NCERT Exemplar Solutions Class 10 Maths Triangles — Exercise 6.3

NCERT Exemplar Problems, Class 10 Mathematics — Triangles, Exercise 6.3. All 15 questions solved.

This exercise splits into four kinds of question, and knowing which kind you are looking at is most of the work.

• Proportionality (Questions 2, 5, 9) — a line parallel to one side of a triangle, so the Basic Proportionality Theorem or its converse applies.
• Area ratios (Questions 4, 8, 10, 12) — the ratio of areas of similar triangles is the square of the ratio of corresponding sides, and every one of these questions is that fact used in one direction or the other.
• Pythagoras and the altitude relations (Questions 1, 11, 15) — including the geometric-mean relation $h^2 = pq$ for the altitude to the hypotenuse.
• Modelling (Questions 14, 15) — shadows and ladders, which reduce to one similar-triangle proportion or one right triangle.

A note on the current syllabus. The area-ratio theorem and Pythagoras’ theorem are both gone from the rationalised Class 10 textbook, whose Triangles chapter ends at the similarity criteria. Questions 1, 4, 8, 10, 11, 12 and 15 therefore rest on material the book no longer proves, though Pythagoras is assumed from Class 9. The proportionality and similarity questions are all on syllabus.

Key insight. In an area question, decide first whether you are going from lengths to areas (square the ratio) or the other way (square-root it). Half the errors in this exercise are the right calculation performed in the wrong direction.

Question 1

In a $\triangle PQR$, $PR^2 – PQ^2 = QR^2$ and $M$ is a point on side $PR$ such that $QM \perp PR$. Prove that $QM^2 = PM \times MR$.

Solution. The given relation rearranges to

$$PR^2 = PQ^2 + QR^2$$

so by the converse of Pythagoras’ theorem, $\triangle PQR$ is right-angled at $Q$, with $PR$ as its hypotenuse. $QM$ is then the altitude from the right angle to the hypotenuse, and what has to be proved is the geometric-mean relation.

In $\triangle PMQ$ and $\triangle QMR$:

• $\angle PMQ = \angle QMR = 90^\circ$;
• $\angle MPQ = 90^\circ – \angle PQM = \angle MQR$, since $\angle PQM + \angle MQR = \angle PQR = 90^\circ$.

So $\triangle PMQ \sim \triangle QMR$ by AA, and matching corresponding sides,

$$\frac{PM}{QM} = \frac{QM}{MR} \implies QM^2 = PM \times MR$$

Question 2

Find the value of $x$ for which $DE \parallel AB$ in Fig. 6.8.

Solution. $DE \parallel AB$ holds exactly when $DE$ divides the two sides of $\triangle CAB$ in the same ratio — the converse of the Basic Proportionality Theorem:

$$\frac{CD}{DA} = \frac{CE}{EB} \implies \frac{x+3}{3x+19} = \frac{x}{3x+4}$$

Cross-multiplying,

$$(x+3)(3x+4) = x(3x+19)$$ $$3x^2 + 13x + 12 = 3x^2 + 19x$$ $$12 = 6x \implies x = 2$$

Check: $CD = 5$, $DA = 25$, so $\frac{CD}{DA} = \frac15$; and $CE = 2$, $EB = 10$, so $\frac{CE}{EB} = \frac15$ ✓.

Question 3

In Fig. 6.9, if $\angle 1 = \angle 2$ and $\triangle NSQ \cong \triangle MTR$, then prove that $\triangle PTS \sim \triangle PRQ$.

The figure shows $\triangle PQR$ with $S$ on $PQ$ and $T$ on $PR$; $\angle 1$ is the angle $PST$ at $S$ and $\angle 2$ the angle $PTS$ at $T$. The points $M$, $N$ lie outside on a line through $Q$ and $R$, with $\triangle NSQ$ and $\triangle MTR$ the two congruent triangles named.

Solution. The congruence $\triangle NSQ \cong \triangle MTR$ gives, in particular,

$$\angle SQN = \angle TRM$$

Since $\angle SQN$ and $\angle PQR$ are the same angle, and $\angle TRM$ and $\angle PRQ$ are the same angle, this says

$$\angle PQR = \angle PRQ \tag{1}$$

Meanwhile $\angle 1 = \angle 2$ means $\angle PST = \angle PTS$, so $\triangle PST$ is isosceles with

$$PS = PT \tag{2}$$

and from (1), $\triangle PQR$ is isosceles with

$$PQ = PR \tag{3}$$

Dividing (2) by (3),

$$\frac{PS}{PQ} = \frac{PT}{PR}$$

Now in $\triangle PTS$ and $\triangle PRQ$: the angle at $P$ is common, and the sides including it are in the same ratio — $\frac{PT}{PR} = \frac{PS}{PQ}$. By SAS similarity,

$$\triangle PTS \sim \triangle PRQ$$

Question 4

Diagonals of a trapezium $PQRS$ intersect each other at the point $O$, $PQ \parallel RS$ and $PQ = 3\,RS$. Find the ratio of the areas of triangles $POQ$ and $ROS$.

Solution. Because $PQ \parallel RS$, the transversal $PR$ gives $\angle OPQ = \angle ORS$ (alternate angles), and $QS$ gives $\angle OQP = \angle OSR$. With two pairs of equal angles,

$$\triangle POQ \sim \triangle ROS$$

The corresponding sides here are $PQ$ and $RS$, in the ratio $3 : 1$. Areas are in the ratio of the squares:

$$\frac{\operatorname{ar}(\triangle POQ)}{\operatorname{ar}(\triangle ROS)} = \left(\frac{PQ}{RS}\right)^2 = 3^2 = 9$$

Question 5

In Fig. 6.10, if $AB \parallel DC$ and $AC$ and $PQ$ intersect each other at the point $O$, prove that $OA \cdot CQ = OC \cdot AP$.

The figure is a trapezium $ABCD$ with $AB \parallel DC$; $P$ lies on $AB$ and $Q$ on $DC$, and the segments $AC$ and $PQ$ cross at $O$.

Solution. Look at $\triangle AOP$ and $\triangle COQ$.

Since $AB \parallel DC$ and $AC$ is a transversal,

$$\angle OAP = \angle OCQ \qquad \text{(alternate angles)}$$

and the angles at $O$ are vertically opposite:

$$\angle AOP = \angle COQ$$

Two pairs of equal angles give $\triangle AOP \sim \triangle COQ$ by AA, so corresponding sides are proportional:

$$\frac{OA}{OC} = \frac{AP}{CQ}$$

Cross-multiplying gives exactly what was asked:

$$OA \cdot CQ = OC \cdot AP$$

Question 6

Find the altitude of an equilateral triangle of side $8$ cm.

Solution. The altitude of an equilateral triangle also bisects the base, so it splits the triangle into two right triangles with hypotenuse $8$ cm and one leg $4$ cm. By Pythagoras,

$$h = \sqrt{8^2 – 4^2} = \sqrt{64 – 16} = \sqrt{48} = 4\sqrt3 \text{ cm}$$

Equivalently, the standard result $h = \dfrac{\sqrt3}{2}\,a$ gives $\dfrac{\sqrt3}{2} \times 8 = 4\sqrt3$.

Question 7

If $\triangle ABC \sim \triangle DEF$, $AB = 4$ cm, $DE = 6$ cm, $EF = 9$ cm and $FD = 12$ cm, find the perimeter of $\triangle ABC$.

Solution. The correspondence $A \leftrightarrow D$, $B \leftrightarrow E$, $C \leftrightarrow F$ gives

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

The known pair fixes the ratio:

$$\frac{AB}{DE} = \frac46 = \frac23$$

so

$$BC = \frac23 \times 9 = 6 \text{ cm}, \qquad CA = \frac23 \times 12 = 8 \text{ cm}$$

$$\text{Perimeter of } \triangle ABC = 4 + 6 + 8 = 18 \text{ cm}$$

A shortcut worth knowing: perimeters of similar triangles are in the same ratio as the sides, so the perimeter could be found directly as $\frac23 \times (6 + 9 + 12) = \frac23 \times 27 = 18$.

Question 8

In Fig. 6.11, if $DE \parallel BC$, find the ratio of $\operatorname{ar}(ADE)$ and $\operatorname{ar}(DECB)$.

Solution. $DE \parallel BC$ gives $\angle ADE = \angle ABC$ (corresponding angles) and $\angle A$ is common, so

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

with ratio $\dfrac{DE}{BC} = \dfrac{6}{12} = \dfrac12$. Hence

$$\frac{\operatorname{ar}(ADE)}{\operatorname{ar}(ABC)} = \left(\frac12\right)^2 = \frac14$$

The quadrilateral $DECB$ is what is left over when $\triangle ADE$ is removed from $\triangle ABC$. Taking $\operatorname{ar}(ADE) = 1$ unit and $\operatorname{ar}(ABC) = 4$ units,

$$\operatorname{ar}(DECB) = 4 – 1 = 3$$

$$\operatorname{ar}(ADE) : \operatorname{ar}(DECB) = 1 : 3$$

The step that is easy to miss is the last one: the question asks for the ratio to the trapezium, not to the whole triangle, so $\frac14$ is not the answer.

Question 9

$ABCD$ is a trapezium in which $AB \parallel DC$ and $P$ and $Q$ are points on $AD$ and $BC$, respectively, such that $PQ \parallel DC$. If $PD = 18$ cm, $BQ = 35$ cm and $QC = 15$ cm, find $AD$.

Solution. With $AB \parallel PQ \parallel DC$, the two non-parallel sides are cut in the same ratio:

$$\frac{AP}{PD} = \frac{BQ}{QC}$$

(To see it, join $AC$ meeting $PQ$ at $O$: in $\triangle ADC$, $PO \parallel DC$ gives $\frac{AP}{PD} = \frac{AO}{OC}$; in $\triangle CAB$, $OQ \parallel AB$ gives $\frac{AO}{OC} = \frac{BQ}{QC}$.)

Substituting the given lengths,

$$\frac{AP}{18} = \frac{35}{15} = \frac73 \implies AP = 18 \times \frac73 = 42 \text{ cm}$$

$$AD = AP + PD = 42 + 18 = 60 \text{ cm}$$

Question 10

Corresponding sides of two similar triangles are in the ratio of $2 : 3$. If the area of the smaller triangle is $48$ cm², find the area of the larger triangle.

Solution. Squaring the side ratio gives the area ratio:

$$\frac{\operatorname{ar}(\text{smaller})}{\operatorname{ar}(\text{larger})} = \left(\frac23\right)^2 = \frac49$$

$$\frac{48}{\operatorname{ar}(\text{larger})} = \frac49 \implies \operatorname{ar}(\text{larger}) = 48 \times \frac94 = 108 \text{ cm}^2$$

Question 11

In a triangle $PQR$, $N$ is a point on $PR$ such that $QN \perp PR$. If $PN \cdot NR = QN^2$, prove that $\angle PQR = 90^\circ$.

Solution. This is the converse of Question 1, and the same pair of triangles does it.

In $\triangle QNP$ and $\triangle RNQ$:

• $\angle QNP = \angle RNQ = 90^\circ$;
• the given relation rearranges to $\dfrac{PN}{QN} = \dfrac{QN}{NR}$ — the sides about those right angles are proportional.

So by SAS similarity, $\triangle QNP \sim \triangle RNQ$, and therefore

$$\angle PQN = \angle QRN \qquad\text{and}\qquad \angle QPN = \angle RQN$$

Now use the angle sum in $\triangle QNP$:

$$\angle QPN + \angle PQN = 90^\circ$$

and substitute $\angle QPN = \angle RQN$:

$$\angle RQN + \angle PQN = 90^\circ$$

But $\angle RQN + \angle PQN$ is exactly $\angle PQR$, since $N$ lies on $PR$ and $QN$ splits the angle at $Q$. Hence $\angle PQR = 90^\circ$.

Question 12

Areas of two similar triangles are $36$ cm² and $100$ cm². If the length of a side of the larger triangle is $20$ cm, find the length of the corresponding side of the smaller triangle.

Solution. Here the areas are given and a length is wanted, so square-root the ratio:

$$\left(\frac{\text{smaller side}}{20}\right)^2 = \frac{36}{100} \implies \frac{\text{smaller side}}{20} = \frac{6}{10} = \frac35$$

$$\text{smaller side} = \frac35 \times 20 = 12 \text{ cm}$$

Question 13

In Fig. 6.12, if $\angle ACB = \angle CDA$, $AC = 8$ cm and $AD = 3$ cm, find $BD$.

Solution. Compare $\triangle ACB$ and $\triangle ADC$:

• $\angle A$ is common to both;
• $\angle ACB = \angle ADC$ is given.

So $\triangle ACB \sim \triangle ADC$ by AA, and the correspondence is $A \leftrightarrow A$, $C \leftrightarrow D$, $B \leftrightarrow C$. Hence

$$\frac{AC}{AD} = \frac{AB}{AC} \implies AC^2 = AD \times AB$$

$$64 = 3 \times AB \implies AB = \frac{64}{3} \text{ cm}$$

Finally $D$ lies on $AB$, so

$$BD = AB – AD = \frac{64}{3} – 3 = \frac{64 – 9}{3} = \frac{55}{3} \text{ cm}$$

Question 14

A $15$ metres high tower casts a shadow $24$ metres long at a certain time and at the same time, a telephone pole casts a shadow $16$ metres long. Find the height of the telephone pole.

Solution. “At the same time” means the sun’s rays make the same angle with the ground in both cases, so the two object-and-shadow right triangles are similar by AA. Corresponding sides are therefore in the same ratio:

$$\frac{\text{height of tower}}{\text{its shadow}} = \frac{\text{height of pole}}{\text{its shadow}}$$

$$\frac{15}{24} = \frac{h}{16} \implies h = \frac{15 \times 16}{24} = 10 \text{ m}$$

Question 15

Foot of a $10$ m long ladder leaning against a vertical wall is $6$ m away from the base of the wall. Find the height of the point on the wall where the top of the ladder reaches.

Solution. The wall is vertical and the ground horizontal, so the ladder, the wall and the ground form a right triangle with the ladder as hypotenuse. By Pythagoras,

$$h = \sqrt{10^2 – 6^2} = \sqrt{100 – 36} = \sqrt{64} = 8 \text{ m}$$

(This is the $6$–$8$–$10$ triangle, which is the $3$–$4$–$5$ triangle doubled.)

Common mistakes

• Answering $1 : 4$ in Question 8. That is $\operatorname{ar}(ADE) : \operatorname{ar}(ABC)$. The question asks for the ratio to the trapezium $DECB$, which is the difference, so the answer is $1 : 3$.
• Squaring when you should square-root. Question 10 goes from lengths to areas (square); Question 12 goes from areas to a length (square-root). Deciding the direction before calculating prevents both errors.
• Comparing $\dfrac{AP}{AD}$ with $\dfrac{BQ}{QC}$ in Question 9. The proportionality is between the two pieces of each side: $\dfrac{AP}{PD} = \dfrac{BQ}{QC}$. And the answer wanted is $AD = AP + PD$, not $AP$.
• Getting the correspondence backwards in Question 13. $\angle ACB = \angle ADC$ with the common $\angle A$ gives $\triangle ACB \sim \triangle ADC$ — so $AC$ is the geometric mean of $AD$ and $AB$, not of $AD$ and $BD$.
• Forgetting that the altitude bisects the base in Question 6. The right triangle has legs $4$ and $h$, not $8$ and $h$.
• Using the full ladder length as a height in Question 15. The ladder is the hypotenuse; the height is the other leg.
• Proving only similarity in Questions 1 and 11. Both ask for a specific relation, and the last line — cross-multiplying the proportion, or adding the two angles at $Q$ — is where the marks are.

Practise next

• Exemplar Exercise 6.4 — the long-answer versions, including more shadow and ladder problems and the proofs of the Basic Proportionality Theorem itself.
• Exemplar Exercise 6.2 for the reasoning behind the criteria used here.
• NCERT textbook Exercise 6.3 — the similarity criteria applied to find lengths, the closest match to Questions 7, 13 and 14.
• NCERT textbook Exercise 6.2 for the Basic Proportionality Theorem, used in Questions 2, 5 and 9.