(AB^{2}-AC^{2})=BC^{2}\Rightarrow 2\,AB^{2}=2\,AC^{2}+BC^{2}.$

Q15. In an equilateral triangle $ABC$, $D$ is a point on side $BC$ such that $BD=\tfrac13 BC.$ Prove that $9\,AD^{2}=7\,AB^{2}.$

Answer: Let $AB=BC=CA=a.$ Drop $AE\perp BC$, so $BE=\tfrac{a}{2}.$ $BD=\tfrac{a}{3}\Rightarrow DE=BE-BD=\tfrac{a}{2}-\tfrac{a}{3}=\tfrac{a}{6}.$ Also $AE^{2}=a^{2}-\left(\tfrac{a}{2}\right)^{2}=\tfrac{3a^{2}}{4}.$ Then $AD^{2}=AE^{2}+DE^{2}=\tfrac{3a^{2}}{4}+\tfrac{a^{2}}{36}=\tfrac{27a^{2}+a^{2}}{36}=\tfrac{28a^{2}}{36}=\tfrac{7a^{2}}{9}.$ Therefore $9\,AD^{2}=7a^{2}=7\,AB^{2}.$

Q16. In an equilateral triangle, prove that three times the square of one side is equal to four times the square of one of its altitudes.

Answer: Side $a$, altitude $h=\tfrac{\sqrt3}{2}a\Rightarrow h^{2}=\tfrac34 a^{2}\Rightarrow 4h^{2}=3a^{2}.$ Hence proved.

Q17. Tick the correct answer and justify: In $\triangle ABC$, $AB=6\sqrt3$ cm, $AC=12$ cm and $BC=6$ cm. The angle $B$ is: (A) $120^{\circ}$ (B) $60^{\circ}$ (C) $90^{\circ}$ (D) $45^{\circ}.$

Answer: $AB^{2}+BC^{2}=108+36=144=AC^{2}.$ By converse of Pythagoras, $\angle B=90^{\circ}.$ Option (C).

Exercise 6.6 (Optional)

Q1. In the figure, $PS$ is the bisector of $\angle QPR$ of $\triangle PQR.$ Prove that $\dfrac{QS}{SR}=\dfrac{PQ}{PR}.$ (Angle Bisector Theorem)

Answer: Through $R$ draw $RT\parallel SP$ meeting $QP$ produced at $T.$ Then $\angle SPR=\angle PRT$ (alternate) and $\angle QPS=\angle PTR$ (corresponding). Since $PS$ bisects $\angle QPR$, $\angle QPS=\angle SPR\Rightarrow \angle PTR=\angle PRT\Rightarrow PT=PR.$ In $\triangle QRT$, $SP\parallel RT\Rightarrow \dfrac{QS}{SR}=\dfrac{QP}{PT}=\dfrac{PQ}{PR}.$

Q2. In the figure, $D$ is a point on hypotenuse $AC$ of $\triangle ABC$, such that $BD\perp AC$, $DM\perp BC$ and $DN\perp AB.$ Prove that: (i) $DM^{2}=DN\cdot MC$ (ii) $DN^{2}=DM\cdot AN.$

Answer: (i) In $\triangle DMC$ and $\triangle BMD$: $\angle DMC=\angle BMD=90^{\circ}$ and $\angle MCD=\angle MDB$ (each $90^{\circ}-\angle DBC$). By AA they are similar, giving $\dfrac{DM}{BM}=\dfrac{MC}{DM}\Rightarrow DM^{2}=BM\cdot MC.$ Now $BMDN$ is a rectangle (three right angles), so $BM=DN.$ Hence $DM^{2}=DN\cdot MC.$
(ii) Similarly $\triangle DNA\sim\triangle BND\Rightarrow DN^{2}=AN\cdot NB=AN\cdot DM.$

Q3. In the figure, $ABC$ is a triangle in which $\angle ABC>90^{\circ}$ and $AD\perp CB$ produced. Prove that $AC^{2}=AB^{2}+BC^{2}+2\,BC\cdot BD.$

Answer: In right $\triangle ADC$: $AC^{2}=AD^{2}+DC^{2}=AD^{2}+(BC+BD)^{2}=AD^{2}+BD^{2}+BC^{2}+2\,BC\cdot BD.$ But in right $\triangle ADB$: $AD^{2}+BD^{2}=AB^{2}.$ Therefore $AC^{2}=AB^{2}+BC^{2}+2\,BC\cdot BD.$

Q4. In the figure, $ABC$ is a triangle in which $\angle ABC<90^{\circ}$ and $AD\perp BC.$ Prove that $AC^{2}=AB^{2}+BC^{2}-2\,BC\cdot BD.$

Answer: $AC^{2}=AD^{2}+DC^{2}=AD^{2}+(BC-BD)^{2}=AD^{2}+BD^{2}+BC^{2}-2\,BC\cdot BD=AB^{2}+BC^{2}-2\,BC\cdot BD.$

Q5. In the figure, $AD$ is a median of a triangle $ABC$ and $AM\perp BC.$ Prove that: (i) $AC^{2}=AD^{2}+BC\cdot DM+\left(\dfrac{BC}{2}\right)^{2}$ (ii) $AB^{2}=AD^{2}-BC\cdot DM+\left(\dfrac{BC}{2}\right)^{2}$ (iii) $AC^{2}+AB^{2}=2\,AD^{2}+\dfrac{1}{2}BC^{2}.$

Answer: Apply Q3 (obtuse case) and Q4 (acute case) results to triangles $ADC$ and $ADB$ noting $D$ is mid-point of $BC.$ Adding (i) and (ii) gives (iii).

Q6. Prove that the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides.

Answer: Using Q5 (iii) on $\triangle ABC$ with median $BD$ and on $\triangle ACD$ with median $DB$, and noting $AC$ and $BD$ are diagonals: $AC^{2}+BD^{2}=AB^{2}+BC^{2}+CD^{2}+DA^{2}.$

Q7. In the figure, two chords $AB$ and $CD$ intersect each other at the point $P.$ Prove that: (i) $\triangle APC\sim\triangle DPB$ (ii) $AP\cdot PB=CP\cdot DP.$

Answer: $\angle APC=\angle DPB$ (vertically opposite), $\angle CAP=\angle BDP$ (angles in same segment $CB$). By AA, $\triangle APC\sim\triangle DPB.$ Hence $\dfrac{AP}{DP}=\dfrac{CP}{BP}\Rightarrow AP\cdot PB=CP\cdot DP.$

Q8. In the figure, two chords $AB$ and $CD$ of a circle intersect each other at the point $P$ (when produced) outside the circle. Prove that: (i) $\triangle PAC\sim\triangle PDB$ (ii) $PA\cdot PB=PC\cdot PD.$

Answer: $\angle P=\angle P$ (common). $\angle PAC=\angle PDB$ (exterior angle of cyclic quadrilateral equals interior opposite). By AA, $\triangle PAC\sim\triangle PDB.$ Hence $PA\cdot PB=PC\cdot PD.$

Q9. In the figure, $D$ is a point on side $BC$ of $\triangle ABC$ such that $\dfrac{BD}{CD}=\dfrac{AB}{AC}.$ Prove that $AD$ is the bisector of $\angle BAC.$

Answer: Produce $BA$ to $E$ such that $AE=AC.$ Then $\angle AEC=\angle ACE.$ Given $\dfrac{BD}{CD}=\dfrac{AB}{AC}=\dfrac{AB}{AE}.$ By converse of BPT in $\triangle BCE$, $AD\parallel EC.$ Hence $\angle BAD=\angle AEC$ (corresponding) and $\angle DAC=\angle ACE$ (alternate). Since $\angle AEC=\angle ACE$, $\angle BAD=\angle DAC$, so $AD$ bisects $\angle BAC.$

Q10. Nazima is fly-fishing in a stream. The tip of her fishing rod is 1.8 m above the surface of the water and the fly at the end of the string rests on the water 3.6 m away and 2.4 m from a point directly under the tip of the rod. If she pulls in the string at the rate of 5 cm per second, what will be the horizontal distance of the fly from her after 12 seconds?

Answer: Initial string length $=\sqrt{(2.4)^{2}+(1.8)^{2}}=\sqrt{5.76+3.24}=\sqrt9=3\ \text{m}.$ String pulled in 12 s $=12\times 5=60$ cm $=0.6$ m. New string length $=3-0.6=2.4$ m. New horizontal distance from rod foot $=\sqrt{(2.4)^{2}-(1.8)^{2}}=\sqrt{5.76-3.24}=\sqrt{2.52}=1.587\ \text{m}.$ Distance of fly from Nazima $=1.587+1.2=2.787\ \text{m}.$

Additional Important Questions

Q1. State Basic Proportionality Theorem.

Answer: If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then the other two sides are divided in the same ratio.

Q2. State the Pythagoras theorem and its converse.

Answer: Pythagoras: In a right triangle, the square of the hypotenuse equals the sum of squares of the other two sides. Converse: If in a triangle the square of one side equals the sum of squares of the other two, the angle opposite the first side is a right angle.

Q3. The areas of two similar triangles are 100 cm² and 49 cm². If the altitude of the bigger triangle is 5 cm, find the corresponding altitude of the smaller one.

Answer: $\dfrac{100}{49}=\left(\dfrac{5}{h}\right)^{2}\Rightarrow \dfrac{10}{7}=\dfrac{5}{h}\Rightarrow h=3.5\ \text{cm}.$

Q4. The hypotenuse of a right triangle is 25 cm. The other two sides are such that one is 5 cm longer than the other. Find the sides.

Answer: Let sides be $x$ and $x+5.$ Then $x^{2}+(x+5)^{2}=25^{2}=625.$ $2x^{2}+10x+25=625\Rightarrow x^{2}+5x-300=0\Rightarrow (x-15)(x+20)=0\Rightarrow x=15.$ Sides are 15 cm and 20 cm.

Q5. If the corresponding sides of two similar triangles are in the ratio 3:5, what is the ratio of their (i) perimeters (ii) areas?

Answer: (i) $3:5$ (perimeter ratio = side ratio for similar figures). (ii) $9:25$ (square of side ratio).

Q6. In a right triangle, the lengths of the two perpendicular sides are $a$ and $b.$ Prove that the length of the perpendicular from the right-angle vertex on the hypotenuse equals $\dfrac{ab}{\sqrt{a^{2}+b^{2}}}.$

Answer: Hypotenuse $c=\sqrt{a^{2}+b^{2}}.$ Area of triangle $=\tfrac12 ab=\tfrac12 c\cdot p\Rightarrow p=\dfrac{ab}{c}=\dfrac{ab}{\sqrt{a^{2}+b^{2}}}.$

Q7. The areas of two similar triangles are in the ratio $25:64.$ Find the ratio of their corresponding medians.

Answer: Ratio of medians $=$ ratio of corresponding sides $=\sqrt{25:64}=5:8.$

Q8. $ABC$ is a triangle in which $AB=AC$ and $D$ is any point on $BC.$ Prove that $AB^{2}-AD^{2}=BD\cdot DC.$

Answer: Drop $AE\perp BC.$ Then $BE=EC.$ $AB^{2}=AE^{2}+BE^{2}$ and $AD^{2}=AE^{2}+ED^{2}.$ So $AB^{2}-AD^{2}=BE^{2}-ED^{2}=(BE+ED)(BE-ED)=BD\cdot(EC-ED)=BD\cdot DC.$

Q9. In a quadrilateral $ABCD$, $\angle B=90^{\circ}$ and $AD^{2}=AB^{2}+BC^{2}+CD^{2}.$ Prove that $\angle ACD=90^{\circ}.$

Answer: In right $\triangle ABC$: $AC^{2}=AB^{2}+BC^{2}.$ Substituting: $AD^{2}=AC^{2}+CD^{2}.$ By converse of Pythagoras in $\triangle ACD$, $\angle ACD=90^{\circ}.$

Q10. The diagonal $BD$ of a parallelogram $ABCD$ intersects $AE$ at $F$, where $E$ is any point on $BC.$ Prove that $DF\cdot EF=FB\cdot FA.$

Answer: In $\triangle DFA$ and $\triangle BFE$: $\angle DFA=\angle BFE$ (vertically opposite). $\angle FDA=\angle FBE$ (alternate, $AD\parallel BC$). By AA, $\triangle DFA\sim\triangle BFE.$ Hence $\dfrac{DF}{BF}=\dfrac{FA}{FE}\Rightarrow DF\cdot EF=FB\cdot FA.$

Q11. State the AAA similarity criterion.

Answer: If in two triangles, the corresponding angles are equal, then the triangles are similar.

Q12. State the SAS similarity criterion.

Answer: If in two triangles, one pair of corresponding angles is equal and the sides including these angles are in the same ratio, the triangles are similar.

Q13. State the SSS similarity criterion.

Answer: If in two triangles, the corresponding sides are proportional, the triangles are similar.

Q14. The legs of a right triangle are in the ratio $3:4$ and the hypotenuse is 20 cm. Find the legs.

Answer: Legs $=3k,\ 4k.$ $(3k)^{2}+(4k)^{2}=400\Rightarrow 25k^{2}=400\Rightarrow k=4.$ Legs are 12 cm and 16 cm.

Q15. The shadow of a tower of height 25 m is 20 m. At the same time the shadow of a tree is 8 m. Find the height of the tree.

Answer: $\dfrac{25}{20}=\dfrac{h}{8}\Rightarrow h=10\ \text{m}.$

More Worked Examples (with Figures)

Worked Example 1. In $\triangle ABC$, $DE\parallel BC$, $AD=4$ cm, $DB=6$ cm and $AE=8$ cm. Find $EC$ and the ratio $\dfrac{\text{ar}(\triangle ADE)}{\text{ar}(\triangle ABC)}.$

Solution. By BPT, $\dfrac{AD}{DB}=\dfrac{AE}{EC}\Rightarrow \dfrac{4}{6}=\dfrac{8}{EC}\Rightarrow EC=12$ cm. Also $\triangle ADE\sim\triangle ABC$ (AA, since $DE\parallel BC$ gives $\angle ADE=\angle ABC$ and $\angle AED=\angle ACB$). Now $\dfrac{AD}{AB}=\dfrac{4}{4+6}=\dfrac{2}{5}.$ Therefore

$$\frac{\text{ar}(\triangle ADE)}{\text{ar}(\triangle ABC)}=\left(\frac{AD}{AB}\right)^{2}=\left(\frac{2}{5}\right)^{2}=\frac{4}{25}.$$

Worked Example 2. In the figure, $\triangle ABC\sim\triangle DEF$ with $AB=6$ cm, $DE=4$ cm and $\text{ar}(\triangle ABC)=72$ cm². Find $\text{ar}(\triangle DEF).$

Solution. $\dfrac{\text{ar}(\triangle ABC)}{\text{ar}(\triangle DEF)}=\left(\dfrac{AB}{DE}\right)^{2}=\left(\dfrac{6}{4}\right)^{2}=\dfrac{9}{4}.$ So $\text{ar}(\triangle DEF)=\dfrac{4}{9}\times 72=32$ cm².

Worked Example 3. In a right triangle, the legs are 9 cm and 40 cm. Find the hypotenuse.

Solution. By Pythagoras, $h^{2}=9^{2}+40^{2}=81+1600=1681\Rightarrow h=\sqrt{1681}=41$ cm.

Worked Example 4. $\triangle ABC$ is right-angled at $B$. $D$ is the foot of the perpendicular from $B$ to $AC.$ If $AB=6$ cm and $BC=8$ cm, find $BD,\ AD,\ DC.$

Solution. $AC=\sqrt{6^{2}+8^{2}}=10$ cm. Area $=\tfrac12\cdot 6\cdot 8=24$, also $=\tfrac12\cdot 10\cdot BD$, so $BD=4.8$ cm. By similar triangles ($\triangle ABD\sim\triangle ABC$): $\dfrac{AD}{AB}=\dfrac{AB}{AC}\Rightarrow AD=\dfrac{36}{10}=3.6$ cm. Then $DC=10-3.6=6.4$ cm.

Worked Example 5. A man goes 10 m due east and then 24 m due north. Find the distance from the starting point.

Solution. $d=\sqrt{10^{2}+24^{2}}=\sqrt{100+576}=\sqrt{676}=26$ m.

MCQ & Short Answer Booster

Q1. (MCQ) Two triangles are similar if their corresponding angles are equal and corresponding sides are:

(A) equal (B) parallel (C) proportional (D) perpendicular.
Answer: (C) proportional.

Q2. (MCQ) If $\triangle ABC\sim\triangle PQR$ and $\dfrac{AB}{PQ}=\dfrac{2}{3}$, then $\dfrac{\text{ar}(\triangle ABC)}{\text{ar}(\triangle PQR)}$ equals:

(A) $\tfrac23$ (B) $\tfrac49$ (C) $\tfrac94$ (D) $\tfrac32.$
Answer: (B) $\tfrac49.$

Q3. (MCQ) If the sides of two similar triangles are in the ratio $2:3$, then the ratio of their perimeters is:

(A) $2:3$ (B) $4:9$ (C) $9:4$ (D) $3:2.$
Answer: (A) $2:3.$

Q4. (MCQ) In $\triangle ABC$, if $AB^{2}+BC^{2}=AC^{2}$, then the right angle is at:

(A) $A$ (B) $B$ (C) $C$ (D) cannot be determined.
Answer: (B) $B$ (the angle opposite the longest side $AC$).

Q5. (MCQ) The length of the altitude of an equilateral triangle of side $a$ is:

(A) $\tfrac{a}{2}$ (B) $\tfrac{a\sqrt2}{2}$ (C) $\tfrac{a\sqrt3}{2}$ (D) $a.$
Answer: (C) $\tfrac{a\sqrt3}{2}.$

Q6. (Fill in) A line drawn parallel to one side of a triangle divides the other two sides in the __________ ratio.
Answer: same.

Q7. (Fill in) If two triangles are similar and the ratio of corresponding sides is $k$, then the ratio of their areas is __________.
Answer: $k^{2}.$

Q8. (Fill in) The Pythagoras theorem holds in a __________-angled triangle.
Answer: right.

Q9. (True/False) All squares are similar — True.

Q10. (True/False) All similar figures are congruent — False (only same-sized similar figures are congruent).

Q11. (True/False) The diagonal of a square divides it into two similar triangles — True (in fact, congruent and so similar).

Q12. (True/False) If the corresponding sides of two triangles are equal, the triangles are similar — True (they are congruent, hence similar).

Theorem-based Long Questions

Q1. State and prove the Basic Proportionality Theorem (BPT).

Answer: See “Theorem 6.1” earlier with full proof and figure.

Q2. State and prove the Pythagoras Theorem using similar triangles.

Answer: See “Theorem 6.7” earlier with full proof and figure.

Q3. Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

Answer: Let $\triangle ABC\sim\triangle PQR.$ Drop altitudes $AM\perp BC$ and $PN\perp QR.$ In $\triangle ABM$ and $\triangle PQN$: $\angle B=\angle Q$ (similar triangles) and $\angle AMB=\angle PNQ=90^{\circ}.$ So by AA, $\triangle ABM\sim\triangle PQN\Rightarrow \dfrac{AM}{PN}=\dfrac{AB}{PQ}.$ Now

$$\frac{\text{ar}(\triangle ABC)}{\text{ar}(\triangle PQR)}=\frac{\tfrac12 BC\cdot AM}{\tfrac12 QR\cdot PN}=\frac{BC}{QR}\cdot\frac{AM}{PN}=\frac{AB}{PQ}\cdot\frac{AB}{PQ}=\left(\frac{AB}{PQ}\right)^{2}.$$

Hence proved.

Q4. State and prove the converse of the Pythagoras theorem.

Answer: Statement. If in a triangle the square of one side equals the sum of squares of the other two, then the angle opposite the first side is a right angle.

Given. $\triangle ABC$ in which $AC^{2}=AB^{2}+BC^{2}.$ To prove. $\angle B=90^{\circ}.$

Construction. Draw $\triangle PQR$ such that $PQ=AB$, $QR=BC$ and $\angle Q=90^{\circ}.$

Proof. By Pythagoras in $\triangle PQR$, $PR^{2}=PQ^{2}+QR^{2}=AB^{2}+BC^{2}=AC^{2}\Rightarrow PR=AC.$ Now in $\triangle ABC$ and $\triangle PQR$: $AB=PQ,\ BC=QR,\ AC=PR.$ By SSS, $\triangle ABC\cong\triangle PQR$, so $\angle B=\angle Q=90^{\circ}.$ Hence proved.

Q5. Prove that in a right triangle, the perpendicular drawn from the right-angle vertex on the hypotenuse divides it into two triangles, each of which is similar to the whole triangle and to each other.

Answer: Let $\triangle ABC$ be right-angled at $B$, $BD\perp AC.$ In $\triangle ABD$ and $\triangle ABC$: $\angle A=\angle A$ (common), $\angle ADB=\angle ABC=90^{\circ}.$ By AA, $\triangle ABD\sim\triangle ABC.$ Similarly, $\triangle BDC\sim\triangle ABC.$ Hence $\triangle ABD\sim\triangle BDC$ (both similar to the same triangle).

Q6. In a triangle, prove that the line joining the mid-points of any two sides is parallel to the third side and equals half of it (Mid-point theorem).

Answer: In $\triangle ABC$, let $D$ and $E$ be mid-points of $AB$ and $AC.$ Then $\dfrac{AD}{DB}=\dfrac{AE}{EC}=1$, so by converse of BPT, $DE\parallel BC.$ Also $\triangle ADE\sim\triangle ABC$ with ratio $\tfrac12$, so $DE=\tfrac12 BC.$

Practice Problems with Hints

P1. In $\triangle ABC$, $DE\parallel BC$ and $AD:DB=2:3.$ If $AE=4$ cm, find $EC.$ Hint: $\dfrac{AD}{DB}=\dfrac{AE}{EC}\Rightarrow \dfrac{2}{3}=\dfrac{4}{EC}\Rightarrow EC=6$ cm.

P2. $\triangle ABC\sim\triangle DEF$ and their perimeters are 30 cm and 18 cm. If $AB=12$ cm, find $DE.$ Hint: Ratio $=30:18=5:3,\ DE=12\times \tfrac35=7.2$ cm.

P3. The areas of two similar triangles are 81 cm² and 49 cm². If the median of the bigger triangle is 9 cm, find the corresponding median of the smaller. Hint: $\dfrac{m}{9}=\sqrt{\tfrac{49}{81}}=\tfrac79\Rightarrow m=7$ cm.

P4. The hypotenuse of a right triangle is $3\sqrt{10}$ cm. If the smaller side is tripled and larger doubled, the new hypotenuse is 9√5 cm. Find the legs. Hint: If legs are $a,b$, $a^2+b^2=90$ and $9a^2+4b^2=405$, giving $a^2=63/5,\ b^2=…$, recompute: solve $a^2+b^2=90$, $(3a)^2+(2b)^2=(9\sqrt5)^2=405\Rightarrow 9a^2+4b^2=405.$ From these $5a^2=405-4(90)=45\Rightarrow a^2=9,\ a=3$ cm; $b^2=81,\ b=9$ cm.

P5. In $\triangle ABC$ right-angled at $C$, if $a,b,c$ are sides opposite to $A,B,C$ respectively, prove $\sin^2 A+\sin^2 B=1.$ Hint: $\sin A=\tfrac{a}{c},\ \sin B=\tfrac{b}{c}\Rightarrow \sin^2 A+\sin^2 B=\tfrac{a^2+b^2}{c^2}=1.$

P6. ABC is a triangle in which AB = AC = 5 cm and BC = 6 cm. Find the length of the altitude from A. Hint: Foot $D$ is mid-point of $BC$, $BD=3$, altitude $=\sqrt{25-9}=4$ cm.

P7. Prove that if a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse, the triangles on each side of the perpendicular are similar to the whole triangle and to each other. Hint: AA similarity using the common angle and the right angles.

Detailed Application Problems with Figures

A1. Shadow / similar-triangle problem. A girl 1.6 m tall stands at a distance of 3.2 m from a lamp post and casts a shadow of 4.8 m on the ground. Find the height of the lamp post.

Solution. $\triangle ABC$ (lamp post + shadow) and $\triangle PQC$ (girl + shadow) are similar (AA). $\dfrac{h}{1.6}=\dfrac{3.2+4.8}{4.8}=\dfrac{8}{4.8}\Rightarrow h=\dfrac{1.6\times 8}{4.8}=\dfrac{12.8}{4.8}=\dfrac{8}{3}\approx 2.67$ m.

A2. Trapezium diagonals. $ABCD$ is a trapezium with $AB\parallel DC$, $AB=20$ cm, $DC=12$ cm. Diagonals $AC$ and $BD$ meet at $O.$ Find the ratio in which $O$ divides $AC$ and the ratio of areas of $\triangle AOB$ and $\triangle COD.$

Solution. $\triangle AOB\sim\triangle COD$ (AA). $\dfrac{AO}{OC}=\dfrac{AB}{CD}=\dfrac{20}{12}=\dfrac{5}{3}.$ Ratio of areas $=\left(\dfrac{20}{12}\right)^{2}=\dfrac{25}{9}.$

A3. Two ladders / right triangle. Two poles of heights 9 m and 14 m stand vertically on a plane ground. If the distance between their feet is 12 m, find the distance between their tops.

Solution. Vertical drop $=14-9=5$ m, horizontal $=12$ m. $d=\sqrt{12^{2}+5^{2}}=\sqrt{144+25}=\sqrt{169}=13$ m.

A4. Streetlight and pedestrian. A man 1.8 m tall walks away from a 9 m tall streetlight at 1 m/s. Find the rate at which his shadow lengthens.

Solution. Let his distance from the lamp be $x$ and shadow length $s.$ Similar triangles give $\dfrac{1.8}{9}=\dfrac{s}{x+s}\Rightarrow 1.8(x+s)=9s\Rightarrow 1.8x=7.2s\Rightarrow s=\tfrac{x}{4}.$ Rate $\dfrac{ds}{dt}=\tfrac14\cdot\dfrac{dx}{dt}=\tfrac14$ m/s.

A5. River crossing problem. A boat is sailing perpendicular to a riverbank with speed 5 m/s while the river current flows along the bank at 12 m/s. Find the speed of the boat as observed from the bank.

Solution. Resultant speed $v=\sqrt{5^{2}+12^{2}}=\sqrt{169}=13$ m/s.

A6. Cyclic figure. Two chords $AB$ and $CD$ of a circle intersect inside the circle at $P.$ If $AP=6$ cm, $PB=4$ cm and $CP=3$ cm, find $PD.$

Solution. $AP\cdot PB=CP\cdot PD\Rightarrow 6\cdot 4=3\cdot PD\Rightarrow PD=8$ cm.

A7. Median ratio. The medians of two similar triangles are in the ratio $4:5.$ Find the ratio of their (i) corresponding sides (ii) areas (iii) perimeters.

Solution. (i) $4:5$ (medians scale with sides). (ii) $4^2:5^2=16:25.$ (iii) $4:5.$

A8. Sides of a right triangle in arithmetic progression. The sides of a right triangle are in AP. Find the ratio of the sides.

Solution. Let sides $=a-d,a,a+d.$ Then $(a+d)^2=a^2+(a-d)^2\Rightarrow a^2+2ad+d^2=a^2+a^2-2ad+d^2\Rightarrow 2ad=a^2-2ad\Rightarrow 4ad=a^2\Rightarrow a=4d.$ Sides $=3d,4d,5d$, i.e., ratio $3:4:5.$

A9. Diagonal of rectangle. The length and breadth of a rectangle are 24 cm and 7 cm. Find its diagonal.

Solution. Diagonal $=\sqrt{24^2+7^2}=\sqrt{576+49}=\sqrt{625}=25$ cm.

A10. Sides of an isosceles right triangle. The hypotenuse of an isosceles right triangle is 12 cm. Find each leg.

Solution. $2x^2=144\Rightarrow x^2=72\Rightarrow x=6\sqrt2$ cm.

Mistakes to Avoid & Tips

Always identify corresponding vertices when writing $\triangle ABC\sim\triangle PQR.$ Order matters: $A\leftrightarrow P, B\leftrightarrow Q, C\leftrightarrow R.$
For BPT, the line must intersect the two sides in distinct interior points, and must be parallel to the third side.
The ratio of areas equals the square of the ratio of corresponding sides — not the ratio itself.
Pythagoras applies only to right triangles. Always check the right angle is identified before using $a^{2}+b^{2}=c^{2}.$
The longest side in a right triangle is always the hypotenuse, and it is opposite the right angle.
Two similar triangles need NOT be congruent. Two congruent triangles ARE similar.
SSA is not a similarity (or congruence) criterion — be careful.
While solving, write step numbers like (1), (2)… and quote the theorem (BPT, AA, etc.) used in each step.
For boards, draw a clear figure even if not asked — it earns marks and clarifies the working.

Quick Recap Formula Sheet

$$\text{BPT:}\quad DE\parallel BC\ \Rightarrow\ \frac{AD}{DB}=\frac{AE}{EC}.$$

$$\text{Areas of similar triangles:}\quad \frac{\text{ar}(\triangle ABC)}{\text{ar}(\triangle PQR)}=\left(\frac{AB}{PQ}\right)^{2}.$$

$$\text{Pythagoras:}\quad (\text{hyp})^{2}=(\text{leg}_1)^{2}+(\text{leg}_2)^{2}.$$

$$\text{Equilateral altitude:}\quad h=\frac{\sqrt3}{2}\cdot a.$$

$$\text{Equilateral area:}\quad A=\frac{\sqrt3}{4}\cdot a^{2}.$$

$$\text{Altitude on hypotenuse:}\quad h=\frac{\text{leg}_1\cdot \text{leg}_2}{\text{hyp}},\quad \text{hyp}^2 \cdot h^2 = (\text{leg}_1\cdot \text{leg}_2)^2.$$

$$\text{Apollonius (median):}\quad AB^{2}+AC^{2}=2\,AD^{2}+\tfrac{1}{2}\,BC^{2}\ \text{(median }AD\text{ to side }BC\text{)}.$$

Board-Style Question Bank (with Full Solutions)

B1. In the figure, $\triangle ABC$ is right-angled at $C$, $D$ is the mid-point of $AB$ and $DE\perp AC.$ Prove that $E$ is the mid-point of $AC$ and $DE=\tfrac12\,BC.$

Answer: $DE\perp AC$ and $BC\perp AC\Rightarrow DE\parallel BC.$ In $\triangle ABC$, since $D$ is mid-point of $AB$ and $DE\parallel BC$, by the converse-of-BPT (mid-point theorem), $E$ is the mid-point of $AC.$ Also $\triangle ADE\sim\triangle ABC$ with ratio $\tfrac12$, so $DE=\tfrac12 BC.$

B2. $ABCD$ is a quadrilateral in which $AD=BC.$ If $P,Q,R,S$ are mid-points of $AB,AC,CD,BD$ respectively, prove that $PQRS$ is a rhombus.

Answer: By mid-point theorem, $PQ\parallel BC$ and $PQ=\tfrac12 BC$, $RS\parallel BC$ and $RS=\tfrac12 BC.$ So $PQ\parallel RS$ and $PQ=RS\Rightarrow PQRS$ parallelogram. Also $QR\parallel AD$ and $QR=\tfrac12 AD=\tfrac12 BC=PQ.$ So adjacent sides equal $\Rightarrow$ rhombus.

B3. In a right triangle, prove that area of the equilateral triangle on the hypotenuse equals the sum of the areas of equilateral triangles on the other two sides.

Answer: Equilateral triangle on side $s$ has area $\tfrac{\sqrt3}{4}s^{2}.$ Sum on legs $=\tfrac{\sqrt3}{4}(a^{2}+b^{2}).$ On hypotenuse $=\tfrac{\sqrt3}{4}c^{2}.$ By Pythagoras, $a^{2}+b^{2}=c^{2}.$ Hence equal.

B4. $P$ and $Q$ are points on sides $AB$ and $AC$ of $\triangle ABC.$ $AP=4$ cm, $PB=8$ cm, $AQ=3$ cm, $QC=9$ cm. Is $PQ\parallel BC$?

Answer: $\dfrac{AP}{PB}=\dfrac{4}{8}=\dfrac12,\ \dfrac{AQ}{QC}=\dfrac{3}{9}=\dfrac13.$ Ratios unequal, so $PQ\not\parallel BC.$

B5. $\triangle ABC$ is an isosceles triangle with $AB=AC.$ The bisector of $\angle B$ meets $AC$ at $D.$ Prove that $\triangle ABD\sim\triangle ABC$ if and only if $\angle BAC=36^{\circ}.$

Answer: Let $\angle A=2x.$ Then $\angle B=\angle C=\tfrac{180-2x}{2}=90-x.$ $\angle ABD=\tfrac12\angle B=\tfrac{90-x}{2}.$ For $\triangle ABD\sim\triangle ABC$ with vertex $A$ common, we need $\angle ABD=\angle ACB=90-x$, giving $\tfrac{90-x}{2}=90-x\Rightarrow 90-x=0$ which is impossible, or correspondence $\triangle ABD\sim\triangle BCA$: $\angle ABD=\angle BCA=90-x$ wrong. Instead AA $\Rightarrow$ $\angle A=\angle A$ and $\angle ABD=\angle BCA.$ This gives $\tfrac{90-x}{2}=90-x\Rightarrow x=90$, contradiction. Take $\triangle BDC$: $\angle BDC=180-\angle ABD-\angle C=180-\tfrac{90-x}{2}-(90-x)=180-\tfrac{90-x}{2}-90+x=\tfrac{180+3x-90}{2}=\tfrac{90+3x}{2}.$ For golden-ratio result $2x=36^{\circ}$ standard.

B6. In $\triangle ABC$, $AD$ is the median to $BC.$ Prove that $AB^{2}+AC^{2}=2(AD^{2}+BD^{2})$ (Apollonius’ theorem).

Answer: Drop $AE\perp BC.$ In right $\triangle ADE$ with $E$ between $D$ and (say) $C$: $AC^{2}=AE^{2}+EC^{2}=AE^{2}+(DC+DE)^{2}$ when $E$ is between $C$ and $D$, and $AB^{2}=AE^{2}+(BD-DE)^{2}.$ Using $BD=DC$, adding:

$AB^{2}+AC^{2}=2AE^{2}+(BD-DE)^{2}+(BD+DE)^{2}=2AE^{2}+2BD^{2}+2DE^{2}=2(AE^{2}+DE^{2})+2BD^{2}=2AD^{2}+2BD^{2}.$

B7. Find the value of $x$ in the figure where $DE\parallel BC$, $AD=x$, $DB=x-2$, $AE=x+2$, $EC=x-1$ (all in cm).

Answer: By BPT, $\dfrac{x}{x-2}=\dfrac{x+2}{x-1}.$ Cross-multiplying: $x(x-1)=(x-2)(x+2)\Rightarrow x^{2}-x=x^{2}-4\Rightarrow -x=-4\Rightarrow x=4$ cm.

B8. The areas of two similar triangles $ABC$ and $PQR$ are in the ratio $9:16.$ If $BC=4.5$ cm, find $QR.$

Answer: $\left(\dfrac{BC}{QR}\right)^{2}=\dfrac{9}{16}\Rightarrow \dfrac{BC}{QR}=\dfrac{3}{4}\Rightarrow QR=\dfrac{4\times 4.5}{3}=6$ cm.

B9. In a right triangle, the altitudes from the two acute angles meet at the right-angle vertex. Comment.

Answer: Yes. In a right triangle, the legs themselves are the altitudes from the acute-angle vertices. Their intersection is the right-angle vertex itself, which is therefore the orthocentre.

B10. If two triangles are equiangular, are they similar? Justify.

Answer: Yes. By the AAA similarity criterion, equiangular triangles are similar — corresponding sides are automatically proportional.

B11. The corresponding sides of two similar triangles are 8 cm and 12 cm. The area of the smaller triangle is 64 cm². Find the area of the larger triangle.

Answer: $\dfrac{\text{ar (smaller)}}{\text{ar (larger)}}=\left(\dfrac{8}{12}\right)^{2}=\dfrac{4}{9}\Rightarrow \text{ar (larger)}=\dfrac{9}{4}\times 64=144$ cm².

B12. Two equilateral triangles have sides in the ratio $5:7.$ Find the ratio of their areas.

Answer: $\left(\dfrac{5}{7}\right)^{2}=\dfrac{25}{49}.$ So areas are in ratio $25:49.$

B13. In $\triangle ABC$, $\angle B=90^{\circ}.$ If $D$ is mid-point of $AC$, prove that $BD=\tfrac12 AC.$

Answer: By Pythagoras, $AC^{2}=AB^{2}+BC^{2}.$ Drop perpendicular from $D$ to $BC$; or note that in a right triangle, the median to the hypotenuse equals half the hypotenuse. (This follows from circumscribing the rectangle on $AB,BC$ — the diagonal of the rectangle passes through $D$ and has length $AC$ with $D$ at the midpoint.)

B14. The two legs of a right triangle are 6 cm and 8 cm. The altitude on the hypotenuse divides the hypotenuse into segments. Find these segments.

Answer: Hypotenuse $=10$ cm. By the relation $a^{2}=p\cdot c$ (leg² = adjacent segment × hypotenuse): $6^{2}=p\cdot 10\Rightarrow p=3.6$ cm and $8^{2}=q\cdot 10\Rightarrow q=6.4$ cm. Check $3.6+6.4=10.$ ✓

B15. In $\triangle PQR$, $S$ on $PQ$ and $T$ on $PR$ with $ST\parallel QR.$ If $PS=3$ cm, $SQ=2$ cm, $ST=4.5$ cm, find $QR.$

Answer: $\triangle PST\sim\triangle PQR$ (AA). $\dfrac{PS}{PQ}=\dfrac{ST}{QR}\Rightarrow \dfrac{3}{5}=\dfrac{4.5}{QR}\Rightarrow QR=7.5$ cm.

B16. $\triangle ABC\sim\triangle DEF$ such that $2\,AB=DE.$ If $BC=8$ cm, find $EF.$

Answer: $\dfrac{AB}{DE}=\dfrac{1}{2}=\dfrac{BC}{EF}\Rightarrow EF=2BC=16$ cm.

B17. The diagonals of a rhombus are 16 cm and 30 cm. Find each side.

Answer: Half-diagonals $=8, 15.$ Side $=\sqrt{8^{2}+15^{2}}=\sqrt{289}=17$ cm.

B18. A right triangle has legs $a$ and $b$ with $a:b=3:4.$ If the hypotenuse is 25 cm, find the legs and the area.

Answer: Legs $=3k,4k.$ $25k^{2}=625\Rightarrow k=5.$ Legs $=15$ cm and $20$ cm. Area $=\tfrac12\cdot 15\cdot 20=150$ cm².

B19. Show that the line segments joining the mid-points of the sides of a triangle divide it into four congruent triangles.

Answer: Let $D, E, F$ be mid-points of $BC, CA, AB.$ By mid-point theorem, $DE\parallel AB$, $EF\parallel BC$, $FD\parallel CA$, with each equal to half the parallel side. So all four small triangles $\triangle AFE, \triangle FBD, \triangle EDC, \triangle DEF$ have sides equal to half of the corresponding sides of $\triangle ABC.$ By SSS each pair is congruent, and each is congruent to the others.

B20. $\triangle ABC$ has sides $a=BC, b=CA, c=AB.$ If $b^{2}+c^{2}=5a^{2}$ and the median $AD$ has length $m_a$, find $m_a$ in terms of $a.$

Answer: By Apollonius’ theorem, $b^{2}+c^{2}=2m_{a}^{2}+\tfrac12 a^{2}.$ So $5a^{2}=2m_{a}^{2}+\tfrac12 a^{2}\Rightarrow 2m_{a}^{2}=\tfrac{9}{2}a^{2}\Rightarrow m_{a}^{2}=\tfrac{9}{4}a^{2}\Rightarrow m_{a}=\tfrac{3a}{2}.$

Conceptual True / False with Reasons

(i) Two triangles having two sides and one angle equal are always similar. False. Only with the angle included by the proportional sides (SAS) does similarity hold.
(ii) The ratio of the areas of two similar triangles is equal to the ratio of corresponding altitudes squared. True.
(iii) All right-angled triangles are similar. False. Only those with an additional matching acute angle are similar.
(iv) A line segment joining the mid-points of two sides of a triangle is half the third side. True (Mid-point theorem).
(v) Pythagoras theorem is valid for an obtuse-angled triangle. False. It applies only to right triangles.
(vi) The triangle formed by joining mid-points of the sides of $\triangle ABC$ has area one-fourth of $\triangle ABC.$ True.
(vii) If $\triangle ABC\sim\triangle DEF$, then perimeters are in the same ratio as corresponding sides. True.
(viii) A right triangle whose sides are 5, 12, 13 is similar to a right triangle whose sides are 10, 24, 26. True (each side doubled).

Higher-Order Thinking (HOTS) Problems

H1. In $\triangle ABC$, $\angle BAC=90^{\circ}$ and $AD\perp BC$ at $D.$ If $BD=5$ cm and $CD=12.8$ cm, find $AB,\ AC$ and $AD.$

Answer: $BC=BD+CD=17.8$ cm. $AD^{2}=BD\cdot CD=5\times 12.8=64\Rightarrow AD=8$ cm. $AB^{2}=BD\cdot BC=5\times 17.8=89\Rightarrow AB=\sqrt{89}\approx 9.43$ cm. $AC^{2}=CD\cdot BC=12.8\times 17.8=227.84\Rightarrow AC\approx 15.09$ cm. (Verify: $AB^{2}+AC^{2}=89+227.84=316.84=17.8^{2}.$)

H2. $ABCD$ is a parallelogram with $E$ the mid-point of $BC.$ $AE$ produced meets $DC$ produced at $F.$ Prove that $\triangle ABE\cong\triangle FCE$ and hence $AD=CF.$

Answer: In $\triangle ABE$ and $\triangle FCE$: $BE=CE$ (mid-point), $\angle AEB=\angle FEC$ (vertically opposite), $\angle ABE=\angle FCE$ (alternate angles, $AB\parallel CD$ since $ABCD$ is a parallelogram). By ASA, $\triangle ABE\cong\triangle FCE.$ So $AB=CF.$ But $AB=DC$ in parallelogram $\Rightarrow CF=DC.$ Hence $DF=2DC.$ Since $AD=BC$ in parallelogram and $E$ mid-point, by similar triangles $AD=CF.$

H3. Sides $AB,\ AC$ and $BC$ of $\triangle ABC$ are 6 cm, 7.5 cm and 4.5 cm respectively. Show that $\triangle ABC$ is right-angled.

Answer: $AB^{2}+BC^{2}=36+20.25=56.25=AC^{2}.$ By converse of Pythagoras, $\angle B=90^{\circ}.$

H4. $D$ and $E$ are points on the sides $AB$ and $AC$ of $\triangle ABC$ such that $DE\parallel BC$ and $\dfrac{AD}{DB}=\dfrac{3}{5}.$ If $AC=4.8$ cm, find $AE.$

Answer: $\dfrac{AE}{EC}=\dfrac{3}{5}\Rightarrow AE=\tfrac{3}{8}AC=\tfrac{3}{8}\times 4.8=1.8$ cm.

H5. If a square is inscribed in a right triangle with legs $a,b$ such that one side of the square lies on the hypotenuse, find the side of the square.

Answer: Let side $=s.$ Hypotenuse $c=\sqrt{a^{2}+b^{2}}.$ The altitude on $c$ is $h=\dfrac{ab}{c}.$ Using similar triangles between the small triangle on top and the whole triangle: $\dfrac{h-s}{h}=\dfrac{s}{c}\Rightarrow s=\dfrac{ch}{c+h}=\dfrac{ab\cdot c}{c\cdot c+ab}=\dfrac{abc}{a^{2}+b^{2}+ab}.$

H6. In $\triangle ABC$, point $D$ is on $BC$ with $BD:DC=2:3.$ If $\triangle ABD$ has area 16 cm², find the area of $\triangle ABC.$

Answer: Triangles $\triangle ABD$ and $\triangle ABC$ share the same altitude from $A$ to $BC.$ So $\dfrac{\text{ar}(\triangle ABD)}{\text{ar}(\triangle ABC)}=\dfrac{BD}{BC}=\dfrac{2}{5}.$ Hence $\text{ar}(\triangle ABC)=\dfrac{5}{2}\times 16=40$ cm².

H7. The bisector of $\angle B$ in $\triangle ABC$ meets $AC$ at $D.$ Given $AB=10$ cm, $BC=14$ cm, $AC=18$ cm. Find $AD$ and $DC.$

Answer: By angle-bisector theorem, $\dfrac{AD}{DC}=\dfrac{AB}{BC}=\dfrac{10}{14}=\dfrac{5}{7}.$ So $AD=\tfrac{5}{12}\times 18=7.5$ cm and $DC=10.5$ cm.

H8. Two similar triangles have areas $A_{1}$ and $A_{2}$ with $A_{1}:A_{2}=1:k.$ Show that the ratio of their corresponding altitudes is $1:\sqrt{k}.$

Answer: Ratio of corresponding sides $=\sqrt{A_{1}:A_{2}}=1:\sqrt k.$ Altitudes scale with sides, so altitudes are also in ratio $1:\sqrt k.$

H9. A vertical pole 8 m high stands on horizontal ground. From a point on the ground 6 m from the foot of the pole, a wire is stretched to the top of the pole. What is the length of the wire?

Answer: $L=\sqrt{8^{2}+6^{2}}=\sqrt{64+36}=\sqrt{100}=10$ m.

H10. In $\triangle ABC$, $D$ on $BC$ with $\angle BAD=\angle CAD$ (i.e., $AD$ bisects $\angle A$). Prove that $AB^{2}-AC^{2}=BD^{2}-DC^{2}\cdot\dfrac{AB^{2}+AC^{2}-BC^{2}}{…}$ — the Stewart-like relation. Sketch: Use the angle bisector length formula $AD^{2}=AB\cdot AC-BD\cdot DC.$

One-Mark Quick Questions

State whether all isosceles triangles are similar. No.
If two triangles are similar with ratio of sides $1:3$, what is the ratio of perimeters? $1:3.$
If $\triangle ABC\sim\triangle DEF$, $AB=4$ cm, $DE=8$ cm and $BC=6$ cm, then $EF=$ ? $12$ cm.
The hypotenuse of a right triangle is 17 cm and one leg is 15 cm. The other leg is? $8$ cm.
The diagonals of a rhombus measure 24 cm and 10 cm. Side $=$ ? $13$ cm.
If two angles of one triangle equal two angles of another, the triangles are __________ . Similar (AA).
The midpoint theorem is a special case of __________ . The Basic Proportionality Theorem.
The longest side of a right triangle is called the __________ . Hypotenuse.
Pythagorean triple $(8,15,?)$. $17.$
Pythagorean triple $(7,24,?)$. $25.$
Pythagorean triple $(20, 21,?)$. $29.$
If a $3:4:5$ triangle is similar to another with shortest side 9 cm, find the longest side. $15$ cm.

Glossary

Similar figures: Figures having the same shape but not necessarily the same size.
Congruent figures: Figures having the same shape and the same size.
Similar triangles: Triangles with equal corresponding angles and proportional corresponding sides.
BPT (Thales’ Theorem): A line parallel to one side of a triangle divides the other two sides in the same ratio.
Converse of BPT: If a line divides any two sides of a triangle in the same ratio, the line is parallel to the third side.
AAA / AA criterion: Equal corresponding angles imply similarity.
SSS criterion: Proportional corresponding sides imply similarity.
SAS criterion: One pair of equal angles and proportional including sides imply similarity.
Areas of similar triangles: Ratio of areas equals the square of the ratio of corresponding sides (or medians, altitudes, perimeters).
Pythagoras Theorem: In a right triangle, $\text{(hypotenuse)}^{2}=\text{(leg}_1)^{2}+\text{(leg}_2)^{2}.$
Converse of Pythagoras: If $a^{2}+b^{2}=c^{2}$ for sides of a triangle, the angle opposite $c$ is right.
Hypotenuse: The side opposite the right angle in a right triangle (longest side).
Median: Line segment from a vertex to the mid-point of the opposite side.
Altitude: Perpendicular from a vertex to the line containing the opposite side.
Angle Bisector Theorem: The internal bisector of an angle divides the opposite side in the ratio of the other two sides.

Common Pythagorean Triples (Useful Memorisation)

$(3, 4, 5)$ and multiples: $(6,8,10),\ (9,12,15),\ (12,16,20),\ (15,20,25),\ (18,24,30).$
$(5, 12, 13)$ and multiples: $(10,24,26),\ (15,36,39),\ (20,48,52).$
$(7, 24, 25)$ and multiples: $(14,48,50).$
$(8, 15, 17),\ (9, 40, 41),\ (11, 60, 61),\ (12, 35, 37),\ (20, 21, 29),\ (28, 45, 53).$

Verify any triple by checking $a^{2}+b^{2}=c^{2}.$ For example $(20, 21, 29)$: $20^{2}+21^{2}=400+441=841=29^{2}.$ ✓

Step-by-Step Strategy for Geometry Proofs

Read the question and identify the figure clearly. Mark all given information.
Draw a neat, well-labelled figure (use a sharp pencil and a ruler in the exam). Add construction lines (extensions, perpendiculars, parallel lines, joining points) where helpful.
Write “Given”, “To prove”, and “Construction” sections clearly.
Identify the relevant theorem (BPT, AA, SSS, SAS, Pythagoras, etc.) and the triangles involved.
Justify each step with a reason: “common angle”, “vertically opposite angles”, “alternate angles since lines are parallel”, “by AA”, “by BPT”, etc.
Conclude with the required result, ending with “Hence proved” or “Hence the result.”
Re-read your proof: every claim should follow from the previous step.

Examination Pattern Tips for Chapter 6

One 1-mark question (often a fill-in or true/false on a definition/criterion).
One 2-mark question (computation using BPT or Pythagoras).
One 3-mark question (similarity proof or area ratio).
One 4 / 5-mark question (proof of BPT, similarity criterion, or Pythagoras Theorem; or combined application).
Total weight typically 10 to 12 marks in the paper, making this chapter critical.

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