NCERT Class 10 Mathematics — Triangles, Exercise 6.1. All 3 questions solved.

Exercise 6.1 comes before any theorem about triangles and tests only the definition that the rest of the chapter is built on. Two figures are similar when they have the same shape, though not necessarily the same size. For two polygons with the same number of sides, “same shape” has a precise meaning in two parts:

• their corresponding angles are equal, and
• their corresponding sides are in the same ratio (proportional).

Congruent figures are the special case in which that common ratio is $1$. So every pair of congruent figures is similar, but similar figures need not be congruent.

Key insight. Similarity needs both conditions, and for polygons with four or more sides neither condition implies the other. A whole family of figures is automatically similar only when a single length fixes its shape: a circle is fixed by its radius, a square by its side, an equilateral triangle by its side. Where the shape can flex, as it can for isosceles triangles, rhombuses and rectangles, the family contains pairs that are not similar. Question 3 is a pair whose sides are proportional but whose angles are not.

Question 1

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

(i) All circles are ________. (congruent, similar)

(ii) All squares are ________. (similar, congruent)

(iii) All ________ triangles are similar. (isosceles, 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)

Solution. Each blank is decided by asking whether the family always has the same shape, and whether it always has the same size.

(i) Similar. Every circle has the same shape; only the radius changes. Enlarging a circle of radius $r_1$ by the factor $\dfrac{r_2}{r_1}$ produces exactly a circle of radius $r_2$. Circles are not all congruent, because circles of different radii have different sizes.

(ii) Similar. Every angle of every square is $90^\circ$, so the angle condition always holds. For squares of sides $a$ and $b$, every ratio of corresponding sides is $\dfrac{a}{b}$, so the side condition holds too. Squares of different sides are not congruent.

(iii) Equilateral. Every angle of an equilateral triangle is $60^\circ$, so any two equilateral triangles have equal corresponding angles, and all their side ratios are the same. Isosceles triangles do not work: one with angles $50^\circ, 50^\circ, 80^\circ$ and one with angles $30^\circ, 30^\circ, 120^\circ$ are both isosceles, but they have different shapes.

(iv) Equal, proportional. This is the definition of similar polygons. The angles must be equal; the sides must be proportional, not equal. Equal sides as well as equal angles would make the polygons congruent, which is more than similarity asks.

Question 2

Give two different examples of pair of

(i) similar figures.    (ii) non-similar figures.

Solution. Any pair that meets both conditions answers (i), and any pair that fails either condition answers (ii). These choices make the reason easy to see.

(i) Similar figures.

• Two circles of different radii, say $2$ cm and $5$ cm. By question 1(i), any two circles are similar.
• Two equilateral triangles with sides $3$ cm and $5$ cm. Every angle is $60^\circ$ in both, and each ratio of corresponding sides is $\dfrac{3}{5}$.

(ii) Non-similar figures.

• A square of side $3$ cm and a rhombus of side $3$ cm with an angle of $60^\circ$. The sides are in the ratio $1 : 1$, but the angles are not equal: $90^\circ$ in the square against $60^\circ$ and $120^\circ$ in the rhombus.
• A square of side $4$ cm and a rectangle measuring $4$ cm by $6$ cm. Every angle is $90^\circ$ in both, but the side ratios $\dfrac{4}{4} = 1$ and $\dfrac{4}{6} = \dfrac{2}{3}$ are different.

The two non-similar pairs are chosen on purpose. The first fails only the angle condition and the second fails only the side condition, which shows that neither condition can be dropped. The textbook makes the same point just before the exercise.

Question 3

State whether the following quadrilaterals are similar or not:

In Fig. 6.8, $PQRS$ has all four sides equal to $1.5$ cm and no right angles marked. It leans over, so it is a rhombus, not a square. $ABCD$ has all four sides equal to $3$ cm and a right angle marked at every corner, so it is a square.

Solution. Check the two conditions of similarity one at a time.

Sides. Matching $P \leftrightarrow A$, $Q \leftrightarrow B$, $R \leftrightarrow C$, $S \leftrightarrow D$,

$$\frac{PQ}{AB} = \frac{QR}{BC} = \frac{RS}{CD} = \frac{SP}{DA} = \frac{1.5}{3} = \frac{1}{2},$$

so the corresponding sides are proportional.

Angles. Every angle of $ABCD$ is $90^\circ$. The angles of $PQRS$ are not right angles: $\angle P$ is acute and $\angle Q$ is obtuse. So $\angle P \ne \angle A$, and no other way of matching the vertices helps, because none of the rhombus’s angles is $90^\circ$.

The side condition holds but the angle condition fails, so the quadrilaterals are not similar. The textbook’s Fig. 6.7 shows the same situation with a rhombus of side $4.2$ cm and a square of side $2.1$ cm. This question uses quadrilaterals because the same thing cannot happen with triangles. For triangles, proportional sides force equal angles (the SSS criterion in Exercise 6.3).

Common mistakes

• Question 1(i), answering “congruent”. Circles of different radii plainly have different sizes. Congruent means same shape and same size, and only “similar” is true of every pair of circles.
• Question 1(iii), choosing “isosceles”. Two isosceles triangles share one feature, a pair of equal sides, but their apex angles can be anything, so their shapes differ. Only fixing all three angles, as an equilateral triangle does, guarantees similarity.
• Question 1(iv), writing “equal” for the sides. Equal angles together with equal sides give congruence, which is too strong. Similarity asks only that the sides share one common ratio.
• Question 3, stopping after the side check. The side ratios all come to $\frac{1}{2}$, and it is tempting to declare the figures similar at that point. For quadrilaterals, proportional sides do not force equal angles, so the angle check cannot be skipped.
• Question 3, assuming a figure with four equal sides is a square. $PQRS$ has four equal sides but no right angles, so it is a rhombus. Reading the angle marks in the figure, and noticing where there are none, is part of the question.
• Question 2, giving two pairs of the same kind. Offering two circles and then another two circles repeats one example rather than giving two different ones. In part (ii) especially, pick one pair that fails the angle test and one that fails the side test.

Practise next

• Exercise 6.2 — the Basic Proportionality Theorem, where proportional sides start being used to calculate lengths and to prove lines parallel.
• Exercise 6.3 — the AA, SSS and SAS criteria, which let you prove two triangles similar by checking only part of what the definition asks for.