NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 6: Triangles
Exercise 6.1

MCQs on similarity, right-triangle properties, and area/ratio results.

Question.  1

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

Right triangle with altitude AD to hypotenuse BC

(A)

\(BD\cdot CD = BC^2\)

(B)

\(AB\cdot AC = BC^2\)

(C)

\(BD\cdot CD = AD^2\)

(D)

\(AB\cdot AC = AD^2\)

Open

Question.  2

2. The diagonals of a rhombus are 16 cm and 12 cm. The side length is

(A)

9 cm

(B)

10 cm

(C)

8 cm

(D)

20 cm

Open

Question.  3

3. If \(\triangle ABC \sim \triangle EDF\) and \(\triangle ABC\) is not similar to \(\triangle DEF\), which is not true?

(A)

\(BC\cdot EF = AC\cdot FD\)

(B)

\(AB\cdot EF = AC\cdot DE\)

(C)

\(BC\cdot DE = AB\cdot EF\)

(D)

\(BC\cdot DE = AB\cdot FD\)

Open

Question.  4

4. If in triangles \(ABC\) and \(PQR\),

\(\dfrac{AB}{QR} = \dfrac{BC}{PR} = \dfrac{CA}{PQ}\), then

(A)

\(\triangle PQR \sim \triangle CAB\)

(B)

\(\triangle PQR \sim \triangle ABC\)

(C)

\(\triangle CBA \sim \triangle PQR\)

(D)

\(\triangle BCA \sim \triangle PQR\)

Open

Question.  5

5. In Fig. 6.3, lines \(AC\) and \(BD\) intersect at \(P\). Given \(PA=6\,\text{cm},\; PB=3\,\text{cm},\; PC=2.5\,\text{cm},\; PD=5\,\text{cm},\; \(\angle APB=50^\circ\) and \(\angle CDP=30^\circ\). Find \(\angle PBA\).

Intersecting chords with lengths and angles at P

(A)

50°

(B)

30°

(C)

60°

(D)

100°

Open

Question.  6

6. In triangles \(DEF\) and \(PQR\), if \(\angle D = \angle Q\) and \(\angle R = \angle E\), which is not true?

(A)

\(\dfrac{EF}{PR} = \dfrac{DF}{PQ}\)

(B)

\(\dfrac{DE}{PQ} = \dfrac{EF}{RP}\)

(C)

\(\dfrac{DE}{QR} = \dfrac{DF}{PQ}\)

(D)

\(\dfrac{EF}{RP} = \dfrac{DE}{QR}\)

Open

Question.  7

7. In triangles \(ABC\) and \(DEF\), if \(\angle B=\angle E\), \(\angle F=\angle C\) and \(AB=3\,DE\), the triangles are

(A)

congruent but not similar

(B)

similar but not congruent

(C)

neither

(D)

congruent as well as similar

Open

Question.  8

8. Given \(\triangle ABC \sim \triangle PQR\) and \(\dfrac{BC}{QR}=\dfrac{1}{3}\), then \(\dfrac{\operatorname{ar}(PRQ)}{\operatorname{ar}(BCA)}\) equals

(A)

9

(B)

3

(C)

\(\dfrac{1}{3}\)

(D)

\(\dfrac{1}{9}\)

Open

Question.  9

9. Given \(\triangle ABC \sim \triangle DFE\) with \(\angle A=30^\circ\), \(\angle C=50^\circ\), \(AB=5\,\text{cm}\), \(AC=8\,\text{cm}\) and \(DF=7.5\,\text{cm}\). Which is true?

(A)

\(DE=12\,\text{cm},\; \angle F=50^\circ\)

(B)

\(DE=12\,\text{cm},\; \angle F=100^\circ\)

(C)

\(EF=12\,\text{cm},\; \angle D=100^\circ\)

(D)

\(EF=12\,\text{cm},\; \angle D=30^\circ\)

Open

Question.  10

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

(A)

\(\angle B=\angle E\)

(B)

\(\angle A=\angle D\)

(C)

\(\angle B=\angle D\)

(D)

\(\angle A=\angle F\)

Open

Question.  11

11. If \(\triangle ABC \sim \triangle QRP\) and \(\dfrac{\operatorname{ar}(ABC)}{\operatorname{ar}(PQR)}=\dfrac{9}{4}\), with \(AB=18\,\text{cm}\) and \(BC=15\,\text{cm}\), then \(PR\) equals

(A)

10 cm

(B)

12 cm

(C)

\(\dfrac{20}{3}\) cm

(D)

8 cm

Open

Question.  12

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

(A)

\(PR\cdot QR = RS^2\)

(B)

\(QS^2 + RS^2 = QR^2\)

(C)

\(PR^2 + QR^2 = PQ^2\)

(D)

\(PS^2 + RS^2 = PR^2\)

Open

NCERT Exemplar Solutions Class 10 – Mathematics – CHAPTER 6: Triangles – Exercise 6.1 | Detailed Answers