Class 7 Mathematics

Chapter 7 Solutions — A Tale of Three Intersecting Lines

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Solutions

Overview

Step-by-step NCERT solutions for A Tale of Three Intersecting Lines (Chapter 7, NCERT Class 7 Mathematics) — every question and answer worked out in full, not just the final result. You can also read the A Tale of Three Intersecting Lines textbook chapter.

Solved

What these solutions cover

All 20 questions in A Tale of Three Intersecting Lines are solved in the PDF. Here's what's inside, exercise by exercise:

Equilateral and Isosceles Triangles

Use the points on the circle and/or the centre to form isosceles triangles. A single circle is given with its centre marked as a dot.
Two diagrams are given, each showing two or three circles of the same size. Diagram 1: Two circles with centres A and B. Diagram 2: Three circles with centres A, B, and C arranged so each circle passes through the other two centres. Use the points on the circles and/or their centres to form isosceles and equilateral triangles.

Triangle Inequality — Are Triangles Possible for Any Lengths?

We checked by construction that there are no triangles having sidelengths 3 cm, 4 cm and 8 cm; and 2 cm, 3 cm and 6 cm. Check if you could have found this without trying to construct the triangle.
2.
Can we say anything about the existence of a triangle for each of the following sets of lengths?
- (a) 10 km, 10 km and 25 km
- (b) 5 mm, 10 mm and 20 mm
- (c) 12 cm, 20 cm and 40 cm
For each set of lengths seen so far, you might have noticed that in at least two of the comparisons, the direct length was less than the sum of the other two (if not, check again!). For example, for the set of lengths 10 cm, 15 cm and 30 cm, there are two comparisons where this happens. Identify them, and identify the comparison where it does not happen.

Triangle Inequality — Which Lengths Form a Triangle?

1.
Which of the following lengths can be the sidelengths of a triangle? Explain your answers. Note that for each set, the three lengths have the same unit of measure.
- (a) 2, 2, 5
- (b) 3, 4, 6
- (c) 2, 4, 8
- (d) 5, 5, 8
- (e) 10, 20, 25
- (f) 10, 20, 35
- (g) 24, 26, 28

Triangle Inequality — Conclusion and Third Side Range

1.
Check if a triangle exists for each of the following set of lengths:
- (a) 1, 100, 100
- (b) 3, 6, 9
- (c) 1, 1, 5
- (d) 5, 10, 12
Does there exist an equilateral triangle with sides 50, 50, 50? In general, does there exist an equilateral triangle of any sidelength? Justify your answer.
3.
For each of the following, give at least 5 possible values for the third length so there exists a triangle having these as sidelengths (decimal values could also be chosen):
- (a) 1, 100
- (b) 5, 5
- (c) 3, 7

Two Sides and Included Angle (SAS)

1.
Construct triangles for the following measurements where the angle is included between the sides:
- (a) 3 cm, 75°, 7 cm
- (b) 6 cm, 25°, 3 cm
- (c) 3 cm, 120°, 8 cm

Two Angles and Included Side (ASA)

1.
Construct triangles for the following measurements (two angles and the included side between them):
- (a) 75°, 5 cm, 75°
- (b) 25°, 3 cm, 60°
- (c) 120°, 6 cm, 30°
2.
For each of the following angles, find another angle for which a triangle is
- (a) possible,
- (b) not possible. Find at least two different angles for each category:
- (a) 30°
- (b) 70°
- (c) 54°
- (d) 144°
3.
Determine which of the following pairs can be the angles of a triangle and which cannot:
- (a) 35°, 150°
- (b) 70°, 30°
- (c) 90°, 85°
- (d) 50°, 150°

Angle Sum Property of Triangles

1.
Find the third angle of a triangle (using a parallel line) when two of the angles are:
- (a) 36°, 72°
- (b) 150°, 15°
- (c) 90°, 30°
- (d) 75°, 45°
Can you construct a triangle all of whose angles are equal to 70°? If two of the angles are 70°, what would the third angle be? If all the angles in a triangle have to be equal, then what must its measure be? Explore and find out.
Here is a triangle in which we know angle_B = angle_C and angle_A = 50°. Can you find angle_B and angle_C? [Figure shows triangle ABC with angle A = 50° at the top and equal angles at B and C.]

Types of Triangles and Altitudes

Construct a triangle ABC with BC = 5 cm, AB = 6 cm, CA = 5 cm. Construct an altitude from A to BC.
Construct a triangle TRY with RY = 4 cm, TR = 7 cm, angle_R = 140°. Construct an altitude from T to RY.
Construct a right-angled triangle ABC with angle_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 angle_A and angle_C take so that the other angle is 90°?]
4.
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.

Sample

Worked examples

3 questions from A Tale of Three Intersecting Lines, solved step by step.

Equilateral and Isosceles Triangles

Use the points on the circle and/or the centre to form isosceles triangles. A single circle is given with its centre marked as a dot.

Recall the definition of an isosceles triangle: a triangle with at least two equal sides.

Key property of a circle: every point on the circle is the same distance (the radius) from the centre.

Strategy 1 — Use two points on the circle and the centre:

Choose any two points P and Q on the circle. Let O be the centre.

Then OP = OQ = radius (both are radii of the same circle).

Triangle OPQ has two equal sides OP and OQ, so it is isosceles.

Strategy 2 — Use three points on the circle:

Choose three points P, Q, R on the circle. Triangle PQR need not be isosceles in general.

However, if you choose P and Q symmetrically about a diameter, then PO = QO, making triangle POQ isosceles as in Strategy 1.

Example construction: Mark any two points A and B on the circle. Join A, B, and the centre O.

Triangle OAB is isosceles with OA = OB (both equal to the radius).

By choosing different pairs of points on the circle, you can form many different isosceles triangles.

Answer: Connect any two points on the circle to the centre. Since both connections are radii (equal lengths), the triangle formed is isosceles. Many such triangles are possible.

Triangle Inequality — Are Triangles Possible for Any Lengths?

We checked by construction that there are no triangles having sidelengths 3 cm, 4 cm and 8 cm; and 2 cm, 3 cm and 6 cm. Check if you could have found this without trying to construct the triangle.

For a triangle to exist, each side must be less than the sum of the other two sides (triangle inequality).

We only need to check the largest side against the sum of the other two.

Set 1: 3 cm, 4 cm, 8 cm

Largest side = 8 cm.

Sum of the other two sides = 3 + 4 = 7 cm.

Since 7 < 8, the sum of the two smaller sides is less than the largest side.

The triangle inequality is violated. So a triangle with these sides does not exist.

Set 2: 2 cm, 3 cm, 6 cm

Largest side = 6 cm.

Sum of the other two sides = 2 + 3 = 5 cm.

Since 5 < 6, the sum of the two smaller sides is less than the largest side.

The triangle inequality is violated. So a triangle with these sides does not exist.

Yes, we could have determined this without construction, simply by checking whether the sum of the two smaller sides exceeds the largest side.

Answer: Yes. For 3, 4, 8: sum of smaller two = 3+4 = 7 < 8, so no triangle. For 2, 3, 6: sum of smaller two = 2+3 = 5 < 6, so no triangle. No construction needed.

Triangle Inequality — Conclusion and Third Side Range

Does there exist an equilateral triangle with sides 50, 50, 50? In general, does there exist an equilateral triangle of any sidelength? Justify your answer.

For an equilateral triangle with sides 50, 50, 50:

Check triangle inequality: sum of any two sides = 50 + 50 = 100 > 50 (the third side).

The triangle inequality is satisfied.

Yes, an equilateral triangle with sides 50, 50, 50 exists.

In general, for an equilateral triangle with any side length a (where a > 0):

Sum of any two sides = a + a = 2a.

Third side = a.

Since 2a > a for any positive a, the triangle inequality is always satisfied.

Therefore, an equilateral triangle exists for any positive sidelength.

Answer: Yes, an equilateral triangle with sides 50, 50, 50 exists (50+50=100>50). In general, for any positive sidelength a, an equilateral triangle exists because a+a=2a>a always holds.

All 20 questions in A Tale of Three Intersecting Lines are solved step by step in the Solutions PDF above.

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Chapter 7 Solutions — A Tale of Three Intersecting Lines

Overview

What these solutions cover

Equilateral and Isosceles Triangles

Triangle Inequality — Are Triangles Possible for Any Lengths?

Triangle Inequality — Which Lengths Form a Triangle?

Triangle Inequality — Conclusion and Third Side Range

Two Sides and Included Angle (SAS)

Two Angles and Included Side (ASA)

Angle Sum Property of Triangles

Types of Triangles and Altitudes

Worked examples

Use the points on the circle and/or the centre to form isosceles triangles. A single circle is given with its centre marked as a dot.

We checked by construction that there are no triangles having sidelengths 3 cm, 4 cm and 8 cm; and 2 cm, 3 cm and 6 cm. Check if you could have found this without trying to construct the triangle.

Does there exist an equilateral triangle with sides 50, 50, 50? In general, does there exist an equilateral triangle of any sidelength? Justify your answer.

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