Q2. Two circles having radii r units intersect each other in such a way that each of them passes through the centre of the other. Then the length of their common chord is
(a) √2r units
(b) √3r units
(c) √5r units
(d) r units
Q3. The chord of a circle is equal to its radius. The angle subtended by this chord at the minor arc of the circle is
(a) 150°
(b) 60°
(c) 75°
(d) 120°
Q4. In ∆ABC, ∠BAC = 90° and AD⊥BC. If BD = 3 cm and CD = 4 cm, then length of AD is
(a) 2√3 cm
(b) 3.5 cm
(c) 6 cm
(d) 5 cm
Q5. The radii of two concentric circles are 17 cm and 25 cm, a straight line PQRS intersects the larger circle at the points P and S and intersects the smaller circle at the points Q and R. If QR = 16 cm, then the length (in cm.) of PS is
(a) 41
(b) 33
(c) 32
(d) 40
Q6. AB is a diameter of a circle with centre O. The tangents at C meets AB produced at Q. If ∠CAB = 34°, then measure of ∠CBA is
(a) 56°
(b) 68°
(c) 34°
(d) 124°
Q7. Among the angles 30°, 36°, 45°, 50° one angles cannot be an exterior angle of a regular polygon. The angle is
(a) 30°
(b) 36°
(c) 45°
(d) 50°
Q8. In ∆ABC, ∠A + ∠B = 65°, ∠B + ∠C = 140°, then find ∠B.
(a) 40°
(b) 25°
(c) 35°
(d) 20°
Q10. In a regular polygon, the exterior and interior angles are in the ratio 1 : 4. The number of sides of the polygon is
(a) 5
(b) 10
(c) 3
(d) 8
Solutions
S7. Ans.(d)
Sol. As we know that
No. of sides =(360°)/(External angle)
No. of sides (I-30°) =(360°)/(30°)=12
No. of sides (II-36°) =(360°)/(36°)=10
No. of sides (III-45°) =(360°)/(45°)=8
Number of sides (IV-50°) =(360°)/(50°)=36/5
∴ 50° cannot be exterior angle
S10. Ans.(b)
Sol. According to question
(Exterior angle)/(Interior angle)=1/4=x/4x
As we know that
Interior angle + Exterior angle = 180°
∴ x + 4x = 180°
5x = 180°
x = 36°
∴ No. of sides
=(360°)/(Exterior angle)
No. of sides =(360°)/(36°)=10
No. of sides = 10
