Chapter 7 exercise answers: A Tale of Three Intersecting Lines

Open NCERT p. 150One way to think about it

1. Mark two points on the circle, say P and Q, and also mark the centre O.
2. Join OP and OQ. Both are radii of the same circle, so OP = OQ.
3. Join PQ to complete triangle OPQ.

In shortTriangle OPQ, formed by joining the centre O to any two points P and Q on the circle, is isosceles, because OP and OQ are both radii of the same circle and so are equal.

Watch this explained “Two equal sides”, 8:08 into Why a compass beats trial and error for building a triangle · हिंदी में देखें

Open NCERT p. 151One way to think about it

1. All the circles are the same size, so they all have the same radius. Call it r. A segment from a centre to any point on its own circle has length r.
2. In the figure, each circle passes through the other circle's centre (B lies on the circle centred at A, and A on the circle centred at B). So AB = r as well.
3. Isosceles: take centre A and any two points P and Q on the circle centred at A. AP = AQ = r, so triangle APQ is isosceles. Every choice of P and Q gives another one.
4. Equilateral: let P be a point where the two circles cross. P is on the circle centred at A, so AP = r. P is on the circle centred at B, so BP = r. And AB = r. So triangle ABP is equilateral. The other crossing point gives a second one.
5. In the three-circle figure, each circle passes through the other two centres, so AB = BC = CA = r and triangle ABC is equilateral. Any two of the centres with a crossing point of their two circles give more equilateral triangles in the same way.

In shortIsosceles: a centre and any two points on its circle, e.g. triangle APQ with AP = AQ = r. Equilateral: two centres and a point where their circles cross, e.g. triangle ABP with AB = AP = BP = r; in the three-circle figure, triangle ABC itself. There are many such triangles; these are examples.

Watch this explained “Two equal sides”, 8:08 into Why a compass beats trial and error for building a triangle · हिंदी में देखें

Open NCERT p. 154One way to think about it

1. For 3 cm, 4 cm, 8 cm: add the two shorter lengths, 3 + 4 = 7 cm.
2. Compare with the longest length: 7 cm is less than 8 cm, so the two shorter sides can never reach far enough to close the triangle.
3. For 2 cm, 3 cm, 6 cm: add the two shorter lengths, 2 + 3 = 5 cm.
4. Compare with the longest length: 5 cm is less than 6 cm, so again the two shorter sides fall short.

In shortYes — adding the two shorter lengths and comparing the sum with the longest length shows both sets fail before any drawing is done: 3 + 4 = 7 < 8, and 2 + 3 = 5 < 6.

Watch this explained “Two that refuse to be built”, 0:00 into The triangle inequality: which three lengths can close · हिंदी में देखें

Open NCERT p. 154Matches NCERT’s answer

(a) 10 km, 10 km and 25 km

1. Add the two smaller lengths: 10 + 10 = 20 km.
2. Compare with the longest length: 20 km is less than 25 km.

AnswerNo triangle exists.

(b) 5 mm, 10 mm and 20 mm

1. Add the two smaller lengths: 5 + 10 = 15 mm.
2. Compare with the longest length: 15 mm is less than 20 mm.

AnswerNo triangle exists.

(c) 12 cm, 20 cm and 40 cm

1. Add the two smaller lengths: 12 + 20 = 32 cm.
2. Compare with the longest length: 32 cm is less than 40 cm.

AnswerNo triangle exists.

Watch this explained “Only the largest can fail”, 4:50 into The triangle inequality: which three lengths can close · हिंदी में देखें

Open NCERT p. 154One way to think about it

1. Check the sets seen so far. 3, 4, 8: 3 < 4 + 8 = 12 and 4 < 3 + 8 = 11, but 8 > 3 + 4 = 7. 2, 3, 6: 2 < 9 and 3 < 8, but 6 > 5. 10, 15, 30: 10 < 45 and 15 < 40, but 30 > 25. Each time at least two comparisons hold.
2. This always happens. Put the three lengths in order: smallest, middle, largest.
3. The smallest length: the largest length on its own is already at least as long, and the middle length is added on top. So the smallest is always less than the sum of the other two.
4. The middle length: the largest on its own is at least as long, and the smallest is added on top. So the middle length is always less than the sum of the other two as well.
5. So the comparisons for the two smaller lengths always hold. Only the comparison for the largest length can fail, as 30 > 10 + 15 does.

In shortYes, it always happens: the smallest and the middle length are always less than the sum of the other two, so at least two comparisons always hold. Only the largest length can fail its comparison, so that is the one comparison to check.

Watch this explained “Only the largest can fail”, 4:50 into The triangle inequality: which three lengths can close · हिंदी में देखें

Open NCERT p. 156Matches NCERT’s answer

(a) 2, 2, 5

1. Add the two smaller lengths: 2 + 2 = 4.
2. Compare with the longest length: 4 is not greater than 5.

AnswerNo, these cannot be the sidelengths of a triangle.

(b) 3, 4, 6

1. Add the two smaller lengths: 3 + 4 = 7.
2. Compare with the longest length: 7 is greater than 6, and the same holds for the other two comparisons.

AnswerYes, these can be the sidelengths of a triangle.

(c) 2, 4, 8

1. Add the two smaller lengths: 2 + 4 = 6.
2. Compare with the longest length: 6 is not greater than 8.

AnswerNo, these cannot be the sidelengths of a triangle.

(d) 5, 5, 8

1. Add the two smaller lengths: 5 + 5 = 10.
2. Compare with the longest length: 10 is greater than 8, and the same holds for the other two comparisons.

AnswerYes, these can be the sidelengths of a triangle.

(e) 10, 20, 25

1. Add the two smaller lengths: 10 + 20 = 30.
2. Compare with the longest length: 30 is greater than 25, and the same holds for the other two comparisons.

AnswerYes, these can be the sidelengths of a triangle.

(f) 10, 20, 35

1. Add the two smaller lengths: 10 + 20 = 30.
2. Compare with the longest length: 30 is not greater than 35.

AnswerNo, these cannot be the sidelengths of a triangle.

(g) 24, 26, 28

1. Add the two smaller lengths: 24 + 26 = 50.
2. Compare with the longest length: 50 is greater than 28, and the same holds for the other two comparisons.

AnswerYes, these can be the sidelengths of a triangle.

Watch this explained “The triangle inequality”, 5:40 into The triangle inequality: which three lengths can close · हिंदी में देखें

Open NCERT p. 159Matches NCERT’s answer

(a) 1, 100, 100

1. Add the two smaller lengths: 1 + 100 = 101.
2. Compare with the longest length: 101 is greater than 100, so the triangle inequality is satisfied.

AnswerYes, a triangle exists.

(b) 3, 6, 9

1. Add the two smaller lengths: 3 + 6 = 9.
2. Compare with the longest length: 9 is not greater than 9, so the triangle inequality fails.

AnswerNo, a triangle does not exist.

(c) 1, 1, 5

1. Add the two smaller lengths: 1 + 1 = 2.
2. Compare with the longest length: 2 is not greater than 5, so the triangle inequality fails.

AnswerNo, a triangle does not exist.

(d) 5, 10, 12

1. Add the two smaller lengths: 5 + 10 = 15.
2. Compare with the longest length: 15 is greater than 12, so the triangle inequality is satisfied.

AnswerYes, a triangle exists.

Watch this explained “The triangle inequality”, 5:40 into The triangle inequality: which three lengths can close · हिंदी में देखें

Open NCERT p. 159Matches NCERT’s answer

1. Check the triangle inequality for 50, 50, 50: 50 + 50 = 100, and 100 is greater than 50, so it holds (and the same check works for every pair of the three equal sides).
2. So a triangle with sides 50, 50, 50 exists.
3. For a general equilateral triangle of side s, every comparison has the same form: s < s + s.
4. s + s = 2s, and 2s is greater than s for every positive s, so the triangle inequality always holds.
5. So an equilateral triangle exists for any positive sidelength.

AnswerYes — a triangle with sides 50, 50, 50 exists, and yes, an equilateral triangle exists for any positive sidelength, because s < s + s is always true.

Watch this explained “The triangle inequality”, 5:40 into The triangle inequality: which three lengths can close · हिंदी में देखें

Open NCERT p. 159Checked by computerAnswers can differ: one example

(a) 1, 100

1. For sides 1 and 100, the third side x must be less than 1 + 100 = 101.
2. Also, 100 must be less than 1 + x, so x must be more than 99.
3. So x must lie strictly between 99 and 101.
4. Five such values: 99.5, 99.75, 100, 100.25, 100.5.

AnswerAny value strictly between 99 and 101, e.g. 99.5, 99.75, 100, 100.25, 100.5.

(b) 5, 5

1. For sides 5 and 5, the third side x must be less than 5 + 5 = 10.
2. Also, 5 must be less than 5 + x, so x must be more than 0.
3. So x must lie strictly between 0 and 10.
4. Five such values: 2, 4, 5, 6, 8.

AnswerAny value strictly between 0 and 10, e.g. 2, 4, 5, 6, 8.

(c) 3, 7

1. For sides 3 and 7, the third side x must be less than 3 + 7 = 10.
2. Also, 7 must be less than 3 + x, so x must be more than 4.
3. So x must lie strictly between 4 and 10.
4. Five such values: 5, 6, 7, 8, 9.

AnswerAny value strictly between 4 and 10, e.g. 5, 6, 7, 8, 9.

Watch this explained “From paths to arithmetic”, 4:09 into The triangle inequality: which three lengths can close · हिंदी में देखें

Open NCERT p. 161One way to think about it

(a) 3 cm, 75°, 7 cm

1. Draw the base AB = 7 cm (the longer of the two given sides).
2. At A, use a protractor to draw a ray making the included angle of 75° with AB.
3. Mark point C on the ray with AC = 3 cm.
4. Join B and C to complete triangle ABC, with the 75° angle included between the 3 cm and 7 cm sides.

In shortTriangle ABC with AB = 7 cm, ∠A = 75°, AC = 3 cm.

(b) 6 cm, 25°, 3 cm

1. Draw the base AB = 6 cm.
2. At A, use a protractor to draw a ray making the included angle of 25° with AB.
3. Mark point C on the ray with AC = 3 cm.
4. Join B and C to complete triangle ABC.

In shortTriangle ABC with AB = 6 cm, ∠A = 25°, AC = 3 cm.

(c) 3 cm, 120°, 8 cm

1. Draw the base AB = 8 cm.
2. At A, use a protractor to draw a ray making the included angle of 120° with AB.
3. Mark point C on the ray with AC = 3 cm.
4. Join B and C to complete triangle ABC.

In shortTriangle ABC with AB = 8 cm, ∠A = 120°, AC = 3 cm.

Watch this explained “Lay the base, then swing the arm”, 2:54 into Constructing from two sides and the angle between them · हिंदी में देखें

Open NCERT p. 162One way to think about it

(a) 75°, 5 cm, 75°

1. Draw the included side AB = 5 cm.
2. At A, set an angle of 75° from AB and draw a long ray.
3. At B, set an angle of 75° from BA, on the same side of AB as the first ray, and draw a long ray.
4. The two rays meet at C, because 75° + 75° = 150° is less than 180°. Triangle ABC is the required triangle.

In shortTriangle ABC with AB = 5 cm, ∠A = 75°, ∠B = 75°.

(b) 25°, 3 cm, 60°

1. Draw the included side AB = 3 cm.
2. At A, set an angle of 25° from AB and draw a long ray.
3. At B, set an angle of 60° from BA, on the same side of AB as the first ray, and draw a long ray.
4. The two rays meet at C, because 25° + 60° = 85° is less than 180°. Triangle ABC is the required triangle.

In shortTriangle ABC with AB = 3 cm, ∠A = 25°, ∠B = 60°.

(c) 120°, 6 cm, 30°

1. Draw the included side AB = 6 cm.
2. At A, set an angle of 120° from AB and draw a long ray.
3. At B, set an angle of 30° from BA, on the same side of AB as the first ray, and draw a long ray.
4. The two rays meet at C, because 120° + 30° = 150° is less than 180°. Triangle ABC is the required triangle.

In shortTriangle ABC with AB = 6 cm, ∠A = 120°, ∠B = 30°.

Watch this explained “Building one properly”, 2:07 into Constructing from two angles and the side between them · हिंदी में देखें

Open NCERT p. 163Checked by computerAnswers can differ: one example

(a) 30°

1. A triangle exists for two angles only when their sum is less than 180°, so the other angle must be less than 180° − 30° = 150°.
2. Any angle below 150°, e.g. 40° and 100°, makes a triangle possible.
3. Any angle of 150° or more, e.g. 150° and 160°, makes a triangle not possible.

AnswerPossible: e.g. 40°, 100° (any angle < 150°). Not possible: e.g. 150°, 160° (any angle ≥ 150°).

(b) 70°

1. A triangle exists for two angles only when their sum is less than 180°, so the other angle must be less than 180° − 70° = 110°.
2. Any angle below 110°, e.g. 30° and 90°, makes a triangle possible.
3. Any angle of 110° or more, e.g. 110° and 150°, makes a triangle not possible.

AnswerPossible: e.g. 30°, 90° (any angle < 110°). Not possible: e.g. 110°, 150° (any angle ≥ 110°).

(c) 54°

1. A triangle exists for two angles only when their sum is less than 180°, so the other angle must be less than 180° − 54° = 126°.
2. Any angle below 126°, e.g. 40° and 100°, makes a triangle possible.
3. Any angle of 126° or more, e.g. 126° and 150°, makes a triangle not possible.

AnswerPossible: e.g. 40°, 100° (any angle < 126°). Not possible: e.g. 126°, 150° (any angle ≥ 126°).

(d) 144°

1. A triangle exists for two angles only when their sum is less than 180°, so the other angle must be less than 180° − 144° = 36°.
2. Any angle below 36°, e.g. 10° and 30°, makes a triangle possible.
3. Any angle of 36° or more, e.g. 36° and 60°, makes a triangle not possible.

AnswerPossible: e.g. 10°, 30° (any angle < 36°). Not possible: e.g. 36°, 60° (any angle ≥ 36°).

Watch this explained “Add them, and compare”, 8:15 into Constructing from two angles and the side between them · हिंदी में देखें

Open NCERT p. 163Matches NCERT’s answer

(a) 35°, 150°

1. Add the two angles: 35° + 150° = 185°.
2. 185° is not less than 180°, so a triangle is not possible.

AnswerNo, these cannot be the angles of a triangle.

(b) 70°, 30°

1. Add the two angles: 70° + 30° = 100°.
2. 100° is less than 180°, so a triangle is possible.

AnswerYes, these can be the angles of a triangle.

(c) 90°, 85°

1. Add the two angles: 90° + 85° = 175°.
2. 175° is less than 180°, so a triangle is possible.

AnswerYes, these can be the angles of a triangle.

(d) 50°, 150°

1. Add the two angles: 50° + 150° = 200°.
2. 200° is not less than 180°, so a triangle is not possible.

AnswerNo, these cannot be the angles of a triangle.

Watch this explained “Add them, and compare”, 8:15 into Constructing from two angles and the side between them · हिंदी में देखें

Open NCERT p. 165Matches NCERT’s answer

(a) 36°, 72°

1. Call the triangle ABC with ∠B = 36° and ∠C = 72°. Through A, draw a line XY parallel to BC.
2. AB and AC are transversals of the parallel lines XY and BC, so the alternate angles are equal: ∠XAB = ∠B = 36° and ∠YAC = ∠C = 72°.
3. ∠XAB, ∠BAC and ∠YAC together make the straight angle along XY, so 36° + ∠BAC + 72° = 180°.
4. So ∠BAC = 180° − 36° − 72° = 72°.

AnswerThe third angle is 72°.

(b) 150°, 15°

1. Draw the line through the third vertex parallel to the opposite side, as in (a). The alternate angles on it are 150° and 15°.
2. These two and the third angle make a straight angle: 150° + third angle + 15° = 180°.
3. So the third angle = 180° − 150° − 15° = 15°.

AnswerThe third angle is 15°.

(c) 90°, 30°

1. Draw the line through the third vertex parallel to the opposite side, as in (a). The alternate angles on it are 90° and 30°.
2. 90° + third angle + 30° = 180°.
3. So the third angle = 180° − 90° − 30° = 60°.

AnswerThe third angle is 60°.

(d) 75°, 45°

1. Draw the line through the third vertex parallel to the opposite side, as in (a). The alternate angles on it are 75° and 45°.
2. 75° + third angle + 45° = 180°.
3. So the third angle = 180° − 75° − 45° = 60°.

AnswerThe third angle is 60°.

Watch this explained “A line that was not there”, 1:41 into Why the three angles of any triangle add to 180° · हिंदी में देखें

Open NCERT p. 165Matches NCERT’s answer

1. If all three angles were 70°, they would add to 70° + 70° + 70° = 210°, but the angle sum of a triangle must be 180°, so this is not possible.
2. If two of the angles are 70° each, the third is 180° − 70° − 70° = 40°.
3. If all three angles are equal, each is 180° ÷ 3 = 60°.

AnswerNo, a triangle cannot have all three angles equal to 70° (they would total 210°, not 180°). If two angles are 70°, the third is 40°. If all three angles are equal, each must be 60°.

Watch this explained “The angle sum property”, 4:44 into Why the three angles of any triangle add to 180° · हिंदी में देखें

Open NCERT p. 165Matches NCERT’s answer

1. The three angles add to 180°: ∠A + ∠B + ∠C = 180°.
2. Since ∠B = ∠C, this becomes 50° + 2∠B = 180°.
3. So 2∠B = 130°, giving ∠B = 65°, and ∠C = ∠B = 65°.

Answer∠B = ∠C = 65°.

Watch this explained “The angle sum property”, 4:44 into Why the three angles of any triangle add to 180° · हिंदी में देखें

Open NCERT p. 170One way to think about it

1. Draw the base BC = 5 cm.
2. From B, draw an arc of radius 6 cm (for AB); from C, draw an arc of radius 5 cm (for CA); the point where they cross is A.
3. Join A to B and A to C to complete triangle ABC.
4. To draw the altitude from A: place a set square with one edge of its right angle along BC, slide it until the other edge passes through A, and draw the perpendicular from A down to BC, marking the foot D.
5. AD is the required altitude from A to BC.

In shortTriangle ABC constructed with BC = 5 cm, AB = 6 cm, CA = 5 cm; the altitude AD is the perpendicular from A to BC, drawn using a set square.

Watch this explained “Align, slide, draw”, 8:06 into What an altitude is, and how to construct one · हिंदी में देखें

Open NCERT p. 170One way to think about it

1. Draw the base RY = 4 cm.
2. At R, set an angle of 140° from RY using a protractor, and draw a ray.
3. Mark point T on this ray such that RT = 7 cm.
4. Join T to Y to complete triangle TRY.
5. Since ∠R = 140° is obtuse, the foot of the altitude from T does not lie on segment RY: extend RY backward past R, then use a set square to drop a perpendicular from T onto this extended line, marking the foot D.
6. Draw TD as a dashed segment, since it lies outside the triangle.

In shortTriangle TRY constructed with RY = 4 cm, ∠R = 140°, RT = 7 cm; because ∠R is obtuse, the altitude from T meets the extension of RY beyond R, and is drawn dashed as it falls outside the triangle.

Watch this explained “When the foot misses the base”, 4:33 into What an altitude is, and how to construct one · हिंदी में देखें

Open NCERT p. 171Checked by computer

1. The right angle is at B, so AC = 5 cm is the side opposite it. As the hint says, take AC as the base.
2. The three angles add to 180° and ∠B = 90°, so ∠A + ∠C = 180° − 90° = 90°.
3. Pick any ∠A between 0° and 90° and set ∠C = 90° − ∠A. For example ∠A = 30°, ∠C = 60°; or ∠A = 45°, ∠C = 45°; or ∠A = 20°, ∠C = 70°.
4. Construct it from two angles and the included side: draw AC = 5 cm, draw ∠A at A and ∠C at C on the same side. They add to 90°, less than 180°, so the arms meet at B, and ∠B = 180° − 90° = 90°.
5. Different choices of ∠A give triangles of different shapes, and there is no end to the angles between 0° and 90° (30°, 30.5°, 30.25°, …).

AnswerInfinitely many different right-angled triangles exist: ∠A can be any angle between 0° and 90°, with ∠C = 90° − ∠A.

Watch this explained “How many is endlessly many”, 9:09 into Two independent ways to classify a triangle: by side and by angle · हिंदी में देखें

Open NCERT p. 171Checked by computer

1. Equilateral: construct one with a compass, say all sides 5 cm, and measure: every angle is 60°. It must be: equal sides face equal angles, so all three angles are equal, and as they add to 180°, each is 180° ÷ 3 = 60°.
2. 60° is less than 90°, so an equilateral triangle has no right angle and no obtuse angle. It can be neither right-angled nor obtuse-angled.
3. Isosceles and right-angled: draw AB = 5 cm, make ∠A = 90° at A, mark AC = 5 cm on the new arm and join BC. AB = AC, so it is isosceles, and ∠A = 90°, so it is right-angled. (The other two angles measure 45° each.)
4. Isosceles and obtuse-angled: do the same with ∠A = 120°. AB = AC = 5 cm, so it is isosceles, and ∠A = 120° is obtuse. (The other two angles measure 30° each.)

AnswerEquilateral and right-angled: no. Equilateral and obtuse-angled: no (every angle of an equilateral triangle is 60°). Isosceles and right-angled: yes, e.g. AB = AC = 5 cm with ∠A = 90°. Isosceles and obtuse-angled: yes, e.g. AB = AC = 5 cm with ∠A = 120°.

Watch this explained “Two labels at once”, 7:09 into Two independent ways to classify a triangle: by side and by angle · हिंदी में देखें