Class 10 Mathematics

Chapter 6 Solutions — Triangles

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Solutions

Overview

Step-by-step NCERT solutions for Triangles (Chapter 6, CBSE Class 10 Mathematics) — every question and answer worked out in full, not just the final result. You can also read the Triangles textbook chapter.

Solved

What these solutions cover

All 29 questions in Triangles are solved in the PDF. Here's what's inside, exercise by exercise:

Similar Figures

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)
2.
Give two different examples of pair of
- (i) similar figures.
- (ii) non-similar figures.
State whether the two quadrilaterals in Fig. 6.8 are similar or not.

Basic Proportionality Theorem (Thales Theorem)

1.
In Fig. 6.17,
- (i) and (ii), DE || BC. Find EC in
- (i) and AD in (ii). [Fig 6.17(i): AD = 1.5 cm, DB = 3 cm, AE = 1 cm. Fig 6.17(ii): DB = 7.2 cm, AE = 1.8 cm, EC = 5.4 cm.]
2.
E and F are points on the sides PQ and PR respectively of a triangle PQR. For each of the following cases, state whether EF || QR:
- (i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm
- (ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm
- (iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm
In Fig. 6.18, if LM || CB and LN || CD, prove that AM/AB = AN/AD.
In Fig. 6.19, DE || AC and DF || AE. Prove that BF/FE = BE/EC.
In Fig. 6.20, DE || OQ and DF || OR. Show that EF || QR.
In Fig. 6.21, A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR. Show that BC || QR.
Using Theorem 6.1, prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side.
Using Theorem 6.2, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side.
ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that AO/BO = CO/DO.
The diagonals of a quadrilateral ABCD intersect each other at the point O such that AO/CO = BO/DO. Show that ABCD is a trapezium.

Criteria for Similarity of Triangles

State which pairs of triangles in Fig. 6.34 are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form.
In Fig. 6.35, triangle ODC ~ triangle OBA, angle BOC = 125° and angle CDO = 70°. Find angle DOC, angle DCO and angle OAB.
Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using a similarity criterion for two triangles, show that OA/OC = OB/OD.
In Fig. 6.36, QR/QS = QT/PR and angle 1 = angle 2. Show that triangle PQS ~ triangle TQR.
S and T are points on sides PR and QR of triangle PQR such that angle P = angle RTS. Show that triangle RPQ ~ triangle RTS.
In Fig. 6.37, if triangle ABE is congruent to triangle ACD, show that triangle ADE ~ triangle ABC.
7.
In Fig. 6.38, altitudes AD and CE of triangle ABC intersect each other at the point P. Show that:
- (i) triangle AEP ~ triangle CDP
- (ii) triangle ABD ~ triangle CBE
- (iii) triangle AEP ~ triangle ADB
- (iv) triangle PDC ~ triangle BEC.
E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that triangle ABE ~ triangle CFB.
9.
In Fig. 6.39, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that:
- (i) triangle ABC ~ triangle AMP
- (ii) CA/PA = BC/MP.
10.
CD and GH are respectively the bisectors of angle ACB and angle EGF such that D and H lie on sides AB and FE of triangle ABC and triangle EFG respectively. If triangle ABC ~ triangle FEG, show that:
- (i) CD/GH = AC/FG
- (ii) triangle DCB ~ triangle HGE
- (iii) triangle DCA ~ triangle HGF.
In Fig. 6.40, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD ⊥ BC and EF ⊥ AC, prove that triangle ABD ~ triangle ECF.
Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of triangle PQR (see Fig. 6.41). Show that triangle ABC ~ triangle PQR.
D is a point on the side BC of a triangle ABC such that angle ADC = angle BAC. Show that CA² = CB · CD.
Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that triangle ABC ~ triangle PQR.
A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.
If AD and PM are medians of triangles ABC and PQR, respectively where triangle ABC ~ triangle PQR, prove that AB/PQ = AD/PM.

Sample

Worked examples

3 questions from Triangles, solved step by step.

Criteria for Similarity of Triangles

S and T are points on sides PR and QR of triangle PQR such that angle P = angle RTS. Show that triangle RPQ ~ triangle RTS.

Given: In triangle PQR, S is on PR, T is on QR, and angle P = angle RTS.

To prove: triangle RPQ ~ triangle RTS.

In triangles RPQ and RTS:

angle R = angle R (common angle)

angle RPQ = angle RTS (given, angle P = angle RTS)

By AA similarity criterion:

triangle RPQ ~ triangle RTS

Hence proved.

Answer: Proved: triangle RPQ ~ triangle RTS (AA: angle R common, angle RPQ = angle RTS given).

Criteria for Similarity of Triangles

D is a point on the side BC of a triangle ABC such that angle ADC = angle BAC. Show that CA² = CB · CD.

Given: In triangle ABC, D is on BC and angle ADC = angle BAC.

To prove: CA² = CB · CD, i.e., CA/CD = CB/CA.

In triangles BAC and ACD:

angle BAC = angle ADC (given)

angle ACB = angle DCA (common angle C)

By AA: triangle BAC ~ triangle ACD.

Therefore corresponding sides are proportional:

CA/CD = CB/CA

⇒ CA² = CB · CD.

Hence proved.

Answer: Proved: CA² = CB · CD (triangles BAC ~ triangle ACD by AA, giving CA/CD = CB/CA).

Criteria for Similarity of Triangles

A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.

Let AB = 6 m be the pole and BC = 4 m be its shadow.

Let DE be the tower and EF = 28 m be its shadow.

At the same time of day, the sun's rays are parallel, so the angles of elevation are equal.

In triangles ABС and DEF:

angle B = angle E = 90° (both pole and tower are vertical)

angle ACB = angle DFE (sun's rays are parallel ⇒ equal angles of elevation/depression)

By AA: triangle ABC ~ triangle DEF.

Therefore: AB/DE = BC/EF

6/DE = 4/28

DE = 6 × 28/4

DE = 6 × 7

DE = 42 m

The height of the tower is 42 m.

Answer: Height of the tower = 42 m

All 29 questions in Triangles are solved step by step in the Solutions PDF above.

Board papers

Triangles in CBSE board papers

Tested in 110 questions across 24 Class 10 board papers (2020–2026). Open a set to see exactly how this chapter is asked:

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Chapter 6 Solutions — Triangles

Overview

What these solutions cover

Similar Figures

Basic Proportionality Theorem (Thales Theorem)

Criteria for Similarity of Triangles

Worked examples

S and T are points on sides PR and QR of triangle PQR such that angle P = angle RTS. Show that triangle RPQ ~ triangle RTS.

D is a point on the side BC of a triangle ABC such that angle ADC = angle BAC. Show that CA² = CB · CD.

A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.

Triangles in CBSE board papers

More solutions in Mathematics

More NCERT Solutions for Class 10