CBSE Class 10 · Boards 2027 · NCERT 2026-27 syllabus
NCERT Solutions Class 10 Maths Chapter 6 Triangles
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Boards begin February 2027
Pre-boards are in December
Triangles is in Geometry, which carries 15 of the 80 marks in the 2026-27 CBSE Maths paper
11 NCERT textbook questions from Chapter 6
Exercise 6.1
NCERT Ex 6.1 Q1(i)1 mark
Fill in the blank: All circles are ______.
Step 1All circles have the same shape (a closed curve at constant distance from a centre).
Step-by-step solution
All circles have the same shape (a closed curve at constant distance from a centre).
Two circles need not have the same radius, so they are not always congruent.
E and F are points on the sides PQ and PR respectively of △PQR. For the case PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm, state whether EF ∥ QR.
Step 1Compute PE/EQ = 3.9/3 = 1.3.
Step-by-step solution
Compute PE/EQ = 3.9/3 = 1.3.
Compute PF/FR = 3.6/2.4 = 1.5.
Compare: 1.3 ≠ 1.5, so PE/EQ ≠ PF/FR.
By converse of BPT, EF ∥ QR would require equal ratios.
Therefore EF is NOT parallel to QR.
Answer:EF is not parallel to QR (ratios 1.3 and 1.5 are unequal).
Using BPT, prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side. (Mid-point Theorem)
Step 1Given: △ABC; D is the mid-point of AB; DE ∥ BC; E lies on AC.
Step-by-step solution
Given: △ABC; D is the mid-point of AB; DE ∥ BC; E lies on AC.
To prove: AE = EC (E is the mid-point of AC).
Proof: DE ∥ BC ⇒ by BPT, AD/DB = AE/EC.
D is the mid-point of AB ⇒ AD = DB ⇒ AD/DB = 1.
Therefore AE/EC = 1 ⇒ AE = EC.
Conclusion: E is the mid-point of AC, so the line through D parallel to BC bisects AC.
Answer:Proved: a line through the mid-point of one side parallel to another side bisects the third side.
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.
Step 1Let the tower's height = h m. Both the pole and the tower are vertical, so each makes a 90° angle with the ground.
Step-by-step solution
Let the tower's height = h m. Both the pole and the tower are vertical, so each makes a 90° angle with the ground.
The sun's rays hit both objects at the same angle of elevation.
By AA similarity, △(pole, shadow, ray) ~ △(tower, shadow, ray).
Diagonals AC and BD of a trapezium ABCD with AB ∥ DC intersect at the point O. Using a similarity criterion for two triangles, show that OA/OC = OB/OD.
Step 1Given: AB ∥ DC; diagonals AC, BD meet at O.
Step-by-step solution
Given: AB ∥ DC; diagonals AC, BD meet at O.
To prove: OA/OC = OB/OD.
Construction: none needed — work in △AOB and △COD.
Proof: ∠AOB = ∠COD (vertically opposite angles).
AB ∥ DC with transversal AC ⇒ ∠OAB = ∠OCD (alternate angles).
By AA similarity criterion, △AOB ~ △COD.
Corresponding sides are proportional ⇒ OA/OC = OB/OD.
Answer:Proved: OA/OC = OB/OD using AA similarity criterion.
CBSE board pattern: Assertion-Reason and case-based questions
Written in the CBSE board format. These are not numbered NCERT/Exemplar questions; some case studies are built on textbook problems. Neither NCERT book has Assertion-Reason questions: they, and case-based questions, are a CBSE board-paper format, here on this chapter's 2026-27 syllabus, with solutions by LazyTopper.
CBSE board pattern · Assertion-Reason1 mark
Assertion (A): All congruent figures are similar, but the converse is not true.
Reason (R): Two figures are similar if they have the same shape; they are congruent if they have the same shape and the same size.
Step 1A: True. Congruent figures share shape and size — same shape ⇒ similar; but two similar figures can differ in size, so similar ⇏ congruent.
Step-by-step solution
A: True. Congruent figures share shape and size — same shape ⇒ similar; but two similar figures can differ in size, so similar ⇏ congruent.
R: True. Similarity = same shape; congruence = same shape AND same size.
R directly explains why congruence implies similarity but not vice versa.
Case study: A girl of height 90 cm walks away from the base of a lamp-post at a speed of 1.2 m/s. The lamp is fixed at 3.6 m above the ground. Answer the following:
(i) After 4 seconds, how far is the girl from the lamp-post (in metres)?
(ii) Using similar triangles, write the relation between the girl's height, the lamp-post's height, the distance of the girl from the post, and the length of her shadow.
(iii) Find the length of the girl's shadow after 4 seconds.
(iv) Which similarity criterion is used to set up the two similar triangles?
Step 1(i) Distance from post in 4 s = speed × time = 1.2 × 4 = 4.8 m.
Step-by-step solution
(i) Distance from post in 4 s = speed × time = 1.2 × 4 = 4.8 m.
(ii) Lamp-post AB and girl CD are both vertical ⇒ ∠B = ∠D = 90°. They share ∠E (the angle at the tip of the shadow). By AA, △ABE ~ △CDE.
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Questions from the NCERT Class 10 Mathematics textbook (the CBSE board-pattern section is in the board format, not numbered book questions); the solutions are written by LazyTopper. LazyTopper is an independent platform and is not affiliated with CBSE or NCERT.