NCERT Exemplar Solutions Class 10 Maths Triangles — Exercise 6.1

NCERT Exemplar Problems, Class 10 Mathematics — Triangles, Exercise 6.1. All 12 multiple-choice questions solved.

Half of this exercise is about correspondence — which vertex of one triangle matches which vertex of the other — and the other half about two results, the area ratio and Pythagoras.

The correspondence is the part that is worth slowing down for. Writing $\triangle ABC \sim \triangle EDF$ is a statement that $A \leftrightarrow E$, $B \leftrightarrow D$, $C \leftrightarrow F$, in that order, and every proportion follows from it:

$$\frac{AB}{ED} = \frac{BC}{DF} = \frac{CA}{FE}$$

Four of the twelve questions here are wrong-answer traps built purely on reading the letters in the wrong order.

The two results used are:

$$\frac{\operatorname{ar}(\triangle_1)}{\operatorname{ar}(\triangle_2)} = \left(\frac{\text{corresponding side}_1}{\text{corresponding side}_2}\right)^2, \qquad \text{and Pythagoras’ theorem}$$

A note on the current syllabus. Both of those have been removed from the Class 10 textbook. The rationalised chapter runs Introduction → Similar Figures → Similarity of Triangles → Criteria for Similarity, and stops: it carries Theorems 6.1 to 6.5 (the Basic Proportionality Theorem, its converse, and AAA, SSS and SAS similarity) and no others. There is no “Areas of Similar Triangles” section and no Pythagoras section — the word “Pythagoras” appears once, in the Introduction, promising a proof the chapter no longer gives. So Questions 1, 2, 8, 11 and 12 here are exemplar-only material. They are still worth doing, and Pythagoras itself is assumed knowledge from Class 9.

Key insight. Never read a similarity statement as “these two triangles are similar”. Read it as a list of three vertex pairings, and write the pairings down. Every question in this exercise that looks subtle is straightforward once they are on paper.

Choose the correct answer from the given four options:

Question 1

In Fig. 6.2, $\angle BAC = 90^\circ$ and $AD \perp BC$. Then,

Solution. The altitude from the right angle creates two smaller triangles, each similar to the original and to each other. In $\triangle BDA$ and $\triangle ADC$:

• $\angle BDA = \angle ADC = 90^\circ$;
• $\angle DBA = 90^\circ – \angle BAD = \angle DAC$ (both are the complement of $\angle BAD$, since $\angle BAC = 90^\circ$).

So $\triangle BDA \sim \triangle ADC$ by AA, and matching corresponding sides:

$$\frac{BD}{AD} = \frac{AD}{CD} \implies AD^2 = BD \cdot CD$$

This is the geometric-mean relation: the altitude to the hypotenuse is the geometric mean of the two segments it cuts the hypotenuse into.

Options (A) and (B) both equate a product of two lengths to $BC^2$, and neither holds — the correct companion relations are $AB^2 = BD \cdot BC$ and $AC^2 = CD \cdot BC$.

Question 2

The lengths of the diagonals of a rhombus are $16$ cm and $12$ cm. Then, the length of the side of the rhombus is

Solution. The diagonals of a rhombus bisect each other at right angles, so they cut it into four congruent right triangles whose legs are the half-diagonals:

$$\frac{16}{2} = 8 \text{ cm} \qquad\text{and}\qquad \frac{12}{2} = 6 \text{ cm}$$

The side of the rhombus is the hypotenuse of one of them:

$$\text{side} = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10 \text{ cm}$$

Using the full diagonals instead gives $\sqrt{16^2 + 12^2} = 20$, which is option (D) — the trap, and the reason that option is there.

Question 3

If $\triangle ABC \sim \triangle EDF$ and $\triangle ABC$ is not similar to $\triangle DEF$, then which of the following is not true?

Solution. Write the correspondence out first: $\triangle ABC \sim \triangle EDF$ means

$$A \leftrightarrow E, \qquad B \leftrightarrow D, \qquad C \leftrightarrow F$$

so the proportion is

$$\frac{AB}{ED} = \frac{BC}{DF} = \frac{CA}{FE}$$

Cross-multiplying the three pairs gives exactly three true products:

• $\dfrac{AB}{ED} = \dfrac{BC}{DF} \implies AB \cdot DF = BC \cdot ED$ — this is option (D), true.
• $\dfrac{BC}{DF} = \dfrac{CA}{FE} \implies BC \cdot FE = CA \cdot DF$ — option (A), true.
• $\dfrac{AB}{ED} = \dfrac{CA}{FE} \implies AB \cdot FE = CA \cdot ED$ — option (B), true.

Option (C) says $BC \cdot DE = AB \cdot EF$. Comparing with (D), which says $BC \cdot DE = AB \cdot FD$, option (C) would require $EF = FD$ — true only in a special case, not in general. So (C) is the one that is not true.

Question 4

If in two triangles $ABC$ and $PQR$, $\ \dfrac{AB}{QR} = \dfrac{BC}{PR} = \dfrac{CA}{PQ}$, then

Solution. Read the pairings off the ratios. $AB$ pairs with $QR$, $BC$ with $PR$, $CA$ with $PQ$, so:

• $A$ and $B$ pair with $Q$ and $R$;
• $B$ and $C$ pair with $P$ and $R$;
• $C$ and $A$ pair with $P$ and $Q$.

$B$ appears in the first two, and the letter common to $\{Q, R\}$ and $\{P, R\}$ is $R$ — so $B \leftrightarrow R$. Similarly $A \leftrightarrow Q$ and $C \leftrightarrow P$.

Written in the first triangle’s order, $\triangle ABC \sim \triangle QRP$. Option (A) reverses it: $\triangle PQR \sim \triangle CAB$ says $P \leftrightarrow C$, $Q \leftrightarrow A$, $R \leftrightarrow B$ — the same three pairings ✓.

Question 5

In Fig. 6.3, two line segments $AC$ and $BD$ intersect each other at the point $P$ such that $PA = 6$ cm, $PB = 3$ cm, $PC = 2.5$ cm, $PD = 5$ cm, $\angle APB = 50^\circ$ and $\angle CDP = 30^\circ$. Then, $\angle PBA$ is equal to

Solution. Test the two triangles $APB$ and $DPC$ for similarity. The sides about $P$ are in the ratios

$$\frac{PA}{PD} = \frac{6}{5}, \qquad \frac{PB}{PC} = \frac{3}{2.5} = \frac{6}{5}$$

equal, and the included angles $\angle APB$ and $\angle DPC$ are vertically opposite, hence equal. So by SAS similarity

$$\triangle APB \sim \triangle DPC$$

and therefore $\angle PBA = \angle PCD$.

Now work in $\triangle DPC$, where two angles are known: $\angle DPC = \angle APB = 50^\circ$ (vertically opposite) and $\angle CDP = 30^\circ$. Hence

$$\angle PCD = 180^\circ – 50^\circ – 30^\circ = 100^\circ$$

so $\angle PBA = 100^\circ$.

The trap is answering $30^\circ$ — the angle corresponding to $\angle PBA$ is $\angle PCD$, not $\angle PDC$.

Question 6

If in two triangles $DEF$ and $PQR$, $\angle D = \angle Q$ and $\angle R = \angle E$, then which of the following is not true?

Solution. The two equal-angle statements give $D \leftrightarrow Q$ and $E \leftrightarrow R$, and the third pairing follows: $F \leftrightarrow P$. So

$$\triangle DEF \sim \triangle QRP \implies \frac{DE}{QR} = \frac{EF}{RP} = \frac{FD}{PQ}$$

Check each option against that chain:

• (A) $\frac{EF}{PR} = \frac{DF}{PQ}$ — the second and third ratios ✓
• (C) $\frac{DE}{QR} = \frac{DF}{PQ}$ — the first and third ✓
• (D) $\frac{EF}{RP} = \frac{DE}{QR}$ — the second and first ✓
• (B) $\frac{DE}{PQ} = \frac{EF}{RP}$ — $DE$ is paired with $QR$, not with $PQ$. ✗

Question 7

In triangles $ABC$ and $DEF$, $\angle B = \angle E$, $\angle F = \angle C$ and $AB = 3\,DE$. Then, the two triangles are

Solution. Two pairs of equal angles give AA similarity, so the triangles are certainly similar ($\triangle ABC \sim \triangle DEF$, since $B \leftrightarrow E$ and $C \leftrightarrow F$ force $A \leftrightarrow D$).

Congruence would additionally need the ratio of corresponding sides to be $1$. Here

$$\frac{AB}{DE} = 3 \ne 1$$

so one triangle is three times the size of the other — similar, not congruent.

Option (A) is impossible in principle: congruent triangles are always similar, with ratio $1$.

Question 8

It is given that $\triangle ABC \sim \triangle PQR$, with $\dfrac{BC}{QR} = \dfrac13$. Then, $\dfrac{\operatorname{ar}(PRQ)}{\operatorname{ar}(BCA)}$ is equal to

Solution. Areas of similar triangles are in the ratio of the squares of corresponding sides. Note carefully which way round the question asks: $\operatorname{ar}(PRQ)$ is the area of $\triangle PQR$ (the order of letters does not change a region), and it is on top. So

$$\frac{\operatorname{ar}(PRQ)}{\operatorname{ar}(BCA)} = \left(\frac{QR}{BC}\right)^2 = 3^2 = 9$$

Option (D) is the same calculation with the ratio inverted, and is what you get by carrying $\frac{BC}{QR}$ straight through without noticing that the areas are the other way round.

Question 9

It is given that $\triangle ABC \sim \triangle DFE$, $\angle A = 30^\circ$, $\angle C = 50^\circ$, $AB = 5$ cm, $AC = 8$ cm and $DF = 7.5$ cm. Then, the following is true:

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

$$\frac{AB}{DF} = \frac{BC}{FE} = \frac{CA}{ED}$$

The angles. $\angle B = 180^\circ – 30^\circ – 50^\circ = 100^\circ$, and $\angle F = \angle B = 100^\circ$. Also $\angle D = \angle A = 30^\circ$ and $\angle E = \angle C = 50^\circ$.

The side. $\frac{AB}{DF} = \frac{5}{7.5} = \frac23$, so

$$\frac{CA}{ED} = \frac23 \implies ED = \frac32 \times 8 = 12 \text{ cm}$$

So $DE = 12$ cm and $\angle F = 100^\circ$ — option (B). Option (A) has the right length with $\angle F$ taken as $\angle C$ rather than $\angle B$.

Question 10

If in triangles $ABC$ and $DEF$, $\ \dfrac{AB}{DE} = \dfrac{BC}{FD}$, then they will be similar, when

Solution. This is the SAS similarity criterion, and the word that matters in it is included: the equal angle must be the one lying between the two sides in each triangle.

• In $\triangle ABC$, the sides $AB$ and $BC$ meet at $B$, so the included angle is $\angle B$.
• In $\triangle DEF$, the sides $DE$ and $FD$ meet at $D$, so the included angle is $\angle D$.

Hence the condition is $\angle B = \angle D$.

Option (A), $\angle B = \angle E$, is the trap for anyone matching $B$ with $E$ because they sit in the same position in the letter order — but $E$ is not where $DE$ and $FD$ meet.

Question 11

If $\triangle ABC \sim \triangle QRP$, $\ \dfrac{\operatorname{ar}(ABC)}{\operatorname{ar}(PQR)} = \dfrac94$, $AB = 18$ cm and $BC = 15$ cm, then $PR$ is equal to

Solution. The correspondence is $A \leftrightarrow Q$, $B \leftrightarrow R$, $C \leftrightarrow P$, so $AB \leftrightarrow QR$ and $BC \leftrightarrow RP$.

The area ratio gives the side ratio by taking the square root:

$$\left(\frac{AB}{QR}\right)^2 = \frac94 \implies \frac{AB}{QR} = \frac32$$

Since the triangles are similar, every pair of corresponding sides is in that ratio, so

$$\frac{BC}{RP} = \frac32 \implies RP = \frac23 \times 15 = 10 \text{ cm}$$

and $PR$ is the same segment as $RP$.

Option (B), $12$ cm, is $QR = \frac23 \times 18$ — the right method applied to the wrong side.

Question 12

If $S$ is a point on side $PQ$ of a $\triangle PQR$ such that $PS = QS = RS$, then

Solution. $PS = QS$ makes $S$ the midpoint of $PQ$, and $RS$ equal to both means $R$ is the same distance from $S$ as $P$ and $Q$ are. So $P$, $Q$ and $R$ all lie on a circle centred at $S$ — whose diameter is $PQ$.

An angle in a semicircle is a right angle, so $\angle PRQ = 90^\circ$, and Pythagoras in $\triangle PQR$ gives

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

Without the circle: $\triangle PSR$ is isosceles, so $\angle SPR = \angle SRP$; likewise $\angle SQR = \angle SRQ$. Their sum is the angle sum of $\triangle PQR$:

$$2\angle SRP + 2\angle SRQ = 180^\circ \implies \angle PRQ = \angle SRP + \angle SRQ = 90^\circ$$

Common mistakes

• Reading a similarity statement by position rather than by pairing. $\triangle ABC \sim \triangle EDF$ pairs $B$ with $D$, not with $D$’s position. Questions 3, 6, 9, 10 and 11 all turn on this; writing the three pairings down before touching a ratio is the fix.
• Using the full diagonals in Question 2. The right triangle inside a rhombus has the half-diagonals as legs. Option (D), $20$ cm, is what the full ones give.
• Inverting the area ratio in Question 8. The question puts $\operatorname{ar}(PRQ)$ on top while giving $\frac{BC}{QR}$, so the ratio has to be turned over before squaring.
• Forgetting to square-root in Question 11. $\frac94$ is an area ratio; the side ratio is $\frac32$.
• Taking the wrong included angle in Question 10. $DE$ and $FD$ meet at $D$, so the included angle is $\angle D$ — regardless of where $D$ sits in the name of the triangle.
• Answering $30^\circ$ in Question 5. $\angle PBA$ corresponds to $\angle PCD$, which has to be found from the angle sum; $\angle PDC = 30^\circ$ corresponds to $\angle PAB$ instead.
• Thinking congruent and similar are alternatives in Question 7. Congruence is similarity with ratio $1$, so option (A) describes something that cannot exist.

Practise next

• Exemplar Exercise 6.2 — the same criteria as reasoning questions, where the justification is the answer.
• Exemplar Exercise 6.3 for the computational versions, and Exercise 6.4 for the long proofs and the ladder and shadow problems.
• NCERT textbook Exercise 6.3 — the similarity criteria as the book sets them, which is what Questions 3 to 11 above assume.
• NCERT textbook Exercise 6.2 for the Basic Proportionality Theorem, and Exercise 6.1 for what similarity of figures means.