Mensuration Model Questions Set 2 Practice Questions Answers Test with Solutions & More Shortcuts
Mensuration PRACTICE TEST [2 - EXERCISES]
Mensuration Model Questions Set 1
Mensuration Model Questions Set 2
Question : 21
In the figure given above, AB is a diameter of the circle with centre O and EC = ED. What is ∠EFO?
a) 25°
b) 15°
c) 20°
d) 30°
Answer »Answer: (c)
The Given,
⇒ ∠EDC = ∠ECD = 35°
Since, ∠OCD = 55°
Then, ∠OCE = 20°
By using then theorem that triangle on the same segment of a circle makes as equal angles.
Here, OE is a segment, which makes a ΔOFE and ΔOCE.
Therefore, ∠OCE = ∠EFO = 20°
Question : 22
If the sum of the circumferences of two circles with radii $R_1$ and $R_2$ is equal to the circumference of a circle of radius R, then
a) $R_1 + R_2$ < R
b) $R_1 + R_2$ = R
c) $R_1 + R_2 > R_3$
d) Nothing definite can be said about the relation among $R_1.R_2$ and R
Answer »Answer: (b)
Question : 23
Assertion (A)
Triangles on the same base and between the same parallel lines are equal in area.
Reason (R)
The distance between two parallel lines is same everywhere.
a) A is true but R is false
b) Both A and R individually true and R is the correct explanation of A
c) Both A and R are individually true but R is not the correct explanation of A
d) A is false but R is true
Answer »Answer: (b)
A. By the properties of triangle, it is true.
R. It is also true that the distance between two parallel lines is same everywhere.
Hence, A and R are true and R is the correct explanation of A.
Question : 24
The cross section of a canal is a trapezium in shape. If the canal is 7 metres wide at the top and 9 metres at the bottom and the area of cross-section is 1280 square metres, find the height of the canal.
a) 154 metres
b) 160 metres
c) 172 metres
d) None of these
Answer »Answer: (b)
Let the height of canal = h.
Then, area of canal = $1/2$ × h(9 + 7)
or = 1280 = $1/2$ h(16)
∴ h = ${1280 × 2}/{16}$ = 160m
Question : 25
In the figure given above, what is ∠CBA?
a) 50°
b) 30°
c) 45°
d) 60°
Answer »Answer: (d)
The sum of opposite angles in cyclic quadrilateral is always 180°.
∴ ∠ACQ + ∠APQ = 180°
75° + ∠APQ = 180°
∴ ∠APQ = 180° – 75° = 105°
∠ACQ + ∠QCR = 180° (∵ Straight line)
75° + ∠QCR = 180°
∠QCR = 180° – 75° = 105°
∠CQR = 180° – 105° – 30° = 45°
Since, ∠APQ + ∠BPQ = 180° (Straight line)
∴ 105° + ∠BPQ = 180°
∠BPQ = 75°
In ΔBPQ ∠B + ∠P + ∠Q = 180°
∠B + 75° + 45° = 180°
⇒ ∠B = 60° ∴ ∠CBA = 60°
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