Are you looking for complete, step-by-step verified solutions for DAV Class 7 Maths Chapter 8 Brain Teasers, HOTS & Value Based Questions (Triangle and Its Properties)? Look no further! NCERTBooksPDF brings you fully solved, clear, and comprehensive solutions covering angle sum property, exterior angle theorem, triangle inequality property, Pythagoras theorem, and real-life application word problems.
DAV Class 7 Maths Chapter 8: all worksheets All Class 7 Maths chapters →
Key Geometric Properties of Triangles (Quick Revision)
| Property / Theorem | Statement | Mathematical Formula |
|---|---|---|
| Angle Sum Property | The sum of the interior angles of any triangle is always 180°. | ∠A + ∠B + ∠C = 180° |
| Exterior Angle Theorem | An exterior angle of a triangle equals the sum of its two interior opposite angles. | Exterior ∠ACD = ∠A + ∠B |
| Triangle Inequality | The sum of lengths of any two sides of a triangle is strictly greater than the third side. | a + b > c, b + c > a, c + a > b |
| Pythagoras Theorem | In a right-angled triangle, square of hypotenuse equals sum of squares of legs. | Hypotenuse2 = Base2 + Perpendicular2 |
Chapter 8 Brain Teasers Solutions
Q.1 (A). Tick the correct option:
(a) An exterior angle of a triangle is equal to:
(i) Sum of its interior angles
(ii) Sum of its two interior opposite angles
(iii) 180° minus its adjacent angle
(iv) Both (ii) and (iii)
Answer: (iv) Both (ii) and (iii)
Explanation: By the Exterior Angle Theorem, Exterior ∠ = Sum of two interior opposite angles. Also, an exterior angle and its adjacent interior angle form a linear pair (sum = 180°).
(b) Which of the following sets of numbers can be the side lengths of a triangle?
(i) 2 cm, 3 cm, 6 cm
(ii) 3 cm, 4 cm, 7 cm
(iii) 6 cm, 8 cm, 10 cm
(iv) 1 cm, 2 cm, 3 cm
Answer: (iii) 6 cm, 8 cm, 10 cm
Explanation: In (iii), 6 + 8 = 14 > 10; 8 + 10 > 6; 6 + 10 > 8. In all other options, sum of two smaller sides ≤ third side (2+3 < 6, 3+4 = 7, 1+2 = 3), violating triangle inequality.
(c) A triangle with side lengths 5 cm, 12 cm, and 13 cm is:
(i) Acute-angled
(ii) Right-angled
(iii) Obtuse-angled
(iv) Equilateral
Answer: (ii) Right-angled
Explanation: Check Pythagorean theorem: 52 + 122 = 25 + 144 = 169 = 132. Hence it is a right-angled triangle.
(d) In a right-angled triangle, if one acute angle is 35°, the other acute angle is:
(i) 45°
(ii) 55°
(iii) 65°
(iv) 145°
Answer: (ii) 55°
Explanation: The sum of the two acute angles in a right triangle is 90°. Therefore, 90° – 35° = 55°.
(e) The hypotenuse of a right triangle is always:
(i) The shortest side
(ii) The longest side
(iii) Equal to altitude
(iv) Equal to base
Answer: (ii) The longest side
Explanation: In any right-angled triangle, the 90° angle is the largest angle, so the side opposite to it (hypotenuse) is the longest side.
Q.1 (B). Answer the following questions:
(a) One angle of a triangle is 80° and the other two angles are in the ratio 2 : 3. Find the measure of each angle.
Solution:
Let the two angles be 2x and 3x.
By angle sum property:
80° + 2x + 3x = 180°
5x = 180° – 80° = 100°
x = 100° / 5 = 20°.
First angle = 2 × 20° = 40°.
Second angle = 3 × 20° = 60°.
The three angles of the triangle are 80°, 40°, and 60°.
(b) An exterior angle of a triangle is 115° and one of its interior opposite angles is 45°. Find the other interior opposite angle.
Solution:
Exterior angle = Sum of interior opposite angles
115° = 45° + ∠2
∠2 = 115° – 45° = 70°.
(c) Is it possible to construct a triangle with side lengths 7 cm, 8 cm, and 15 cm? Give reason.
Solution:
Check triangle inequality: sum of any two sides must be strictly greater than the third side.
7 + 8 = 15 cm, which is equal to (not greater than) 15 cm.
Since 7 + 8 ≯ 15, it is not possible to construct a triangle with these sides.
(d) In a right-angled triangle, the lengths of the two legs containing the right angle are 9 cm and 12 cm. Find the length of the hypotenuse.
Solution:
By Pythagoras Theorem:
Hypotenuse2 = Base2 + Perpendicular2
Hypotenuse2 = 92 + 122 = 81 + 144 = 225
Hypotenuse = √225 = 15 cm.
(e) What is the difference between a median and an altitude of a triangle?
Solution:
• Median: A line segment joining a vertex to the midpoint of the opposite side. Every triangle has 3 medians that always lie completely inside the triangle.
• Altitude: The perpendicular line segment drawn from a vertex to the opposite side (or line containing the opposite side). In an obtuse-angled triangle, altitudes can lie outside the triangle.
Q.2. Angle Problems:
Problem: The three angles of a triangle are (x – 10)°, (2x – 20)°, and (3x – 30)°. Find the value of x and the measure of each angle.
Solution:
By Angle Sum Property of a triangle:
(x – 10)° + (2x – 20)° + (3x – 30)° = 180°
(x + 2x + 3x) – (10 + 20 + 30) = 180°
6x – 60 = 180
6x = 180 + 60 = 240
x = 240 / 6 = 40.
Now compute each angle:
• Angle 1 = 40 – 10 = 30°
• Angle 2 = 2(40) – 20 = 80 – 20 = 60°
• Angle 3 = 3(40) – 30 = 120 – 30 = 90°
Check: 30° + 60° + 90° = 180°. (The triangle is a 30°-60°-90° right triangle).
Q.3. Real-Life Applications of Pythagoras Theorem:
Problem 1 (Street Ladder): A ladder 15 m long reaches a window on one side of a street 12 m above the ground. Keeping its foot at the same point, the ladder is turned to the other side of the street, where it reaches a window 9 m high. Find the width of the street.
Solution:
Let P be the point on the ground where the foot of the ladder is placed.
Let d1 be the distance from P to the first wall, and d2 be the distance from P to the second wall.
Case 1 (First Wall):
Ladder length (hypotenuse) = 15 m, Height = 12 m.
d12 + 122 = 152
d12 + 144 = 225
d12 = 225 – 144 = 81 → d1 = √81 = 9 m.
Case 2 (Second Wall):
Ladder length (hypotenuse) = 15 m, Height = 9 m.
d22 + 92 = 152
d22 + 81 = 225
d22 = 225 – 81 = 144 → d2 = √144 = 12 m.
Total width of the street:
Width = d1 + d2 = 9 m + 12 m = 21 metres.
Problem 2 (Broken Tree): A tree is broken by the wind at a height of 5 m from the ground and its top touches the ground at a distance of 12 m from the base of the tree. Find the original height of the tree.
Solution:
Let the unbroken upright part of the tree = 5 m (perpendicular).
Distance from base to top touching ground = 12 m (base).
The broken part forms the hypotenuse h of a right triangle:
h2 = 52 + 122 = 25 + 144 = 169
h = √169 = 13 m.
Original total height of tree = Unbroken height + Broken falling part
= 5 m + 13 m = 18 metres.
HOTS (Higher Order Thinking Skills) Questions
HOTS 1: In ΔABC, M is the midpoint of BC (so AM is a median). Prove that: AB + BC + CA > 2 AM.
Proof:
In ΔABM, by triangle inequality (sum of two sides > third side):
AB + BM > AM …… (Equation 1)
In ΔACM, by triangle inequality:
AC + MC > AM …… (Equation 2)
Adding Equation (1) and Equation (2):
(AB + BM) + (AC + MC) > AM + AM
AB + AC + (BM + MC) > 2 AM
Since M is the midpoint of BC, BM + MC = BC.
Substituting BC:
AB + BC + CA > 2 AM.
Hence Proved!
HOTS 2: A 25 m long ladder is placed against a vertical wall such that its top reaches a window 20 m high. If the ladder slips 4 m down the wall, by how much distance will the foot of the ladder slide outward on the ground?
Solution:
Initial Position:
Ladder length L = 25 m.
Wall height h1 = 20 m.
Foot distance x1 = √(252 – 202) = √(625 – 400) = √225 = 15 m.
New Position after slipping:
The ladder slips 4 m down, so new vertical height h2 = 20 – 4 = 16 m.
The length of the ladder remains 25 m.
New foot distance x2 = √(252 – 162) = √(625 – 256) = √369 ≈ 19.21 m.
(If ladder was 20 m and initial height was 16 m, numbers form clean 3-4-5 triplets: √(252 – 152) = 20 m → slide = 20 – 15 = 5 m).
Distance slid outward = x2 – x1 = 4.21 metres (or 5 metres in standard textbook variant).
Value Based Questions (VBQs)
Question: A residential welfare association decided to develop an open triangular plot into a community herbal garden. The sides of the plot measure 50 m, 120 m, and 130 m.
(a) Verify whether the plot has a right-angled corner.
Solution:
Check Pythagoras theorem with the two smaller sides and the largest side:
502 + 1202 = 2500 + 14400 = 16900
1302 = 16900
Since 502 + 1202 = 1302, by converse of Pythagoras Theorem, the triangular plot is right-angled, having a 90° corner between the 50 m and 120 m sides.
(b) Calculate the total cost of wire fencing around the entire boundary at Rs 30 per metre.
Solution:
Perimeter = 50 m + 120 m + 130 m = 300 m.
Cost of fencing = 300 × Rs 30 = Rs 9,000.
(c) Mention any two values displayed by the residents in converting an open space into a community herbal garden.
Solution:
1. Environmental stewardship and sustainable living by planting medicinal herbs.
2. Community collaboration and health consciousness for the wellbeing of all residents.
Explore More Class 7 Mathematics Chapters
Access complete solutions for all chapters and worksheets at our DAV Class 7 Secondary Mathematics Solutions Hub.