CBSE Important Questions Class 10 Maths Chapter 6 – Triangles

Class 10 Maths Chapter 6 Triangles is part of the CBSE Class 10 Maths syllabus. It explains similar figures and similar triangles, the Basic Proportionality Theorem (Thales theorem), and the criteria for similarity: AAA, SSS and SAS. Students also learn how the areas of two similar triangles are related to the squares of their corresponding sides, and they study the Pythagoras theorem and its converse.

These CBSE important questions for Class 10 Maths Chapter 6 cover numerical problems on a ladder leaning against a wall, the Baudhayan theorem, proving the Pythagoras theorem, finding a side using similar triangles, the ratio of the perimeters of similar triangles, and applying the midpoint and proportionality theorems. Every question has a clear step-by-step solution in simple English with figures, so students can practise on their own, parents can check the working, and teachers can use them for class tests.

Q1. If the hypotenuse of a right-angled triangle is 41 cm and the area of the triangle is 180 sq. cm, then the difference between the lengths of the legs of the triangle must be
(a) 22 cm
(b) 25 cm
(c) 27 cm
(d) 31 cm

Answer: (d) 31 cm
Let the two legs be a and b.
Area = ½ab = 180, so ab = 360.
By the Pythagoras theorem: a² + b² = 41² = 1681.
Now, (a − b)² = a² + b² − 2ab = 1681 − 720 = 961
a − b = √961 = 31 cm

Q2. A 13 m long ladder reaches a ventilator 12 m high from the ground when placed against a wall. Find the distance of the foot of the ladder from the wall.

Answer: Let AB be the wall up to the ventilator (12 m) and AC the ladder (13 m). BC is the distance of the foot from the wall, and angle ABC = 90°.
By the Pythagoras theorem:
AC² = AB² + BC²
(13)² = (12)² + BC²
169 = 144 + BC²
BC² = 25
BC = 5 m
So the foot of the ladder is 5 m away from the wall.

Q3. A ladder is placed against a wall such that its foot is 5.5 m from the wall and its top reaches a window 9 m above the ground. The length of the ladder is
(a) 10.24 m
(b) 10.54 m
(c) 11 m
(d) 11.9 m

Answer: (b) 10.54 m
By the Pythagoras theorem:
Length² = (5.5)² + (9)²
= 30.25 + 81
= 111.25
Length = √111.25 ≈ 10.54 m

Q4. State the Baudhayan theorem.

Answer: The Baudhayan theorem states that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
This is the same result that is known in the rest of the world as the Pythagoras theorem. It was written down by the Indian mathematician Baudhayana in the Sulba Sutras, several centuries before Pythagoras.

Q5. In the given figure the lengths of the line segments are in cm. It is given that △CAB ~ △CED. Prove the similarity and find the value of x.

Answer: In △CAB and △CED:
∠C is common to both triangles, and ∠CAB = ∠CED (given, as they are corresponding angles).
So by the AA similarity criterion, △CAB ~ △CED.
Since the triangles are similar, their corresponding sides are in the same ratio:
CA/CE = CB/CD = AB/ED
Substituting the given lengths and solving the proportion gives x = 6 cm.

Q6. ABC is a right triangle, right-angled at B. Let D and E be any points on AB and BC respectively. Prove that AE² + CD² = AC² + DE².

Answer: In △ABE, which is right-angled at B:
AE² = AB² + BE² ...(i)
In △DBC, which is right-angled at B:
CD² = BD² + BC² ...(ii)
Adding (i) and (ii):
AE² + CD² = AB² + BE² + BD² + BC²
Now regroup the terms:
= (AB² + BC²) + (BD² + BE²)
In △ABC, right-angled at B: AB² + BC² = AC²
In △DBE, right-angled at B: BD² + BE² = DE²
Therefore AE² + CD² = AC² + DE². Hence proved.

Q7. Prove that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Answer: Given: A right triangle ABC, right-angled at B.
To prove: AC² = AB² + BC²
Construction: Draw BD ⊥ AC.
Proof: In △ADB and △ABC, ∠A is common and ∠ADB = ∠ABC = 90°.
So △ADB ~ △ABC (AA similarity).
Therefore AD/AB = AB/AC, which gives AB² = AD × AC ...(i)
Similarly, in △BDC and △ABC, ∠C is common and ∠BDC = ∠ABC = 90°.
So △BDC ~ △ABC, which gives CD/BC = BC/AC, so BC² = CD × AC ...(ii)
Adding (i) and (ii):
AB² + BC² = AD × AC + CD × AC
= AC (AD + CD)
= AC × AC  (since AD + CD = AC)
AB² + BC² = AC². Hence proved.

Q8. The hypotenuse of a right triangle is 6 m more than twice the shortest side. If the third side is 2 m less than the hypotenuse, find the sides of the triangle.

Answer: Let the shortest side be x metres.
Hypotenuse = (2x + 6) m and the third side = (2x + 6) − 2 = (2x + 4) m.
By the Pythagoras theorem:
(2x + 6)² = x² + (2x + 4)²
4x² + 24x + 36 = x² + 4x² + 16x + 16
4x² + 24x + 36 = 5x² + 16x + 16
0 = x² − 8x − 20
(x − 10)(x + 2) = 0
So x = 10 or x = −2. A length cannot be negative, so x = 10.
The sides are 10 m, 24 m and 26 m.

Q9. A ladder 25 m long reaches a window of a building 20 m above the ground. The distance of the foot of the ladder from the building is
(a) 5 m
(b) 10 m
(c) 15 m
(d) 20 m

Answer: (c) 15 m
By the Pythagoras theorem:
25² = 20² + d²
625 = 400 + d²
d² = 225
d = 15 m

Q10. State and prove the Basic Proportionality Theorem (Thales theorem).

Answer: Statement: If a line is drawn parallel to one side of a triangle to intersect the other two sides at distinct points, then the other two sides are divided in the same ratio.
Given: △ABC with DE ∥ BC, where D lies on AB and E on AC.
To prove: AD/DB = AE/EC
Construction: Join BE and CD. Draw EM ⊥ AB and DN ⊥ AC.
Proof: area(△ADE)/area(△BDE) = AD/DB (same height EM)
area(△ADE)/area(△DEC) = AE/EC (same height DN)
But △BDE and △DEC are on the same base DE and between the same parallels DE and BC, so their areas are equal.
Therefore AD/DB = AE/EC. Hence proved.

Q11. The perimeters of two similar triangles are 36 cm and 48 cm respectively. If one side of the first triangle is 9 cm, what is the corresponding side of the other triangle?

Answer: In two similar triangles, the ratio of the perimeters is equal to the ratio of any pair of corresponding sides.
Perimeter of first / Perimeter of second = side of first / corresponding side of second
36/48 = 9/x
36x = 48 × 9 = 432
x = 432/36
x = 12 cm

Q12. The areas of two similar triangles are 81 cm² and 49 cm². If the altitude of the first triangle is 6.3 cm, find the corresponding altitude of the second triangle.

Answer: If two triangles are similar, the ratio of their areas equals the ratio of the squares of their corresponding altitudes.
area₁/area₂ = (altitude₁)² / (altitude₂)²
81/49 = (6.3)² / h²
Taking the square root of both sides:
9/7 = 6.3/h
9h = 7 × 6.3 = 44.1
h = 4.9 cm

Related Study Material on Extramarks

• CBSE Important Questions Class 10 Maths Chapter 1
• CBSE Important Questions Class 10 Maths Chapter 2
• CBSE Important Questions Class 10 Maths Chapter 3
• CBSE Important Questions Class 10 Maths Chapter 4
• CBSE Important Questions Class 10 Maths Chapter 5
• CBSE Important Questions Class 10 Maths Chapter 7
• CBSE Important Questions Class 10 Maths Chapter 8
• CBSE Important Questions Class 10 Maths Chapter 9
• CBSE Important Questions Class 10 Maths Chapter 10
• CBSE Important Questions Class 10 Maths Chapter 11
• CBSE Important Questions Class 10 Maths Chapter 12
• CBSE Important Questions Class 10 Maths Chapter 13
• CBSE Important Questions Class 10 Maths Chapter 14