Geometry (Part-8)

Total Questions: 50

1. If two tangents inclined at an angle 120° are drawn to a circle of radius 6 cm, then what is the length (in cm) of each tangent? [SSC CHSL 04/08/2023 (1st Shift)]

Correct Answer: (b) 2√3
Solution:

2. PS and PT are two tangents from a point P outside the circle with centre O. If S and T are points on the circle such that ∠SPT = 130º, then the degree measure of ∠OST is equal to: [SSC CHSL 04/08/2023 (2nd Shift)]

Correct Answer: (c) 65°
Solution:

3. In a rhombus STUV, S and U are joined ∠SUV = 44º, ∠STU = 92º, what is the degree measure of 4∠SVU - 3∠TSU? [CHSL 04/08/2023 (2nd Shift)]

Correct Answer: (d) 236°
Solution:

4. MN is the diameter of a circle with centre O. P and S are two points on the circumference of the circle on either side of MN, such that ∠PMN = 50º and ∠MNS = 35º. What is the degree measure of the difference of ∠PMS and ∠PNS? [SSC CHSL 04/08/2023 (3rd Shift)]

Correct Answer: (d) 30°
Solution:

5. ∆ABC ~ ∆PQR, AB = 12cm PQ = 18 cm and the perimeter of ∆ABC is 45 cm. Find the perimeter of ∆PQR. [SSC CHSL 04/08/2023 (3rd Shift)]

Correct Answer: (c) 67.5 cm
Solution:

6. The sides of similar triangle ∆ABC and ∆DEF are in the ratio of √3/√5. If the area of ∆ABC is 90 cm², then the area of ∆DEF (in cm²) is: [SSC CHSL 04/08/2023 (4th Shift)]

Correct Answer: (a) 150
Solution:

7. The radius of a circle is 21 cm. What will be the length (in cm) of an arc of the circle that subtends a 22.5° angle at the center? (user π = 22/7 ) [SSC CHSL 04/08/2023 (4th Shift)]

Correct Answer: (a) 8.25
Solution:

8. Explain it- [SSC CHSL 07/08/2023 (1st Shift)]

Correct Answer: (d) 123°
Solution:

Required obtuse angle = 180º - (2 × 28.5º) = 123º

9. In ΔABC and ΔPQR, B = Q, C = R and AB = 2PQ, then the two triangles are ________. [SSC CHSL 07/08/2023 (1st Shift)]

Correct Answer: (c) similar but not congruent
Solution:

10. If the angles P, Q and R of ∆PQR satisfy the relation 2 R - P = Q - R, then find the measure of ∠R. [SSC CHSL 07/08/2023 (1st Shift)]

Correct Answer: (a) 45°
Solution: