CBSE Class 7 Mathematics A Tale of Three Intersecting Lines Practice Paper Worksheet Set 02 with Answers

CBSE Class 7 Maths Ganita Prakash Chapter 7 A Tale of Three Intersecting Lines Practice Paper 19 is a 40-mark test of 20 questions arranged in Sections A to E, given here with step-by-step answers. It checks whether three given lengths can make a triangle (the triangle inequality), classifying triangles by their sides and angles, the angle sum and exterior angle properties, the altitudes of a right-angled triangle, and constructing a triangle from two sides and the included angle.

Section A: Multiple Choice Questions (1 Mark Each)

Section B: Short Answer Questions (2 Marks Each)

Question 11. State the angle sum property of a triangle.
Answer:
The three interior angles of a triangle always add up to 180°.
If the angles of △ABC are ∠A, ∠B and ∠C, then ∠A + ∠B + ∠C = 180°.
For example, a triangle with angles 50° and 60° must have the third angle 180° − 110° = 70°.
In simple words: Whatever the shape of a triangle, its three corners together make 180°.

Teacher's Note:
1. Write the property in words and also as a formula.
2. This property holds for every triangle: acute, right or obtuse.

Question 12. Find the third angle if two angles are 58° and 62°.
Answer:
By the angle sum property, the three angles add up to 180°.
Sum of the given angles = 58° + 62° = 120°.
Third angle = 180° − 120° = 60°.
In simple words: Add the two angles and subtract the total from 180°.

Teacher's Note:
1. Check: 58° + 62° + 60° = 180°.
2. Add first, then subtract once; this avoids mistakes.

Question 13. Name the triangle whose all sides are equal.
Answer:
A triangle with all three sides equal is called an equilateral triangle.
All its angles are also equal, each 60°.
In simple words: Three equal sides make an equilateral triangle.

Teacher's Note:
1. Equilateral: 3 equal sides; isosceles: at least 2 equal sides; scalene: no equal sides.
2. Every equilateral triangle is also isosceles.

Question 14. Write one real-life use of triangle inequality.
Answer:
Before building a triangular frame (a roof truss, a bridge support or a bicycle frame), the builder checks that the sum of any two rod lengths is greater than the third.
If this condition fails, the rods cannot be joined into a triangle, so material and effort would be wasted.
Use: checking whether three given lengths of rods, beams or sticks can be joined to make a triangular frame.
Another use: the straight path between two places is always shorter than going via a third place, so a direct road saves distance.
In simple words: Triangle inequality tells us in advance whether three pieces can be joined into a triangle.

Teacher's Note:
1. One clear example with a reason is enough for 2 marks.
2. Shortest-path examples also use the same idea.

Section C: Short Answer Questions (3 Marks Each)

Question 15. Check whether a triangle can be formed with sides 9 cm, 11 cm and 20 cm.
Answer:
Longest side = 20 cm.
Sum of the other two sides = 9 + 11 = 20 cm.
For a triangle, this sum must be greater than the longest side. Here 20 cm is equal to 20 cm, not greater.
The 9 cm and 11 cm sides would lie flat along the 20 cm side and meet only on it.
Hence a triangle cannot be formed.
In simple words: 9 + 11 only equals 20, it is not bigger, so the sides cannot rise up to make a corner.

Teacher's Note:
1. Equal is not enough: the sum must be strictly greater.
2. Comparing the two shorter sides with the longest side is enough.

Question 16. Find the exterior angle of a triangle if the interior opposite angles are 46° and 54°.
Answer:
Exterior angle property: an exterior angle equals the sum of the two interior opposite angles.
Exterior angle = 46° + 54° = 100°.
Check: the third interior angle = 180° − 100° = 80°, and 46° + 54° + 80° = 180°. Also 80° + 100° = 180° (linear pair).
In simple words: Just add the two far-away inside angles to get the outside angle.

Teacher's Note:
1. Do not subtract from 180° here; that gives the interior angle next to it.
2. Checking with the angle sum is a good habit.

Question 17. Classify a triangle with angles 60°, 60° and 60°.
Answer:
By angles: every angle is 60°, which is less than 90°, so the triangle is acute-angled.
By sides: all angles are equal, so all sides are equal too; the triangle is equilateral.
Hence it is an equilateral (and acute-angled) triangle.
In simple words: Three 60° angles give a triangle with all sides equal and all corners sharp.

Teacher's Note:
1. Give both classifications to earn full marks.
2. Equal angles always face equal sides.

Section D: Long Answer Question (5 Marks)

Question 18. Construct a triangle when two sides are 5 cm and 7 cm and the included angle is 60°. Explain the steps.
Answer:
Steps of construction:
Step 1: Draw a line segment BC = 7 cm.
Step 2: With B as centre and any convenient radius, draw an arc that cuts BC at a point P and goes up above BC.
Step 3: With P as centre and the same radius, draw an arc cutting the first arc at Q. Draw the ray BQ; then ∠QBC = 60°. (A protractor can also be used to mark 60°.)
Step 4: With B as centre and radius 5 cm, cut the ray BQ at A, so that AB = 5 cm.
Step 5: Join AC.
△ABC is the required triangle with BC = 7 cm, AB = 5 cm and ∠B = 60°.

In simple words: Draw the 7 cm side, make a 60° angle at one end, mark 5 cm along it and join the last side.

Teacher's Note:
1. The given angle must be the angle between the two given sides, so it is drawn at B, the common end.
2. With the same radius, the arc from P cuts the first arc exactly where an equilateral triangle corner would be, which gives 60°.
3. Label all given measures on your figure.

Section E: Case Study Based Questions (4 Marks Each)

Question 19. Case Study 1: Roof Truss Design
A civil engineer designs a triangular roof truss using wooden beams of lengths 9 m, 12 m and 15 m. Before fixing the structure, the engineer verifies mathematically whether the beams can form a triangle. This verification ensures strength and prevents structural failure during heavy rainfall.
Answer the following questions:
(a) Identify the longest beam.
(b) Find the sum of the other two beams.
(c) Can the triangle be formed?
(d) Name the rule applied.
Answer:
(a) The lengths are 9 m, 12 m and 15 m, so the longest beam is 15 m.
(b) Sum of the other two beams = 9 m + 12 m = 21 m.
(c) 21 m is greater than 15 m, so yes, the triangle can be formed. (The other checks, 9 + 15 = 24 > 12 and 12 + 15 = 27 > 9, are also true.)
(d) The rule applied is the triangle inequality: the sum of any two sides of a triangle is greater than the third side.
In simple words: The two shorter beams together (21 m) are longer than the longest beam (15 m), so they make a triangle.

Teacher's Note:
1. Checking the longest side against the other two is enough.
2. Units must be the same (all in metres) before comparing.

Question 20. Case Study 2: Art Installation with Angles
An artist creates a triangular art installation using two angles measuring 48° and 72°. To maintain symmetry, the third angle must be calculated without using tools. The artist applies a fundamental property of triangles to complete the design accurately.
Answer the following questions:
(a) State the property used.
(b) Find the third angle.
(c) Classify the triangle by angles.
(d) State whether it is scalene or isosceles.
Answer:
(a) Angle sum property of a triangle: the three angles of a triangle add up to 180°.
(b) Third angle = 180° − (48° + 72°) = 180° − 120° = 60°.
(c) The angles are 48°, 72° and 60°; all are less than 90°, so it is an acute-angled triangle.
(d) No two angles are equal, so no two sides are equal. It is a scalene triangle.
In simple words: The missing angle is 60°; all angles are sharp and all different, so the triangle is acute-angled and scalene.

Teacher's Note:
1. Equal angles face equal sides, so three different angles mean three different sides.
2. Check: 48° + 72° + 60° = 180°.