CBSE Class 7 Mathematics A Tale of Three Intersecting Lines Assignment Set 01

Short Answer Questions

Question 1. A triangle is named △PQR.
(i) Name its three vertices and its three sides.
(ii) Write its three angles using three letters each, and also in the short form.
Answer:
(i) Vertices: P, Q and R. Sides: the line segments PQ, QR and RP.
(ii) Angles: ∠QPR, ∠PQR and ∠QRP; in short form ∠P, ∠Q and ∠R.
The vertices may be named in any order, so the same triangle may also be called △RQP.
In simple words: A triangle PQR has corners P, Q, R, sides PQ, QR, RP, and angles ∠P, ∠Q, ∠R.

Teacher's Note:
1. The vertex letter is always in the middle of a three-letter angle name.
2. Order of letters in the triangle's name does not matter.

Question 2. State the triangle inequality in your own words. Using it, check whether a triangle can be drawn with sides 6 cm, 7 cm and 12 cm.
Answer:
Triangle inequality: for three lengths to form a triangle, each length must be smaller than the sum of the other two. The direct path between two corners is always shorter than the roundabout path through the third corner.
Checking 6, 7, 12:
12 < 6 + 7 = 13 ✓, 7 < 6 + 12 = 18 ✓, 6 < 7 + 12 = 19 ✓
All three comparisons hold, so a triangle with these sides can be drawn.
In simple words: Each side must be less than the other two put together, and 6, 7, 12 passes all three tests.

Teacher's Note:
1. The comparison with the longest side (12 < 13) is the only close one.
2. The other two always hold when the lengths are in order.

Question 3. Write the steps you would follow to construct an equilateral triangle ABC of side 5 cm using only a ruler and a compass.
Answer:
Step 1: Draw the base AB = 5 cm with the ruler.
Step 2: Open the compass to 5 cm. With A as centre, draw a long arc above AB.
Step 3: Keeping the same opening, with B as centre, draw another arc cutting the first arc at C.
Step 4: Join AC and BC. Then △ABC is equilateral, since AC = BC = AB = 5 cm.
In simple words: Draw a 5 cm base, swing 5 cm arcs from both ends, and join the crossing point to the ends.

Teacher's Note:
1. Do not change the compass opening between the two arcs.
2. C lies on both arcs, so it is 5 cm from A and from B.

Question 4. Two angles of a triangle measure 40° and 65°.
(i) Find the third angle.
(ii) Name the type of this triangle according to its angle measures, with a reason.
Answer:
(i) By the angle sum property, third angle = 180° − (40° + 65°) = 180° − 105° = 75°.
(ii) The angles are 40°, 65° and 75°. Each is less than 90°, so all three are acute. It is an acute-angled triangle.
In simple words: The third angle is 75°, and since all angles are below 90°, the triangle is acute-angled.

Teacher's Note:
1. Find the third angle before classifying.
2. All three angles must be checked for an acute-angled triangle.

Question 5. (i) What is an altitude of a triangle?
(ii) How many altitudes can be drawn in a triangle, and from where to where is each one drawn?
Answer:
(i) An altitude is the perpendicular line segment drawn from a vertex to the side opposite to it. Its length is the height of that vertex from that side.
(ii) A triangle has three altitudes, one from each vertex. Each is drawn from a vertex to the opposite side (or to that side extended), meeting it at 90°.
In simple words: An altitude is a 90° line from a corner to the opposite side, and each triangle has three.

Teacher's Note:
1. The words 'perpendicular' and 'opposite side' are the key.
2. In an obtuse-angled triangle, two of the altitudes need the side extended.

Question 6. Classify each triangle by the lengths of its sides and give the reason for your answer.
(i) 5 cm, 5 cm, 5 cm
(ii) 6 cm, 6 cm, 9 cm
(iii) 4 cm, 7 cm, 9 cm
Answer:
(i) Equilateral triangle, because all three sides are equal.
(ii) Isosceles triangle, because exactly two sides (6 cm and 6 cm) are equal.
(iii) Scalene triangle, because the sides 4 cm, 7 cm and 9 cm are all different.
Each set also satisfies the triangle inequality (5 < 10, 9 < 12, 9 < 11), so all three triangles can really be drawn.
In simple words: Three equal sides make equilateral, two make isosceles, and none make scalene.

Teacher's Note:
1. Count how many sides are equal.
2. It is good practice to check that the triangle exists too.

Question 7. Without constructing, check whether a triangle exists for each set of lengths (the three lengths in a set have the same unit). Show the comparison you use.
(i) 4, 4, 9
(ii) 7, 8, 10
Answer:
(i) Compare the longest length with the sum of the other two: 4 + 4 = 8, and 8 < 9. The longest length is greater than the sum, so the triangle inequality fails. No triangle exists.
(ii) 7 + 8 = 15 > 10. The longest length is less than the sum of the other two, so the inequality holds. A triangle exists.
In simple words: 4, 4, 9 fails because 8 is less than 9, while 7, 8, 10 works because 15 is more than 10.

Teacher's Note:
1. Only the longest length needs to be compared.
2. Write the comparison clearly as the question asks.

Question 8. Can 100° and 80° be two of the angles of a triangle? Give a reason, and then state the rule that tells us when two given angles can be the angles of a triangle.
Answer:
No. 100° + 80° = 180°, leaving 0° for the third angle. The arms drawn at the two ends of the base would be parallel and would never meet.
Rule: two given angles can be angles of a triangle only if their sum is less than 180°. If the sum is 180° or more, no triangle exists.
In simple words: 100° and 80° already use up 180°, so no third angle is left; two angles must total less than 180°.

Teacher's Note:
1. Sum exactly 180° gives parallel arms.
2. Sum more than 180° gives arms bending apart.

Question 9. Write the steps of construction of △ABC in which AB = 6 cm, ∠A = 60° and AC = 4.5 cm. Also say which kind of data is given here: two sides with the included angle, or two angles with the included side.
Answer:
Step 1: Draw the base AB = 6 cm.
Step 2: At A, draw a ray making an angle of 60° with AB, using a protractor.
Step 3: On this ray, mark the point C such that AC = 4.5 cm.
Step 4: Join BC to complete △ABC.
The sides AB and AC and the angle ∠A between them are given, so the data is two sides and the included angle.
In simple words: Draw AB, make 60° at A, mark C 4.5 cm along that arm and join BC; this is side-angle-side data.

Teacher's Note:
1. The included angle sits at the common vertex of the two sides.
2. Measure AC along the new arm, not along AB.

Question 10. In the figure, the side BC of △ABC is extended up to D. If ∠A = 55° and ∠B = 65°, find the exterior angle x = ∠ACD.

Answer:
In △ABC, by the angle sum property:
∠A + ∠B + ∠ACB = 180°
55° + 65° + ∠ACB = 180°, so ∠ACB = 180° − 120° = 60°.
∠ACB and ∠ACD together form a straight angle along BD.
So x = ∠ACD = 180° − 60° = 120°.
Note that 120° = 55° + 65°: the exterior angle equals the sum of the two opposite interior angles.
In simple words: The angle at C inside is 60°, so the outside angle x on the straight line is 120°.

Teacher's Note:
1. Exterior angle = sum of the two opposite interior angles.
2. This gives a quick check of the answer.

Question 11. The sides of a triangle are 9 cm, 12 cm and x cm, where x is a whole number. Find the smallest and the largest value that x can take.
Answer:
The third side x must satisfy the triangle inequality with 9 and 12.
x < 9 + 12, so x < 21.
12 < 9 + x, so x > 3.
So x lies strictly between 3 and 21. As x is a whole number, the smallest value is 4 cm and the largest value is 20 cm.
In simple words: The third side must be more than 12 − 9 = 3 and less than 12 + 9 = 21, so it runs from 4 to 20.

Teacher's Note:
1. Third side lies between the difference and the sum of the other two.
2. The end values 3 and 21 themselves are not allowed.

Question 12. In △ABC, ∠B = ∠C and ∠A = 40°. Find ∠B and ∠C, and name the type of the triangle according to its angles.
Answer:
Let ∠B = ∠C = y. By the angle sum property,
40° + y + y = 180°
2y = 140°, so y = 70°.
Hence ∠B = ∠C = 70°. The angles 40°, 70° and 70° are all less than 90°, so it is an acute-angled triangle.
In simple words: The two equal angles share 140°, giving 70° each, and all angles are acute.

Teacher's Note:
1. Use one letter for the two equal angles.
2. Check: 40° + 70° + 70° = 180°.

Question 13. A triangle is constructed on an included side of 6 cm with an angle of 70° at each end of that side.
(i) Find the third angle of the triangle.
(ii) Classify this triangle according to its angle measures.
Answer:
(i) The two angles at the ends of the included side add up to 70° + 70° = 140°, which is less than 180°, so the triangle exists.
Third angle = 180° − 140° = 40°.
(ii) The angles are 70°, 70° and 40°. All are acute, so it is an acute-angled triangle.
In simple words: The two 70° angles leave 40° for the third, and every angle is acute.

Teacher's Note:
1. The 6 cm length does not affect the angles.
2. Two equal angles also make the triangle isosceles.

Question 14. In △ABC, ∠B = 90°.
(i) Name the altitude drawn from A and the altitude drawn from C.
(ii) Explain why these two altitudes are also sides of the triangle.
Answer:
(i) The altitude from A is the side AB, and the altitude from C is the side CB.
(ii) Since ∠B = 90°, the side AB is already perpendicular to BC. So the perpendicular from A to the opposite side BC is AB itself.
In the same way, CB is perpendicular to AB, so it is the altitude from C. In a right-angled triangle, the two sides forming the right angle are themselves altitudes.
In simple words: With a right angle at B, AB and CB already stand at 90° to each other, so they are the altitudes from A and C.

Teacher's Note:
1. The foot of both these altitudes is the vertex B.
2. Only the altitude from B needs to be drawn.

Question 15. To test the lengths 3, 5 and 8 (same unit), a student draws the longest length as the base AB and then draws a circle of radius 3 with centre A and a circle of radius 5 with centre B. Describe what happens to the two circles and what this tells us about the triangle.
Answer:
The base AB = 8, the longest length. The radii are 3 and 5, and 3 + 5 = 8 = AB.
So the two circles just touch each other at a single point lying on AB itself. They do not cross at two points, so no third vertex is obtained off the base.
Hence no triangle exists for the lengths 3, 5 and 8.
In simple words: Because 3 + 5 equals 8 exactly, the circles only touch on the base, so no triangle forms.

Teacher's Note:
1. Equality is not enough; the sum must be greater.
2. The touching point would make the 'triangle' a flat line.

Question 16. Using the angle sum property, explain why a triangle cannot have
(i) two right angles
(ii) two obtuse angles.
Answer:
(i) Two right angles would add up to 90° + 90° = 180° by themselves. The third angle would have to be 0°, which is impossible for a triangle.
(ii) An obtuse angle is more than 90°, so two obtuse angles would add up to more than 180°. This already exceeds the total of 180° allowed, so such a triangle cannot exist.
In simple words: Two right angles use up all 180°, and two obtuse angles go beyond it, so neither is possible.

Teacher's Note:
1. A triangle has at most one angle of 90° or more.
2. So at least two angles are always acute.

Question 17. The three angles of a triangle are in the ratio 2 : 3 : 4. Find the three angles and classify the triangle by its angle measures.
Answer:
Let the angles be 2x, 3x and 4x. By the angle sum property,
2x + 3x + 4x = 180°
9x = 180°, so x = 20°.
The angles are 2 × 20° = 40°, 3 × 20° = 60° and 4 × 20° = 80°.
Check: 40° + 60° + 80° = 180°. All three are less than 90°, so the triangle is acute-angled.
In simple words: Splitting 180° in the ratio 2 : 3 : 4 gives 40°, 60°, 80°, all acute.

Teacher's Note:
1. Add the ratio parts: 2 + 3 + 4 = 9.
2. Each part is 180° ÷ 9 = 20°.

Question 18. An isosceles triangle has two of its side lengths equal to 4 cm and 9 cm. Decide which pair of sides must be the equal pair, justify your choice using the triangle inequality, and find the perimeter of the triangle.
Answer:
If the equal sides were 4 cm and 4 cm, the sides would be 4, 4, 9. But 4 + 4 = 8 < 9, so the triangle inequality fails.
So the equal pair must be 9 cm and 9 cm, giving sides 4, 9, 9. Check: 9 < 4 + 9 = 13 and 4 < 9 + 9 = 18, so the triangle exists.
Perimeter = 4 + 9 + 9 = 22 cm.
In simple words: Two 4 cm sides cannot reach across 9 cm, so the sides must be 4, 9, 9 with perimeter 22 cm.

Teacher's Note:
1. Test both possible equal pairs.
2. Only one choice passes the triangle inequality.

Question 19. In △ABC, ∠B is twice ∠A and ∠C is three times ∠A.
(i) Find all three angles.
(ii) Name the two altitudes of this triangle that are also sides of it.
Answer:
(i) Let ∠A = x. Then ∠B = 2x and ∠C = 3x.
x + 2x + 3x = 180°, so 6x = 180° and x = 30°.
Hence ∠A = 30°, ∠B = 60° and ∠C = 90°.
(ii) Since ∠C = 90°, the sides CA and CB meet at a right angle. So CA is the altitude from A to BC, and CB is the altitude from B to AC.
In simple words: The angles are 30°, 60°, 90°, and the two sides meeting at the right angle C are altitudes.

Teacher's Note:
1. Express all angles in terms of one unknown.
2. The right angle tells you which sides are altitudes.

Question 20. A base AB is drawn and an arm is drawn at A making ∠A = 40°. Using the idea of parallel lines, explain why the arm drawn at B will never meet this arm when ∠B = 140° or more, whatever the length of AB may be.
Answer:
Draw the base AB and the arm l at A making 40° with it. Now turn the arm at B further and further.
The arm at B with the least turn that still fails to meet l is the one parallel to l.
AB is a transversal of these two parallel lines, so the interior angles on the same side add up to 180°. That arm makes ∠B = 180° − 40° = 140°.
For ∠B = 140° the arms are parallel and never meet; for ∠B more than 140° they spread further apart. The length of AB plays no part in this.
In simple words: At 140° the arm at B is parallel to the arm at A, and beyond that it moves away, whatever the base length.

Teacher's Note:
1. Co-interior angles between parallel lines add up to 180°.
2. This is why two given angles must total less than 180°.