# QUESTIONS ON CIRCLE (PART-4)

#### QUESTIONS ON CIRCLE (PART-4)

Most of these questions are taken from the previous year’s examinations conducted by the Staff Selection Commission (SSC) on Geometry (Circle) of the Quantitative Aptitude (Maths) section of the following exams as well as other similar exams. They are all solved and supported by detailed explanation.

1. Combined Graduate Level (CGL) Exam Tier-I & Tier-II

2. Combined Higher Secondary (10+2) Exam (CHSL) Tier-I

3. SI in Delhi Police and CPO Exam Paper-I & Paper-II

#### QUERY 31

Two circle touch externally. The sum of their areas is 130π cm². and the distance between their centres is 14 cm. Find the radii of those two circles.

A) 13,    1
B) 12,    2
C) 11,    3
D) 10,    4

MAHA GUPTA
Some questions are better done applying hit and trial approach; it’s like that.

As the circles are touching externally and the difference of their radii is 14 cm, so possible pairs of radii in natural numbers: (13, 1), (12, 2), (11, 3), (10, 4) and like that.

First two pairs are not possible at all as square of even one exceeds 130; let’s try 11, 3. 11² + 3² = 130; the process ends. Now no need to  try another

So 11 and 3 are two required radii (option ‘C’)

NOTE: It will be done more quickly with the help of given answer options.

#### QUERY 32

Find the ratio of area of a square inscribed in a semicircle to the area of square inscribed in the circle in case of radius is r.

A) 1 : 2
B) 1 : 4
C) 2 : 5
D) 3 : 4

MAHA GUPTA
Let ABCD is the square inscribed in the semi-circle with center ‘O’, and the side CD lies on the diameter of the semi-circle; and also let the point M the mid-point of AB.

Now Join OB and OM to get the right triangle OMB with OB as the hypotenuse.

Therefore OM² + MB² = OB²
Obviously OM = side of the square inscribed in the semi-circle, MB = half of the side; and OB = radius

Now letting the side of the square as ‘a’
a² + (a/2)² = r²
By solving the above a² = 4r²/5
But a² is the area of the square we know
Hence area of the square inscribed in a circle = 4r²/5

Now, diagonal of the square inscribed in the circle must be equal to 2r
Therefore its area = 2r²/2 = 2r²                       ……area of the square = (diagonal²)/2

Hence the required ratio = 4r²/5 : 2r² = 2/5 : 1 = 2:5 (option ‘C’)

#### QUERY 33

Three circles of radii 3.5 cm each are placed in such a way that each touches the other two. The area of the portion enclosed by the circles is?

A) 1.975 cm²
B) 1.967 cm²
C) 19.67 cm²
D) 21.21 cm²

SHIKHA CHAUN We know that if all the centers are joined of such circles, we get an equilateral triangle; thus ΔABC is an equilateral triangle.

Therefore, area of ΔABC = (√3/4)a²
= √3/4 × 7²
= 21.217 cm²

Now we find the area of the 3 sectors of these circles
Area of a sector = θ/360 × πr²
= 60/360 × 22/7 × 3.5 × 3.5
Therefore area of all the 3 sectors = 3(60/360 × 22/7 × 3.5 × 3.5)
=19.25

Now, we have to find the area of the shaded region
Area of the shaded region = Area ΔABC – Area of the 3 sectors
= 21.217 – 19.25
= 1.967 cm² (option ‘B’)

#### QUERY 34

If a chord of a circle is equal to the radius of the circle, the the angle subtended by the chord at a point on the minor arc is?

A) 30°
B) 150°
C) 60°
D) 120°

MAHA GUPTA
Let O be the center and AB the chord
Obviously ΔOAB will be an equilateral triangle; means ∠AOB = 60

We know that angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle; so the angle subtended by the chord AB in the major arc = 30

Hence angle subtended by it in the minor arc = 180 – 30 = 150 (option ‘B’)

#### QUERY 35

C1 and C2 are two concentric circles with centers at O. Their radii are 12 cm and 3 cm respectively. B and C are the points of contact of two tangents drawn to C2 from a point A lying on the circle C1. Then the area of the quadrilateral ABOC is? SHIV KISHOR AC = √(AO² – OC²)
= √(144 – 9)
= √135
= 3√15

Now the area of ABOC = 2 × area ΔACO
= 2 × (1/2 × OC × AC)
= 2 × 1/2 × 3 × 3√15
= 9√15 (option ‘C’)

#### QUERY 36

Three equal circles of unit radius touch each other. Then, the area of the circle circumscribing the three circles is?

A) 6π(2 + √3)²
B) π6(2 + √3)²
C) π3(2 + √3)²
D) 3π(2 + √3)²

SHIV KISHOR Radius of the circle circumscribing these three circles = ON
= OM + MN
= 1 × sec 30° + 1
= (2 + √3)/√3

Therefore the required area = π[(2 + √3)√3
π3(2 + √3)² (option ‘C’)

#### QUERY 37

There are 3 circles each with radius 14 cm, each circle touching other two. If all these circles covered with a rubber band then what is the total length of that rubber band?

A) 160 cm
B) 165 cm
C) 172 cm
D) 205 cm On observing you’ll see that every covered part of the circle is 1/3 of its circumference, so three covered parts of the circle = circumference of 1 circle

Therefore the total length of that rubber band = 3*28 + circumference of one circle
= 84 + 2*(22/7)*14
= 84 + 88
= 172 cm (option ‘C’)

#### QUERY 38

On a semicircle with diameter AD, chord BC is parallel to the diameter AD. Also, AB = CD = 2, while AD=8, what is the length of BC?

A) 6.5
B) 7
C) 5
D) 5.75

SHIV KISHOR (Option ‘B’)

MAHA GUPTA We know that any angle of the semi-circle is = 90

Thus, ∠ACD is a right angle, and ∆ACD is a right triangle.

We also know that in any right triangle, a height drawn through the right angle forms similar triangles.
Thus, in ∆ACD, height CE forms the following similar triangles:
∆AEC, ∆CED and ∆ACD.

Corresponding sides of similar triangles are always in the same ratio.
Thus, in ∆ACD and ∆CED:
CD/DA = ED/DC
=> 2/8 = ED/2
=> ED = 1/2.

Similarly, AF = 1/2.

Now, FE = AD – (AF + DE) = 8 – (1/2 + 1/2) = 7.
Since BC || AD, quadrilateral BCEF is a rectangle, implying that BC = FE.
Thus, BC = FE = 7 (option ‘B’)

#### QUERY 39

If the area of region bounded by the inscribed and the circumscribed circles of a square is 9π, then area of square is ?

A) 6π
B) 5π
C) 25
D) 36

MAHA GUPTA
Area of the circumscribed circle of a square is double the area of the circle inscribed in it; and area of the inscribed circle is π(a/2)², where a = side of the square

Therefore, 2π(a/2)² – π(a/2)² = 9π
On solving a = 6

Therefore, area of the square = 6² = 36 (option ‘D’)

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