Q1.
The sides of a triangle are 13 cm, 14 cm, and 15 cm. What is its area?
- 84 sq cm
- 90 sq cm
- 72 sq cm
- 105 sq cm
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Answer: A. 84 sq cm
We use Heron's formula to find the area of the triangle. First, calculate the semi-perimeter (s) = (a+b+c)/2. s = (13+14+15)/2 = 42/2 = 21 cm. Area = sqrt(s(s-a)(s-b)(s-c)) = sqrt(21 * (21-13) * (21-14) * (21-15)) = sqrt(21 * 8 * 7 * 6) = sqrt(7056) = 84 sq cm.
Q2.
In a triangle ABC, if angle A = 60 degrees and angle B = 75 degrees, what is the measure of angle C?
- 45 degrees
- 55 degrees
- 60 degrees
- 75 degrees
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Answer: A. 45 degrees
The sum of angles in a triangle is 180 degrees. So, Angle C = 180 - (Angle A + Angle B) = 180 - (60 + 75) = 180 - 135 = 45 degrees.
Q3.
A triangle has sides of length 5 cm, 12 cm, and 13 cm. What type of triangle is it?
- Equilateral triangle
- Isosceles triangle
- Right-angled triangle
- Obtuse-angled triangle
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Answer: C. Right-angled triangle
Check if the Pythagorean theorem (a^2 + b^2 = c^2) holds for the given side lengths. 5^2 + 12^2 = 25 + 144 = 169. Also, 13^2 = 169. Since the sum of the squares of the two shorter sides equals the square of the longest side, it is a right-angled triangle.
Q4.
The perimeter of a right-angled triangle is 60 cm. If its hypotenuse is 25 cm, what is the area of the triangle?
- 150 sq cm
- 120 sq cm
- 180 sq cm
- 100 sq cm
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Answer: A. 150 sq cm
Let the perpendicular sides be 'a' and 'b', and the hypotenuse 'c'. Given c = 25 cm. Perimeter = a + b + c = 60 cm. So, a + b = 60 - 25 = 35 cm. By Pythagorean theorem, a^2 + b^2 = c^2 = 25^2 = 625. We know (a+b)^2 = a^2 + b^2 + 2ab. Substitute values: 35^2 = 625 + 2ab => 1225 = 625 + 2ab => 2ab = 600 => ab = 300. The area of a right-angled triangle = (1/2) * base * height = (1/2) * ab = (1/2) * 300 = 150 sq cm.
Q5.
In triangle ABC, AD, BE, and CF are the medians intersecting at G. If the length of AD is 12 cm, what is the length of AG?
- 6 cm
- 8 cm
- 9 cm
- 4 cm
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Answer: B. 8 cm
The medians of a triangle intersect at the centroid (G). The centroid divides each median in the ratio 2:1, with the longer part towards the vertex. So, AG:GD = 2:1. Given AD = 12 cm. Therefore, AG = (2/3) * AD = (2/3) * 12 = 8 cm.
Q6.
Two similar triangles have areas of 64 sq cm and 100 sq cm respectively. If a side of the first triangle is 8 cm, what is the length of the corresponding side of the second triangle?
- 10 cm
- 12.5 cm
- 15 cm
- 12 cm
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Answer: A. 10 cm
For similar triangles, the ratio of their areas is equal to the square of the ratio of their corresponding sides. Area1/Area2 = (Side1/Side2)^2. Given Area1 = 64 sq cm, Area2 = 100 sq cm, Side1 = 8 cm. So, 64/100 = (8/Side2)^2. Taking the square root of both sides: sqrt(64/100) = 8/Side2 => 8/10 = 8/Side2. Therefore, Side2 = 10 cm.
Q7.
In triangle ABC, AD is the angle bisector of angle A, meeting BC at D. If AB = 8 cm, AC = 12 cm, and BC = 10 cm, what is the length of BD?
- 4 cm
- 5 cm
- 6 cm
- 4.5 cm
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Answer: A. 4 cm
According to the Angle Bisector Theorem, if AD is the angle bisector of angle A, then it divides the opposite side BC in the ratio of the other two sides. So, AB/AC = BD/DC. Given AB = 8 cm, AC = 12 cm. So, BD/DC = 8/12 = 2/3. Since BC = BD + DC = 10 cm, we can write BD = (2/(2+3)) * BC = (2/5) * 10 = 4 cm.
Q8.
In triangle ABC, O is the orthocenter. If angle BAC = 70 degrees, what is the measure of angle BOC?
- 110 degrees
- 100 degrees
- 120 degrees
- 90 degrees
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Answer: A. 110 degrees
The orthocenter (O) is the intersection point of the altitudes of a triangle. The angle formed by the orthocenter with two vertices is supplementary to the angle at the third vertex. So, Angle BOC = 180 degrees - Angle BAC. Given Angle BAC = 70 degrees. Therefore, Angle BOC = 180 - 70 = 110 degrees.
Q9.
In triangle PQR, I is the incenter. If angle QIR = 125 degrees, what is the measure of angle P?
- 70 degrees
- 75 degrees
- 80 degrees
- 60 degrees
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Answer: A. 70 degrees
The incenter (I) is the intersection point of the angle bisectors of a triangle. The angle formed by the incenter with two vertices is given by the formula: Angle QIR = 90 degrees + (Angle P)/2. Given Angle QIR = 125 degrees. So, 125 = 90 + (Angle P)/2. Subtract 90 from both sides: 125 - 90 = (Angle P)/2 => 35 = (Angle P)/2. Therefore, Angle P = 35 * 2 = 70 degrees.
Q10.
In triangle ABC, D is a point on BC such that BD:DC = 2:3. E is a point on AD such that AE:ED = 1:2. If the area of triangle ABC is 90 sq cm, what is the area of triangle ABE?
- 12 sq cm
- 18 sq cm
- 24 sq cm
- 30 sq cm
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Answer: A. 12 sq cm
The area of a triangle can be divided proportionally by a line segment from a vertex to the opposite side.
1. Consider triangle ABC and line AD. Since D is on BC such that BD:DC = 2:3, the ratio of the areas of triangle ABD and triangle ADC is equal to the ratio of their bases (BD:DC), as they share the same height from A. Area(ABD) = (BD / BC) * Area(ABC) = (2 / (2+3)) * 90 = (2/5) * 90 = 36 sq cm.
2. Now consider triangle ABD and line BE. Since E is on AD such that AE:ED = 1:2, the ratio of the areas of triangle ABE and triangle EBD is equal to the ratio of their bases (AE:ED), as they share the same height from B. Area(ABE) = (AE / AD) * Area(ABD) = (1 / (1+2)) * 36 = (1/3) * 36 = 12 sq cm.
