Figure it Out - 7.1 (Page 366)
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1. Use the points on the circle and/or the centre to form isosceles triangles.
Triangles formed using radii from the center to points on the circle are isosceles triangles because radii are equal in length.
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2. Use the points on the circles and/or their centres to form isosceles and equilateral triangles. The circles are of the same size.
Equilateral and isosceles triangles can be constructed by connecting intersection points of equal circles and their centers.
Figure it Out - 7.2 (Pages 372-374)
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1. We checked by construction that there are no triangles having side lengths 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.
Yes, by checking the triangle inequality. For 3, 4, and 8, $3 + 4 = 7$ which is not greater than 8. For 2, 3, and 6, $2 + 3 = 5$ which is not greater than 6.
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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
(a) 10 km, 10 km, 25 km: Does not exist ($10 + 10 < 25$)
(b) 5 mm, 10 mm, 20 mm: Does not exist ($5 + 10 < 20$)
(c) 12 cm, 20 cm, 40 cm: Does not exist ($12 + 20 < 40$)
Figure it Out - 7.3 (Page 378)
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1. Which of the following lengths can be the side lengths of a triangle? Explain your answers. Note that for each set, the three lengths have the same unit of measure.
(a) 2, 2, 5: No ($2 + 2 \not> 5$)
(b) 3, 4, 6: Yes ($3 + 4 > 6$)
(c) 2, 4, 8: No ($2 + 4 \not> 8$)
(d) 5, 5, 8: Yes ($5 + 5 > 8$)
(e) 10, 20, 25: Yes ($10 + 20 > 25$)
(f) 10, 20, 35: No ($10 + 20 \not> 35$)
(g) 24, 26, 28: Yes ($24 + 26 > 28$)
Figure it Out - 7.4 (Page 382)
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1. Check if a triangle exists for each of the following set of lengths:
(a) 1, 100, 100: Yes ($1 + 100 > 100$)
(b) 3, 6, 9: No ($3 + 6 = 9$)
(c) 1, 1, 5: No ($1 + 1 < 5$)
(d) 5, 10, 12: Yes ($5 + 10 > 12$) -
2. 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.
Yes, because $50 + 50 > 50$. In general, equilateral triangles exist for any positive sidelength since $s + s > s$ ($2s > s$) is always true for any positive number $s$.
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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: Possible values for third side $x$: 100, 100.5, 99.5, 100.2, 99.8 (strictly between 99 and 101)
(b) 5, 5: Possible values: 2, 4, 6, 8, 9 (any value strictly between 0 and 10)
(c) 3, 7: Possible values: 5, 6, 7, 8, 9 (any value strictly between 4 and 10)
Figure it Out - 7.5 (Page 386)
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1. Construct triangles for the following measurements where the angle is included between the sides:
(a) 3 cm, $75^{\circ}$, 7 cm
(b) 6 cm, $25^{\circ}$, 3 cm
(c) 3 cm, $120^{\circ}$, 8 cm
[Triangles can be successfully constructed for all valid side-angle-side combinations]
Figure it Out - 7.6 (Page 388)
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1. Construct triangles for the following measurements:
(a) $75^{\circ}$, 5 cm, $75^{\circ}$
(b) $25^{\circ}$, 3 cm, $60^{\circ}$
(c) $120^{\circ}$, 6 cm, $30^{\circ}$
[Triangles can be constructed when the sum of the two given angles is less than $180^{\circ}$]
Figure it Out - 7.7 (Pages 390-392)
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1. 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) $30^{\circ}$: (a) Possible: $50^{\circ}, 60^{\circ}$; (b) Not possible: $150^{\circ}, 160^{\circ}$
(b) $70^{\circ}$: (a) Possible: $40^{\circ}, 50^{\circ}$; (b) Not possible: $110^{\circ}, 120^{\circ}$
(c) $54^{\circ}$: (a) Possible: $60^{\circ}, 70^{\circ}$; (b) Not possible: $130^{\circ}, 140^{\circ}$
(d) $144^{\circ}$: (a) Possible: $20^{\circ}, 30^{\circ}$; (b) Not possible: $40^{\circ}, 50^{\circ}$ -
2. Determine which of the following pairs can be the angles of a triangle and which cannot:
(a) $35^{\circ}$, $150^{\circ}$: Cannot ($35^{\circ} + 150^{\circ} = 185^{\circ} \not< 180^{\circ}$)
(b) $70^{\circ}$, $30^{\circ}$: Can ($70^{\circ} + 30^{\circ} = 100^{\circ} < 180^{\circ}$)
(c) $90^{\circ}$, $85^{\circ}$: Can ($90^{\circ} + 85^{\circ} = 175^{\circ} < 180^{\circ}$)
(d) $50^{\circ}$, $150^{\circ}$: Cannot ($50^{\circ} + 150^{\circ} = 200^{\circ} \not< 180^{\circ}$)
Figure it Out - 7.8 (Pages 394-396)
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1. Find the third angle of a triangle (using a parallel line) when two of the angles are:
(a) $36^{\circ}$, $72^{\circ}$: $180^{\circ} - (36^{\circ} + 72^{\circ}) = 72^{\circ}$
(b) $150^{\circ}$, $15^{\circ}$: $180^{\circ} - (150^{\circ} + 15^{\circ}) = 15^{\circ}$
(c) $90^{\circ}$, $30^{\circ}$: $180^{\circ} - (90^{\circ} + 30^{\circ}) = 60^{\circ}$
(d) $75^{\circ}$, $45^{\circ}$: $180^{\circ} - (75^{\circ} + 45^{\circ}) = 60^{\circ}$ -
2. Can you construct a triangle all of whose angles are equal to $70^{\circ}$? If two of the angles are $70^{\circ}$ 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.
No, because $70^{\circ} + 70^{\circ} + 70^{\circ} = 210^{\circ} \neq 180^{\circ}$. If two angles are $70^{\circ}$, the third angle is $180^{\circ} - 140^{\circ} = 40^{\circ}$. If all angles are equal, each must measure $180^{\circ} \div 3 = 60^{\circ}$.
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3. Here is a triangle in which we know $\angle B=\angle C$ and $\angle A=50^{\circ}$. Can you find $\angle B$ and $\angle C$?
$\angle B + \angle C = 180^{\circ} - 50^{\circ} = 130^{\circ}$. Since $\angle B = \angle C$, each angle is $130^{\circ} \div 2 = 65^{\circ}$.
Figure it Out - 7.9 (Pages 406-408)
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1. Construct a triangle ABC with $BC=5$ cm, $AB=6$ cm, $CA=5$ cm. Construct an altitude from A to BC.
[Triangle ABC constructed successfully and perpendicular altitude drawn from vertex A to base BC using a set square]
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2. Construct a triangle TRY with $RY=4$ cm, $TR=7$ cm, $\angle R=140^{\circ}$. Construct an altitude from T to RY.
[Obtuse triangle TRY constructed and altitude dropped from vertex T to extended line RY]
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3. Construct a right-angled triangle ABC with $\angle B=90^{\circ}$, $AC=5$ cm. How many different triangles exist with these measurements?
Infinitely many different right-angled triangles exist with hypotenuse AC = 5 cm, as vertex B can lie anywhere on the semicircle with diameter AC.
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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.
(i) Equilateral right-angled/obtuse-angled: Not possible (all angles of an equilateral triangle are fixed at $60^{\circ}$).
(ii) Isosceles right-angled: Possible (angles $90^{\circ}, 45^{\circ}, 45^{\circ}$).
(iii) Isosceles obtuse-angled: Possible (e.g., angles $100^{\circ}, 40^{\circ}, 40^{\circ}$).
