Similarity and Trigonometry Basics for Competitive Exams

Similarity and Trigonometry Basics - Problems on Triangles - Applications

Q. If the area of triangle XYZ is 24 cm² and the base is 8 cm, what is the height of the triangle?

A. 6 cm
B. 8 cm
C. 4 cm
D. 3 cm

Solution

Area = 1/2 * base * height. Thus, 24 = 1/2 * 8 * height. Therefore, height = 24 / 4 = 6 cm.

Correct Answer: A — 6 cm

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Q. If triangle GHI is similar to triangle JKL and the ratio of their corresponding sides is 3:5, what is the ratio of their areas?

A. 3:5
B. 9:25
C. 15:25
D. 5:3

Solution

The ratio of the areas of similar triangles is the square of the ratio of their corresponding sides. Therefore, (3/5)² = 9/25.

Correct Answer: B — 9:25

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Q. If triangle PQR is similar to triangle STU and the length of side PQ is 9 cm while side ST is 15 cm, what is the ratio of PQ to ST?

A. 3:5
B. 5:3
C. 9:15
D. 15:9

Solution

The ratio of corresponding sides of similar triangles is PQ:ST = 9:15, which simplifies to 3:5.

Correct Answer: A — 3:5

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Q. If two triangles are similar, and the lengths of the sides of the first triangle are 3, 4, and 5, what are the lengths of the corresponding sides of the second triangle if the shortest side is 6?

A. 6, 8, 10
B. 9, 12, 15
C. 12, 16, 20
D. 15, 20, 25

Solution

The ratio of the sides of similar triangles is constant. If the shortest side of the first triangle (3) corresponds to 6, then the sides are scaled by a factor of 2. Thus, the other sides are 4*2=8 and 5*2=10.

Correct Answer: A — 6, 8, 10

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Q. In a right triangle, if one leg is 8 cm and the hypotenuse is 10 cm, what is the length of the other leg?

A. 6 cm
B. 7 cm
C. 8 cm
D. 9 cm

Solution

Using the Pythagorean theorem: a² + b² = c². Here, 8² + b² = 10². Thus, 64 + b² = 100, so b² = 36, and b = 6 cm.

Correct Answer: A — 6 cm

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Q. In triangle DEF, if DE = 10 cm, DF = 15 cm, and angle E = 45 degrees, what is the length of EF using the Law of Cosines?

A. 5√2 cm
B. 10√2 cm
C. 15√2 cm
D. 20 cm

Solution

Using the Law of Cosines: EF² = DE² + DF² - 2 * DE * DF * cos(E). EF² = 10² + 15² - 2 * 10 * 15 * cos(45°) = 100 + 225 - 150√2. Thus, EF = 5√2 cm.

Correct Answer: A — 5√2 cm

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Q. In triangle MNO, if MN = 12 cm, NO = 16 cm, and angle M = 60 degrees, what is the length of MO using the Law of Cosines?

A. 8 cm
B. 10 cm
C. 12 cm
D. 14 cm

Solution

Using the Law of Cosines: MO² = MN² + NO² - 2 * MN * NO * cos(M). MO² = 12² + 16² - 2 * 12 * 16 * cos(60°) = 144 + 256 - 192 = 208. Thus, MO = √208 = 10 cm.

Correct Answer: B — 10 cm

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Similarity and Trigonometry Basics - Problems on Triangles - Applications

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