Figure it Out
Q1. Use the points on the circle and/or the centre to form isosceles triangles.
Answer:
To form isosceles triangles, choose any point on the circle and join it to the centre and to another point on the circle. Since the radius from the centre to any point on the circle is equal, the triangle formed will have at least two equal sides (the radii).
Q2. Use the points on the circles and/or their centres to form isosceles and equilateral triangles. The circles are of the same size.
Answer:
For isosceles triangles: Choose one centre and one point on the other circle, and join to the other centre. Since radii are equal, two sides will be equal. For equilateral triangles: Use the intersection points of the two circles and both centres. Since all radii are equal (circles same size), the triangle formed by the two centres and an intersection point will have all sides equal.
Q1. 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
Answer:
For each triangle, use the given side lengths and included angle. (a) Draw base 3 cm, construct 75° at one endpoint, mark 7 cm on the other arm, join. (b) Draw base 6 cm, construct 25°, mark 3 cm, join. (c) Draw base 3 cm, construct 120°, mark 8 cm, join.
Q2. We have seen that triangles do not exist for all sets of sidelengths. Is there a combination of measurements in the case of two sides and the included angle where a triangle is not possible? Justify your answer using what you observe during construction.
Answer:
A triangle is always possible for any two sides and the included angle, because you can always draw one side, then the angle at one endpoint, and then mark the other side length on the other arm, and join the endpoints. The third side naturally forms. So no combination of two sides and the included angle makes a triangle impossible, since the two sides determine the third side uniquely.
Q1. Construct triangles for the following measurements: (a) 75°, 5 cm, 75° (b) 25°, 3 cm, 60° (c) 120°, 6 cm, 30°
Answer:
These are two angles and included side. (a) Draw base 5 cm, construct 75° at both endpoints, the arms meet to form the triangle. (b) Draw base 3 cm, construct 25° at one end and 60° at the other, arms meet. (c) Draw base 6 cm, construct 120° at one end and 30° at the other, arms meet.
Q2. Do triangles always exist? Do triangles exist for every combination of two angles and their included side? Explore.
Answer:
No. If the sum of the two given angles is 180° or more, the two arms will be parallel or diverge, so they never meet. Also if one angle is 90° or more and the other is also large, they may not meet. A triangle exists only if the sum of the two angles is less than 180°.
Q3. Find examples of measurements of two angles with the included side where a triangle is not possible.
Answer:
Examples: (a) 90° and 90° with any included side: sum = 180°, lines are parallel. (b) 100° and 90°: sum = 190° > 180°, lines diverge. (c) 150° and 40°: sum = 190° > 180°. (d) 120° and 60°: sum = 180°, lines are parallel.
Q4. It is clear that if the line from B is "inclined" sufficiently to the right, then it will not meet the line l. (a) Try to find a possible ?B (marked in the figure) for this to happen. (b) What could be smallest value of ?B for the lines to not meet?
Answer:
From the hint: when the line through B is parallel to the line through A, the angle ?B equals 180° - 40° = 140°. (a) Any ?B > 140° will cause the lines to not meet. (b) The smallest ?B for which lines do not meet is 140° (when the line is parallel). For ?B less than 140°, the lines will meet and form a triangle.
Q5. 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°
Answer:
A triangle is possible when sum of two angles < 180°.
(a) 30°: Possible: 80° (sum=110°), 100° (sum=130°). Not possible: 150° (sum=180°), 160° (sum=190°).
(b) 70°: Possible: 60° (sum=130°), 80° (sum=150°). Not possible: 110° (sum=180°), 120° (sum=190°).
(c) 54°: Possible: 80° (sum=134°), 90° (sum=144°). Not possible: 126° (sum=180°), 130° (sum=184°).
(d) 144°: Possible: 20° (sum=164°), 30° (sum=174°). Not possible: 36° (sum=180°), 40° (sum=184°).
Q6. 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°
Answer:
(a) 35° + 150° = 185° > 180°, cannot be angles of a triangle.
(b) 70° + 30° = 100° < 180°, can be angles of a triangle.
(c) 90° + 85° = 175° < 180°, can be angles of a triangle.
(d) 50° + 150° = 200° > 180°, cannot be angles of a triangle.
Q7. Like the triangle inequality, can you form a rule that describes the two angles for which a triangle is possible? Can the sum of the two angles be used for framing this rule?
Answer:
Yes. The rule: A triangle is possible with two given angles if and only if the sum of the two angles is less than 180°. If the sum is equal to or greater than 180°, no triangle exists. This is because the three angles must sum to 180°, so the third angle would be 180° - sum of the two, which must be positive.
Q8. 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°
Answer:
Using angle sum property: third angle = 180° - (sum of two given angles).
(a) 180° - (36° + 72°) = 180° - 108° = 72°
(b) 180° - (150° + 15°) = 180° - 165° = 15°
(c) 180° - (90° + 30°) = 180° - 120° = 60°
(d) 180° - (75° + 45°) = 180° - 120° = 60°
Q9. 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.
Answer:
No, a triangle cannot have all angles 70° because sum would be 210° > 180°. If two angles are 70°, third angle = 180° - 140° = 40°. For all angles to be equal, each must be 180°/3 = 60° (equilateral triangle).
Q10. Here is a triangle in which we know ?B = ?C and ?A = 50°. Can you find ?B and ?C?
Answer:
Let ?B = ?C = x. Using angle sum property: 50° + x + x = 180° => 50° + 2x = 180° => 2x = 130° => x = 65°. So ?B = ?C = 65°.
Q1. Construct a triangle ABC with BC = 5 cm, AB = 6 cm, CA = 5 cm. Construct an altitude from A to BC.
Answer:
Step 1: Construct triangle ABC with sides BC=5 cm, AB=6 cm, CA=5 cm. Step 2: To construct altitude from A to BC, place ruler along BC. Place set square with one edge on ruler and slide until vertical edge touches A. Draw the perpendicular from A meeting BC (or its extension) at D. AD is the altitude.
Q2. Construct a triangle TRY with RY = 4 cm, TR = 7 cm, ?R = 140°. Construct an altitude from T to RY.
Answer:
Step 1: Draw base RY = 4 cm. At R, construct angle 140°. Mark T such that TR = 7 cm. Join TY. Step 2: To construct altitude from T to RY, extend RY if needed. Place ruler along RY, set square with vertical edge touching T, draw perpendicular from T to line RY. The foot of the perpendicular lies on the extension of RY since angle R is obtuse.
Q3. Construct a right-angled triangle ?ABC with ?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 ?A and ?C take so that the other angle is 90°?]
Answer:
AC is the hypotenuse (side opposite the right angle). Draw AC = 5 cm. At A and C, any pair of acute angles that sum to 90° (since ?A + ?C = 90° because ?B = 90°) will work. So there are infinitely many such triangles because ?A can be any acute angle between 0° and 90°, and then ?C = 90° - ?A. Each choice gives a different triangle.
Q4. 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.
Answer:
(i) An equilateral triangle cannot be right-angled because all angles are 60°, not 90°. (ii) An equilateral triangle cannot be obtuse-angled because all angles are 60°, all acute.
An isosceles triangle can be right-angled: e.g., sides 3 cm, 3 cm, 3√2 cm (45-45-90 triangle). An isosceles triangle can be obtuse-angled: e.g., sides 5 cm, 5 cm, 9 cm (with obtuse angle between the equal sides). Construct these by drawing the base, then constructing the required angles at the base endpoints.