NCERT Exemplar Solutions Class 10 Maths Triangles — Exercise 6.2

NCERT Exemplar Problems, Class 10 Mathematics — Triangles, Exercise 6.2. All 12 questions solved.

Every question here asks a yes/no question and then asks why, and the “why” is always the name of a criterion plus the reason it applies. The criteria available are:

• AA — two pairs of equal angles;
• SSS — all three pairs of sides proportional;
• SAS — two pairs of sides proportional and the included angles equal;
• the Basic Proportionality Theorem and its converse, for a line parallel to one side.

Three of the twelve questions are false for the same structural reason, and it is worth naming in advance: a criterion with one condition weakened is not a criterion. Equal angles without proportional sides fails for quadrilaterals (Question 6); proportional sides with a non-included equal angle fails for triangles (Question 12); and the right similarity statement written with the vertices in the wrong order is simply a different, false statement (Questions 2 and 5).

Key insight. When a similarity claim is false, the useful answer is not “no” but “no — and here is the correct statement”. Questions 2 and 5 both have a true version one letter-swap away, and giving it is what shows you understand the correspondence rather than having guessed.

Question 1

Is the triangle with sides $25$ cm, $5$ cm and $24$ cm a right triangle? Give reasons for your answer.

Solution. No. By the converse of Pythagoras’ theorem, a triangle is right-angled exactly when the square of the longest side equals the sum of the squares of the other two. Here the longest side is $25$ cm, and

$$25^2 = 625, \qquad 5^2 + 24^2 = 25 + 576 = 601$$

Since $625 \ne 601$, the triangle is not right-angled.

The numbers are chosen to look like the familiar triple $7$, $24$, $25$ — for which $49 + 576 = 625$ works exactly. Replacing $7$ by $5$ breaks it, and the shortfall of $24$ is small enough to be missed if the squares are not actually computed.

Question 2

It is given that $\triangle DEF \sim \triangle RPQ$. Is it true to say that $\angle D = \angle R$ and $\angle F = \angle P$? Why?

Solution. Partly — the first is true and the second is false, so the statement as a whole is not true.

The similarity $\triangle DEF \sim \triangle RPQ$ pairs the vertices in order:

$$D \leftrightarrow R, \qquad E \leftrightarrow P, \qquad F \leftrightarrow Q$$

So $\angle D = \angle R$ ✓, but $\angle F = \angle Q$, not $\angle P$. The correct companion statement is $\angle E = \angle P$.

Question 3

$A$ and $B$ are respectively the points on the sides $PQ$ and $PR$ of a triangle $PQR$ such that $PQ = 12.5$ cm, $PA = 5$ cm, $BR = 6$ cm and $PB = 4$ cm. Is $AB \parallel QR$? Give reasons for your answer.

Solution. Yes. First get the piece of $PQ$ that the question does not give directly:

$$AQ = PQ – PA = 12.5 – 5 = 7.5 \text{ cm}$$

Now compare the two ratios that the converse of the Basic Proportionality Theorem needs:

$$\frac{PA}{AQ} = \frac{5}{7.5} = \frac23, \qquad \frac{PB}{BR} = \frac46 = \frac23$$

They are equal, so $AB$ divides $PQ$ and $PR$ in the same ratio, and by the converse of the BPT, $AB \parallel QR$.

The one thing to watch is that the ratio must be $\frac{\text{near piece}}{\text{far piece}}$ in both — mixing $\frac{PA}{PQ}$ with $\frac{PB}{BR}$ compares different things.

Question 4

In Fig. 6.4, $BD$ and $CE$ intersect each other at the point $P$. Is $\triangle PBC \sim \triangle PDE$? Why?

Solution. Yes, by the SAS similarity criterion. Compare the sides about the point of intersection:

$$\frac{PB}{PD} = \frac{5}{10} = \frac12, \qquad \frac{PC}{PE} = \frac{6}{12} = \frac12$$

and the angles between those sides, $\angle BPC$ and $\angle DPE$, are vertically opposite, hence equal.

Two pairs of proportional sides with the included angles equal is exactly SAS similarity, so

$$\triangle PBC \sim \triangle PDE$$

Question 5

In triangles $PQR$ and $MST$, $\angle P = 55^\circ$, $\angle Q = 25^\circ$, $\angle M = 100^\circ$ and $\angle S = 25^\circ$. Is $\triangle QPR \sim \triangle TSM$? Why?

Solution. No — although the two triangles are similar, the correspondence written is the wrong one.

Fill in the missing angles first:

$$\angle R = 180^\circ – 55^\circ – 25^\circ = 100^\circ, \qquad \angle T = 180^\circ – 100^\circ – 25^\circ = 55^\circ$$

So the angles are $P = 55^\circ$, $Q = 25^\circ$, $R = 100^\circ$ and $M = 100^\circ$, $S = 25^\circ$, $T = 55^\circ$. Matching equal angles:

$$P \leftrightarrow T, \qquad Q \leftrightarrow S, \qquad R \leftrightarrow M$$

The statement $\triangle QPR \sim \triangle TSM$ claims $Q \leftrightarrow T$, but $\angle Q = 25^\circ$ and $\angle T = 55^\circ$. The correct version, in the same starting order, is

$$\triangle QPR \sim \triangle STM$$

Question 6

Is the following statement true? Why? “Two quadrilaterals are similar, if their corresponding angles are equal.”

Solution. Not true. For polygons with more than three sides, equal angles are not enough — the corresponding sides must also be in the same ratio.

A single example settles it: a square and a rectangle that is not a square. Every angle of each is $90^\circ$, so all four pairs of corresponding angles are equal; but the sides of the square are in the ratio $1 : 1$ while the rectangle’s are, say, $2 : 1$. They are plainly not the same shape.

Triangles are the exception — for them, equal angles do force proportional sides, which is why AA similarity exists for triangles and has no analogue for quadrilaterals. The extra freedom a fourth vertex brings is exactly what breaks it.

Question 7

Two sides and the perimeter of one triangle are respectively three times the corresponding sides and the perimeter of the other triangle. Are the two triangles similar? Why?

Solution. Yes, by SSS similarity — and the point is that the third pair of sides is forced, not assumed.

Let the smaller triangle have sides $a$, $b$, $c$. We are told the larger has sides $3a$ and $3b$ — call its third side $z$ — and that its perimeter is three times the smaller one’s:

$$3a + 3b + z = 3(a + b + c)$$

The $3a$ and $3b$ cancel, leaving

$$z = 3c$$

So all three pairs of sides are in the ratio $3 : 1$, and the triangles are similar by SSS.

Question 8

If in two right triangles, one of the acute angles of one triangle is equal to an acute angle of the other triangle, can you say that the two triangles will be similar? Why?

Solution. Yes. Each triangle already has a right angle, so that is one pair of equal angles for free; the given equal acute angles are a second pair. Two pairs of equal angles is the AA criterion, so the triangles are similar.

(The third pair then agrees automatically, since the angles of each triangle sum to $180^\circ$ — which is why AA is enough and AAA is never needed as a separate test.)

Question 9

The ratio of the corresponding altitudes of two similar triangles is $\dfrac35$. Is it correct to say that the ratio of their areas is $\dfrac65$? Why?

Solution. No. In similar triangles every pair of corresponding lengths — sides, altitudes, medians, perimeters — is in the same ratio, and the areas are in the ratio of the squares of that:

$$\frac{\operatorname{ar}_1}{\operatorname{ar}_2} = \left(\frac35\right)^2 = \frac{9}{25}$$

The proposed $\frac65$ comes from doubling the numerator instead of squaring the fraction — $\frac{3 \times 2}{5}$ rather than $\frac{3^2}{5^2}$. Areas scale as the square of a length in two dimensions, so both numerator and denominator must be squared.

Question 10

$D$ is a point on side $QR$ of $\triangle PQR$ such that $PD \perp QR$. Will it be correct to say that $\triangle PQD \sim \triangle RPD$? Why?

Solution. No, not in general. The two triangles share the right angle at $D$, so a single further pair of equal angles would settle it — but nothing in the question supplies one.

The similarity $\triangle PQD \sim \triangle RPD$ would require $\angle QPD = \angle PRD$. Adding $\angle DPR$ to both sides, that says

$$\angle QPR = \angle PRD + \angle DPR = 180^\circ – 90^\circ = 90^\circ$$

using the angle sum of $\triangle PDR$. So the claim holds exactly when $\angle P = 90^\circ$ — the configuration of Question 1 of Exercise 6.1. For a general triangle $PQR$ it is false.

Question 11

In Fig. 6.5, if $\angle D = \angle C$, then is it true that $\triangle ADE \sim \triangle ACB$? Why?

A note on the book. The exemplar refers to Fig. 6.5 here but does not print it — the space beside this question, which the indented text of Question 12 leaves room for, is blank, and no figure numbered 6.5 appears anywhere in the unit. The configuration is nevertheless fixed by the statement itself, and the figure below is drawn from it: for $\triangle ADE \sim \triangle ACB$ to mean anything, the correspondence must be $A \leftrightarrow A$, $D \leftrightarrow C$, $E \leftrightarrow B$, which places $D$ on $AC$ and $E$ on $AB$, with $DE$ joining them.

Solution. Yes, by the AA criterion.

In $\triangle ADE$ and $\triangle ACB$:

• $\angle A = \angle A$ — the two triangles share the angle at $A$;
• $\angle ADE = \angle ACB$ — this is the given $\angle D = \angle C$.

Two pairs of equal angles, so

$$\triangle ADE \sim \triangle ACB$$

and consequently $\dfrac{AD}{AC} = \dfrac{AE}{AB} = \dfrac{DE}{CB}$.

The order of the letters is doing real work in this question: the claim is not $\triangle ADE \sim \triangle ABC$. $D$ pairs with $C$ and $E$ with $B$, which is why the given angle equality is the one that helps.

Question 12

Is it true to say that if in two triangles, an angle of one triangle is equal to an angle of another triangle and two sides of one triangle are proportional to the two sides of the other triangle, then the triangles are similar? Give reasons for your answer.

Solution. Not true as stated, because the word included is missing.

The SAS similarity criterion requires the equal angles to be the ones lying between the two pairs of proportional sides. If the equal angle is not the included one, the conclusion can fail — this is the similarity version of the familiar fact that “SSA” is not a congruence criterion.

Concretely: take $\triangle ABC$ with $AB = 4$, $BC = 6$ and $\angle A = 40^\circ$, and $\triangle DEF$ with $DE = 8$, $EF = 12$ and $\angle D = 40^\circ$. The sides are proportional ($1 : 2$) and an angle of each is $40^\circ$ — but $\angle A$ lies between $AB$ and $AC$, not between $AB$ and $BC$, so nothing forces the triangles to have the same shape.

State it correctly and it is a theorem: if one angle of a triangle is equal to one angle of another and the sides including those angles are proportional, the triangles are similar.

Common mistakes

• Saying only “no” to Questions 2 and 5. Both have a correct version a letter-swap away, and writing it is what demonstrates the correspondence has been understood.
• Comparing $\dfrac{PA}{PQ}$ with $\dfrac{PB}{BR}$ in Question 3. The converse of the BPT compares the two pieces of each side, so it is $\dfrac{PA}{AQ}$ against $\dfrac{PB}{BR}$. Computing $AQ = 7.5$ first removes the temptation.
• Doubling instead of squaring in Question 9. $\left(\frac35\right)^2 = \frac{9}{25}$. The offered $\frac65$ is a length ratio scaled the wrong way.
• Assuming the right angle at $D$ is enough in Question 10. One pair of equal angles never gives similarity; the claim holds only in the special case $\angle P = 90^\circ$.
• Forgetting the perimeter forces the third side in Question 7. Without that step the answer is “only two sides are known to be proportional”, which is not SSS. One line of algebra supplies the third.
• Dropping “included” in Question 12. It is the whole content of the question, and the same word is what Question 10 of Exercise 6.1 tests from the other direction.
• Recognising $5$, $24$, $25$ as a Pythagorean triple in Question 1. The triple is $7$, $24$, $25$. Squaring the numbers takes five seconds and settles it.

Practise next

• Exemplar Exercise 6.3 — the same criteria used to compute lengths, areas and ratios.
• Exemplar Exercise 6.1 for the objective versions, where the correspondence traps of Questions 2 and 5 appear as options.
• NCERT textbook Exercise 6.2 — the Basic Proportionality Theorem and its converse, which Question 3 applies.
• NCERT textbook Exercise 6.3 for AA, SSS and SAS similarity on the book’s own figures.