NCERT Solutions Ganita Prakash (Part 1) Chapter 7 –1717.5 Types of Triangles — Figure it Out
Book page 170 Updated on2026-09-05
Q1.
Construct a triangle ABC with BC = 5 cm, AB = 6 cm, CA = 5 cm. Construct an altitude from A to BC.
Answer
First check the triangle inequality, then build it and drop the perpendicular.
Longest side = 6 cm 6 < 5 + 5 = 10 ✓ → the triangle exists
Draw BC = 5 cm.
With B as centre, draw an arc of radius 6 cm (this is AB).
With C as centre, draw an arc of radius 5 cm (this is CA), cutting the first arc at A.
Join AB and AC.
Align a ruler along BC, slide a set square on it until its vertical edge touches A, and draw AD ⟂ BC.
∆ABC with BC = 5 cm, AB = 6 cm, CA = 5 cm, and the altitude AD drawn to BC.
Check it yourself: CA = CB = 5 cm, so the triangle is isosceles with apex C. Measuring AD gives about 4.8 cm, and D falls inside BC.
Q2.
Construct a triangle TRY with RY = 4 cm, TR = 7 cm, ∠R = 140°. Construct an altitude from T to RY.
Answer
∠R is the included angle between RY and TR, so build the angle at R.
Draw RY = 4 cm.
At R, draw an arm making ∠R = 140° with RY, on one side of it.
Mark T on that arm with TR = 7 cm.
Join TY. ∆TRY is the required triangle.
For the altitude, extend YR beyond R, then drop the perpendicular from T to that extended line; call its foot D. TD is the altitude.
∠R = 140° is obtuse, so the foot D of the altitude from T lies outside the segment RY.
Why it happens: Because ∠R = 140° is obtuse, T leans away past R. The perpendicular from T therefore lands on the extension of YR, not on the segment RY itself — exactly the situation studied on page 168.
Q3.
Construct a right-angled triangle ∆ABC with ∠B = 90°, AC = 5 cm. How many different triangles exist with these measurements? [Hint: Note that the other measurements can take any values. Take AC as the base. What values can ∠A and ∠C take so that the other angle is 90°?]
Answer
Infinitely many different triangles fit these two conditions.
∠A + ∠B + ∠C = 180° ∠A + 90° + ∠C = 180° ∠A + ∠C = 90° So any pair with ∠A + ∠C = 90° works
∠A
∠C
∠B
Valid?
30°
60°
90°
Yes
45°
45°
90°
Yes (isosceles)
10°
80°
90°
Yes
72.5°
17.5°
90°
Yes
How to construct one: draw AC = 5 cm, choose any ∠A between 0° and 90°, make ∠C = 90° – ∠A, and let the two arms meet at B.
Every position of B on the circle with AC as diameter gives a right angle at B.
Why it happens: Fixing AC and ∠B = 90° still leaves the shape free. The right-angle vertex B can sit anywhere on the circle drawn with AC as diameter, and each position gives a genuinely different triangle — so there are endlessly many of them, all sharing the same 5 cm side.
Q4.
Through construction, explore if it is possible to construct an equilateral triangle that is (i) right-angled (ii) obtuse-angled. Also construct an isosceles triangle that is (i) right-angled (ii) obtuse-angled.
Answer
Equilateral: neither is possible. Isosceles: both are possible.
(i) Equilateral and right-angled — impossible
All three angles equal → each = 180° ÷ 3 = 60° 60° ≠ 90° → no right angle can appear
(ii) Equilateral and obtuse-angled — impossible
Each angle is 60°, and 60° < 90° So an equilateral triangle is always acute-angled
Isosceles and right-angled — possible
Take ∠B = 90° and the two equal sides AB = BC Then ∠A = ∠C = (180° – 90°) ÷ 2 = 45° Angles: 90°, 45°, 45°
To draw it: make ∠B = 90°, mark AB = BC = 4 cm along the two arms, and join AC.
Isosceles and obtuse-angled — possible
Take the apex angle = 120° Then the two base angles = (180° – 120°) ÷ 2 = 30° each Angles: 120°, 30°, 30°
To draw it: make ∠B = 120° with BA = BC = 4 cm along the arms, then join AC.
Why it happens: An equilateral triangle has all its angles pinned at 60°, leaving no freedom at all. An isosceles triangle only pins the two base angles equal to each other; the apex angle is still free to be 90°, 120° or anything less than 180°, so both kinds exist.