Q1. An equilateral ∆ABC is inscribed in circle with centre O. Then ∠BOC is equal to-
(a) 120°
(b) 75°
(c) 180°
(d) 60°
(We know that the angle subtended by an arc of a circle at the centre is twice the angle subtended by the arc at any other point on the circumference of the circle.
∴∠BOC=2∠BAC
=2×60°
=120° )
Q2. In a triangle, an exterior angle equals the sum of the two interior-
(a) Two right angles
(b) Two acute angles
(c) Two interior opposite angles
(d) Adjacent angles
( In a triangle an exterior angle equals the sum of the two interior opposite angles)
Q3. If two cubes of dimensions 2 cm by 2 cm by 2 cm are placed side by side, what would be the dimensions of the resulting cuboid?
(a) 4, 2, 2
(b) 4, 3, 2
(c) 4, 4, 2
(d) 4, 1. 2
( Length of cuboid = 2 + 2 = 4 cm
Breadth of cuboid = 2 cm
Height of cuboid = 2 cm )
Q4. The number of triangles in the given figure is-
(a) 5
(b) 6
(c) 9
(d) 10
(The total number of triangles are 10)
Q5. The measure of an angle is twice the measure of its supplementary angle. Then, its measure is-
(a) 120°
(b) 60°
(c) 100°
(d) 90°
( Let the angle be x, then its supplementary be (180° – x).
∴x=2(180°-x)
⇒3x=360°
⇒x=120° )
Q6. Lines l and m intersect at O, forming angles as shown in figure. If x = 45° , then values of y, z and u are-
(a) 45°, 135°, 135°
(b) 135°, 135°, 45°
(c) 135°, 45°, 135°
(d) 115°, 45°, 115°
( x = z [∴ Vertically opposite angles]
⇒ x = 45° ⇒ z = 45°
Also, y + x = 180° [∴ Linear pair]
⇒ y = 180° – 45° ⇒ y = 135°
Also y = u
[∴ Vertically opposite angles]
⇒ u = 135° )
Q7. In figure, determine the value of y.
(a) 25°
(b) 35°
(c) 15°
(d) 40°
(Since OA and OB are opposite rays.
⇒ ∠AOC+∠COF+∠FOB=180°
⇒ 5y + 5y + 2y = 180°
⇒ 12y = 180°
⇒ y=(180°)/12=15°)
Q8. In the given figure if PQ∥RS,∠QPT=115° and ∠PTR=20° then ∠SRT is-
(a) 155°
(b) 150°
(c) 135°
(d) 95°
( Since, ∠QPL+∠PLR=180° [∴ Linear pair]
⇒ ∠PLR=180°-115°=65°
And ∠RLP=∠LRT+∠RTL
⇒ ∠LRT=∠RLP-∠RTL
= 65° – 20° = 45°
Also, ∠SRT+∠LRT=180°
⇒ ∠SRT=180°-∠LRT=180°-45°=135°
∴ ∠SRT=135° )
Q9. AB and CD are two parallel lines. PQ cuts AB and CD at E and F, respectively. EL is the bisector of ∠EFB . If ∠LEB=35°, then ∠CFQ is-
(a) 130°
(b) 85°
(c) 110°
(d) 95°
( Given ∠LEB=35°
⇒ ∠FEB=2×∠LEB=70°
⇒ ∠CFE=∠FEB=70° [∴ Alternate angles]
∴ ∠CFQ=180°-∠CFE=180°-70°=110° )
Q10. AB and CD are two parallel lines. The points B and C are joined such that ∠ABC=65°. A line CE is drawn making angle of 35° with the line CB, EF is drawn parallel to AB, 3 as shown in figure, then ∠CEF is-
(a) 160°
(b) 155°
(c) 150°
(d) 145°
(Since, AB∥CD
⇒ ∠BCD=∠ABC=65°
⇒∠ECD=65°-∠BCE
=65°-35°=30°
Now, ∠CEF+∠ECD=180° [∴ Linear pair]
∴ ∠CEF=180°-30°=150° )
