NCERT Exemplar Solutions
Class 10 - Mathematics
CHAPTER 6: Triangles

NCERT Exemplar Class 10 Mathematics Unit 6 (Triangles) complete answers and solutions.

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

Exercise 6.2

Conceptual questions on similarity of triangles (reasons required throughout).

Question. 1

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

Answer:

No.

Handwritten Notes

Open

Question. 2

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

Answer:

No. \(\angle D=\angle R\) is true, but \(\angle F=\angle P\) is false..

Open

Question. 3

Points \(A\) and \(B\) lie on sides \(PQ\) and \(PR\) of \(\triangle PQR\) such that \(PQ=12.5\,\text{cm}\), \(PA=5\,\text{cm}\), \(PB=4\,\text{cm}\) and \(BR=6\,\text{cm}\). Is \(AB\parallel QR\)? Give reasons.

Answer:

Yes.

Open

Question. 4

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

Fig. 6.4: Two transversals BD and CE intersecting at P with given segments

Answer:

Yes.

Open

Question. 5

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

Answer:

No. But \(\triangle QPR \sim \triangle STM\).

Open

Question. 6

Is the following statement true? Why?

“Two quadrilaterals are similar if their corresponding angles are equal.”

Answer:

False.

Open

Question. 7

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

Answer:

Yes.

Open

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?

Answer:

Yes.

Open

Question. 9

The ratio of the corresponding altitudes of two similar triangles is \(\dfrac{3}{5}\). Is it correct to say that ratio of their areas is \(\dfrac{6}{5}\)? Why?

Answer:

No. The area ratio is \(\left(\dfrac{3}{5}\right)^2=\dfrac{9}{25}\).

Open

Question. 10

Point \(D\) is on side \(QR\) of \(\triangle PQR\) with \(PD\perp QR\). Is \(\triangle PQD \sim \triangle RPD\)? Why?

Answer:

No (not necessarily).

Open

Question. 11

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

Answer:

Yes.

Open

Question. 12

Is the following statement always true? “If an angle of one triangle equals an angle of another and two sides of one triangle are proportional to the corresponding two sides of the other triangle, then the triangles are similar.” Give reasons.

Answer:

Not always.

Open

Exercise 6.3

Proof and application problems on similarity.

Question. 1

In \(\triangle PQR\), suppose \(PR^2 - PQ^2 = QR^2\) and \(M\) lies on \(PR\) with \(QM \perp PR\). Prove that \(QM^2 = PM \cdot MR\).

Answer:

Proved.

Open

Question. 2

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

Fig. 6.8: Inverted triangle ABC with points D on AC and E on BC. Segments: AD=3x+19, DC=x+3;  BE=3x+4, EC=x.

Answer:

\(x=2\)

Open

Question. 3

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

Fig. 6.9: Large triangle PNR with base MQRN, cevians through S and T to Q and R.

Answer:

Proved.

Open

Question. 4

Diagonals of a trapezium \(PQRS\) intersect at \(O\). If \(PQ\parallel RS\) and \(PQ=3\,RS\), find \(\dfrac{\operatorname{ar}(\triangle POQ)}{\operatorname{ar}(\triangle ROS)}\).

Answer:

\(9:1\)

Open

Question. 5

In Fig. 6.10, if \(AB\parallel DC\) and lines \(AC\) and \(PQ\) meet at \(O\), prove that \(OA\cdot CQ = OC\cdot AP\).

Fig. 6.10: Trapezium ABCD with AB ∥ DC, diagonal AC and a line PQ through A and D meeting BC at Q; AC and PQ intersect at O.

Answer:

Proved.

Open

Question. 6

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

Answer:

\(4\sqrt{3}\,\text{cm}\)

Open

Question. 7

If \(\triangle ABC \sim \triangle DEF\), with \(AB=4\,\text{cm}\), \(DE=6\,\text{cm}\), \(EF=9\,\text{cm}\) and \(FD=12\,\text{cm}\), find the perimeter of \(\triangle ABC\).

Answer:

18 cm

Open

Question. 8

In Fig. 6.11, if \(DE\parallel BC\), find \(\operatorname{ar}(ADE):\operatorname{ar}(DECB)\).

Fig. 6.11: Triangle ABC with DE ∥ BC, DE=6 cm, BC=12 cm.

Answer:

\(1:3\)

Open

Question. 9

In trapezium \(ABCD\) with \(AB\parallel DC\), points \(P\) and \(Q\) lie on \(AD\) and \(BC\) respectively, with \(PQ\parallel DC\). If \(PD=18\,\text{cm}\), \(BQ=35\,\text{cm}\) and \(QC=15\,\text{cm}\), find \(AD\).

Answer:

60 cm

Open

Question. 10

Corresponding sides of two similar triangles are in the ratio \(2:3\). If the area of the smaller is \(48\,\text{cm}^2\), find the area of the larger triangle.

Answer:

108 cm²

Open

Question. 11

In \(\triangle PQR\), point \(N\) lies on \(PR\) with \(QN\perp PR\). If \(PN\cdot NR = QN^2\), prove that \(\angle PQR = 90^\circ\).

Answer:

Proved.

Open

Question. 12

Areas of two similar triangles are \(36\,\text{cm}^2\) and \(100\,\text{cm}^2\). If a corresponding side of the larger is 20 cm, find the corresponding side of the smaller.

Answer:

12 cm

Open

Question. 13

In Fig. 6.12, if \(\angle ACB = \angle CDA\), \(AC=8\,\text{cm}\) and \(AD=3\,\text{cm}\), find \(BD\).

Fig. 6.12: Triangle ABC with point D on AB and segment CD drawn.

Answer:

\(\dfrac{55}{3}\,\text{cm}\)

Open

Question. 14

A 15 m tower casts a 24 m shadow. At the same time a telephone pole casts a 16 m shadow. Find the pole’s height.

Answer:

10 m

Open

Question. 15

A 10 m ladder leans against a vertical wall with its foot 6 m from the wall. Find the height reached on the wall.

Answer:

8 m

Open

Exercise 6.4

Similar triangles: applications, ladders & shadows, mid-point theorems. Figures referenced using placeholders—add src later.

Question. 1

In Fig. 6.16, if \(\angle A = \angle C\), \(AB = 6\,\text{cm}\), \(BP = 15\,\text{cm}\), \(AP = 12\,\text{cm}\) and \(CP = 4\,\text{cm}\), find the lengths \(PD\) and \(CD\).

Fig. 6.16

Answer:

\(PD = 5\,\text{cm},\; CD = 2\,\text{cm}.\)

Open

Question. 2

Given \(\triangle ABC \sim \triangle EDF\) with \(AB=5\,\text{cm}\), \(AC=7\,\text{cm}\), \(DE=12\,\text{cm}\), \(DF=15\,\text{cm}\). Find the remaining sides.

Answer:

\(BC=\dfrac{25}{4}\,\text{cm}=6.25\,\text{cm},\; EF=\dfrac{84}{5}\,\text{cm}=16.8\,\text{cm}.\)

Open

Question. 3

Prove: If a line is drawn parallel to one side of a triangle to meet the other two sides, then it divides those sides in the same ratio.

Answer:

Proved.

Open

Question. 4

In Fig. 6.17, \(PQRS\) is a parallelogram and \(AB\parallel PS\). Prove that \(OC\parallel SR\).

Fig. 6.17

Answer:

Proved.

Open

Question. 5

A 5 m ladder reaches a wall at height 4 m. If the foot is moved \(1.6\,\text{m}\) towards the wall, by how much does the top slide up?

Answer:

\(0.8\,\text{m}\)

Open

Question. 6

City route: \(AC\perp CB\), \(AC=2x\) km, \(CB=2(x+7)\) km. A direct highway \(AB=26\) km is planned. How many km are saved?

Answer:

\(8\,\text{km}\)

Open

Question. 7

A flag pole 18 m high casts a 9.6 m shadow. Find the distance from the top of the pole to the far end of the shadow.

Answer:

\(20.4\,\text{m}\)

Open

Question. 8

A lamp is on a 6 m pole. A 1.5 m woman casts a 3 m shadow. How far is she from the pole?

Answer:

\(9\,\text{m}\)

Open

Question. 9

In Fig. 6.18, \(\triangle ABC\) is right-angled at \(B\) and \(BD\perp AC\). If \(AD=4\,\text{cm}\) and \(CD=5\,\text{cm}\), find \(BD\) and \(AB\).

Fig. 6.18

Answer:

\(BD=2\sqrt{5}\,\text{cm},\; AB=6\,\text{cm}.\)

Open

Question. 10

In Fig. 6.19, \(\triangle PQR\) is right-angled at \(Q\) and \(QS\perp PR\). If \(PQ=6\,\text{cm}\), \(PS=4\,\text{cm}\), find \(QS\), \(RS\) and \(QR\).

Fig. 6.19

Answer:

\(QS=2\sqrt5\,\text{cm},\; RS=5\,\text{cm},\; QR=3\sqrt5\,\text{cm}.\)

Open

Question. 11

In \(\triangle PQR\), let \(PD\perp QR\) with \(D\in QR\). If \(PQ=a\), \(PR=b\), \(QD=c\) and \(DR=d\), prove

\[(a+b)(a-b)=(c+d)(c-d).\]

Answer:

Proved.

Open

Question. 12

In a quadrilateral \(ABCD\), \(\angle A+\angle D=90^\circ\). Prove that

\[AC^2+BD^2=AD^2+BC^2.\]

Answer:

Proved.

Open

Question. 13

In Fig. 6.20, \(\ell\parallel m\) and the segments \(AB,\,CD,\,EF\) meet at \(P\). Prove

\[\dfrac{AE}{BF}=\dfrac{AC}{BD}=\dfrac{CE}{FD}.\]

Fig. 6.20

Answer:

Proved.

Open

Question. 14

In Fig. 6.21, \(PA, QB, RC, SD\) are all perpendicular to a line \(\ell\). Given \(AB=6\,\text{cm}\), \(BC=9\,\text{cm}\), \(CD=12\,\text{cm}\) and \(SP=36\,\text{cm}\). Find \(PQ, QR, RS\).

Fig. 6.21

Answer:

\(PQ=4\,\text{cm},\; QR=4\,\text{cm},\; RS=20\,\text{cm}.\)

Open

Question. 15

In a trapezium \(ABCD\) with \(AB\parallel DC\), diagonals \(AC\) and \(BD\) meet at \(O\). Through \(O\), draw \(PQ\parallel AB\) meeting \(AD\) at \(P\) and \(BC\) at \(Q\). Prove that \(PO=QO\).

Answer:

Proved.

Open

Question. 16

In Fig. 6.22, segment \(DF\) meets \(AC\) at \(E\) in \(\triangle ABC\), where \(E\) is the midpoint of \(CA\) and \(\angle AEF=\angle AFE\). Prove

\[\dfrac{BD}{CD}=\dfrac{BF}{CE}.\]

Fig. 6.22

Answer:

Proved.

Open

Question. 17

Prove that the area of the semicircle on the hypotenuse of a right triangle equals the sum of the areas of the semicircles on the other two sides.

Answer:

Proved.

Open

Question. 18

Prove that the area of the equilateral triangle on the hypotenuse of a right triangle equals the sum of the areas of the equilateral triangles on the other two sides.

Answer:

Proved.

Open

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