Circle-RRB

Circle

A circle is the set of all points in a plane equidistant from a fixed point. Fixed point is called centre of the circle and the distance from centre to any where of the line is called radius of the circle.

Let r is the radius of circle, then diameter ‘d’ = 2r

1. Circumference of a circle = 2 $\pi$ r

2. Area of a circle = $\pi$ r²

3. Length of an arc = $\frac{ ^{2\pi r}}{360} \theta$ where $\theta$ is the central angle

4. Area of a sector $\frac{1}{2}$ (arc × r) = $\frac{\pi r^{2}}{360}$ $\theta$

5. Circumference of a semi-circle = $\pi$ r

6. Area of semi-circle = $\frac{\pi r^{2}}{2}$

Question – 1

Given that a “12-inch pizza” means circular pizza with a diameter of 12 inches, changing from an 8-inch pizza to a 12-inch pizza gives you approximately what percent increase in the total amount of pizza?

A. 33
B. 50
C. 67
D. 80
E. 125

Solution: (E) 125

The 8-inch pizza has a radius of r = 4, so the area is $\pi$ r $^{2}$ = 16 $\pi$ . That area is how much pizza you get. The 12-inch pizza has a radius of r = 6 and an area of 36 $\pi$ . When you change from 16 to 36, what is the percentage change? That’s more than double, so it must be a percent greater than 100%. The only answer choice greater than 100% is answer E.

Question – 2

What is the diameter of circle Q?

Statement I — the circumference of Q is C = 12 $\pi$

Statement II — the area of Q is A = 36 $\pi$

Solution: D

Statement I: if you know the circumference, then you can use C = 2 $\pi$ r to solve for the radius, or C = $\pi$ d to solve for the diameter. Either way, you can find the diameter, so this statement by itself is sufficient.

Statement II: if you know the area, you can find the radius, and then double that to get the diameter. This statement by itself is also sufficient. Both statements alone are sufficient. Answer = D.

Question – 3

The ratio between the length and the breadth of a rectangular park is 3 : 2. If a man cycling along the boundary of the park at the speed of 12 km/hr completes one round in 8 minutes, then the area of the park (in sq. m) is:

A. 15360

B. 153600

C. 30720

D. 307200

Answer: Option B

Explanation:

Perimeter = Distance covered in 8 min = $\frac{1200}{60}$ × 8 = 1600 m

Let length = 3x metres and breadth = 2x metres.
Then, 2(3x + 2x) = 1600 or x = 160.
Length = 480 m and Breadth = 320 m.
Area = (480 x 320) $m^{2}$ = 153600 $m^{2}$ .(Ans)

Question – 4

An error 2% in excess is made while measuring the side of a square. The percentage of error in the calculated area of the square is:

A. 2%

B. 2.02%

C. 4%

D. 4.04%

Answer: Option D

Explanation:

100 cm is read as 102 cm.
$A_{{1}}$ = (100 x 100) $cm^{2}$ and $A_{{2}}$ (102 x 102) $cm^{2}$ .

$A_{{2}}$ – $A_{{1}}$ = (102)²– (100)²

= (102 + 100) x (102 – 100)

= 404 $cm^{{2}}$ .

Percentage error = 4.04%

Question – 5

The ratio between the perimeter and the breadth of a rectangle is 5 : 1. If the area of the rectangle is 216 sq. cm, what is the length of the rectangle?

A. 16 cm

B. 18 cm

C. 24 cm

D. Data inadequate

E. None of these

Answer: Option B

Explanation:

$\frac{2(l + b)}{B}=\frac{5}{1}$

$\Rightarrow2l + 2b = 5b$

$\Rightarrow3b = 2l$

b = (2/3) l

Then, Area = 216 $cm^{2}$

$\Rightarrow l\times b= 216$

$\Rightarrow \rho = 234$

$\Rightarrow l = 18 cm.(Ans)$

Question – 6

The percentage increase in the area of a rectangle, if each of its sides is increased by 20% is:

A. 40%

B. 42%

C. 44%

D. 46%

Answer: Option C

Explanation:

Let original length = x meters and original breadth = y meters.

Original area = (xy) square meter

New length = ( $\frac{120}{100}$ )x meter = ( $\frac{6}{5}$ )x meter

New breadth = ( $\frac{120}{100}$ )y meter = ( $\frac{6}{5}$ ) y meter

New Area = ( $\frac{6}{5}$ )x meter × ( $\frac{6}{5}$ )y meter = ( $\frac{36}{25}$ ) sq. meter

The difference between the original area = xy and new-area $\frac{36}{25}$ xy is

= ( $\frac{36}{25}$ )xy – xy

= xy( $\frac{36}{25}$ – 1)

= xy( $\frac{11}{25}$ ) or ( $\frac{11}{25}$ )xy

Increase % =(( $\frac{11}{25}$ )xy÷ xy) × 100 = 44%

Question – 7

A rectangular park 60 m long and 40 m wide has two concrete crossroads running in the middle of the park and rest of the park has been used as a lawn. If the area of the lawn is 2109 sq. m, then what is the width of the road?

A. 2.91 m

B. 3 m

C. 5.82 m

D. None of these

Answer: Option B

Explanation:

Area of the park = (60 x 40) $m^{{2}}$ = 2400 $m^{{2}}$ .

Area of the lawn = 2109 $m^{{2}}$ .

Area of the crossroads = (2400 – 2109) $m^{{2}}$ = 291 $m^{{2}}$ .

Let the width of the road be x meters. Then,

60x + 40x – $x^{{2}}$ = 291

$\Rightarrow$ $x^{2}$ – 100x + 291 = 0

$\Rightarrow$ (x – 97)(x – 3) = 0

$\Rightarrow$ x = 3.

Question 8:

The diagonal of the floor of a rectangular closet is 7 $\frac{1}{2}$ feet. The shorter side of the closet is 4 $\frac{1}{2}$ feet. What is the area of the closet in square feet?

A. $5_{4}^{1}$

B. $13_{2}^{1}$

C. 27

D. 37

Answer: Option C

Explanation:

Other side = $(\frac{15}{2})^{2} - (\frac{9}{2})^{2} ft$
= $\frac{225}{4} - \frac{81}{4} ft$

= $\frac{144}{4} ft$

=6 ft

Area of closet = (6 x 4.5) sq. ft = 27 sq. ft.

Question 9:

What is the area of the hall?

I. Material cost of flooring per square meter is Rs. 2.50
II. Labour cost of flooring the hall is Rs. 3500
III. Total cost of flooring the hall is Rs. 14,500.

A. I and II only

B. II and III only

C. All I, II and III

D. Any two of the three

E. None of these

Answer: Option C

Explanation:

I. Material cost = Rs. 2.50 per $m^{2}$
II. Labour cost = Rs. 3500.
III. Total cost = Rs. 14,500.

Let the area be A sq. metres.

Material cost = Rs. (14500 – 3500) = Rs. 11,000.

$\frac{5A}{2} = 11000 \Leftrightarrow A = (\frac{11000\times2}{5})$ = 4400 $m^{2}$ .

Thus, all I, II and III are needed to get the answer.

Correct answer is (C). (Ans)

Question 10 :

What is the area of a right-angled triangle?

I. The perimeter of the triangle is 30 cm.

II. The ratio between the base and the height of the triangle is 5 : 12.

III. The area of the triangle is equal to the area of a rectangle of length 10 cm.

A. I and II only

B. II and III only

C. I and III only

D. III, and either I or II only

E. None of these

Answer: Option A

Explanation:

From II, base : height = 5 : 12.

Let base = 5x and height = 12x.

Then, hypotenuse = $(5x)^{{2}}$ + $(12x)^{{2}}$ = 13x.

From I, perimeter of the triangle = 30 cm.

5x + 12x + 13x = 30 $\Leftrightarrow$ x = 1.

So, base = 5x = 5 cm, height = 12x = 12 cm.

Area = $\ (_{2}^{1}\times5\times12) _{{cm^{2}}}$ = 30 ${{cm^{2}}}$ .

Thus, I and II together give the answer.

Clearly III is redundant, since the breadth of the rectangle is not given.

Correct answer is (A).

Question 11:

What is the area of rectangular field?

I. The perimeter of the field is 110 meters.

II. The length is 5 meters more than the width.

III. The ratio between length and width is 6 : 5 respectively.

A. I and II only

B. Any two of the three

C. All I, II and III

D. I, and either II or III only

E. None of these

Answer: Option B

Explanation:

I. 2(l + b) = 110 $\Rightarrow$ l + b = 55.

II. l = (b + 5) $\Rightarrow$ l – b = 5.
III. $\frac{l}{b}=\frac{6}{5}\Rightarrow$ 5l – 6b = 0.

These are three equations in l and b. We may solve them pairwise.

Any two of the three will give the answer.
Correct answer is (B). (Ans)

Question 12:

What is the area of the given rectangle?

I. Perimeter of the rectangle is 60 cm.

II. Breadth of the rectangle is 12 cm.

III. Sum of two adjacent sides is 30 cm.

A. I only

B. II only

C. I and II only

D. II and III only

E. II and either I or III

Answer: Option E

Explanation:

From I and II, we can find the length and breadth of the rectangle and therefore the area can be obtained.

So, III is redundant.

Also, from II and III, we can find the length and breadth and therefore the area can be obtained.

So, I is redundant.

Correct answer is “II and either I or III”.

Question 13:

What is the cost painting the two adjacent walls of a hall at Rs. 5 per $m^{2}$ . which has no windows or doors?

I. The area of the hall is 24 sq. m.

II. The breadth, length and height of the hall are in the ratio of 4 : 6 : 5 respectively.

III. Area of one wall is 30 sq. m.

A. I only

B. II only

C. III only

D. Either I or III

E. All I, II and III are required.

Answer: Option C

Explanation:

From II, let l = 4x, b = 6x and h = 5x.

Then, area of the hall = (24 $x^{2}$ ) $m^{2}$ .

From I. Area of the hall = 24 $m^{2}$ .

From II and I, we get 24 $x^{2}$ = 24 $\Leftrightarrow$ x = 1.
l = 4 m, b = 6 and h = 5 m.Thus, area of two adjacent walls = [(l x h) + (b x h)] $m^{2}$ can be found out and so the cost of painting two adjacent walls may be found out. Thus, III is redundant.
Answer: Option C

Example 14:

The largest chord of a circle is known to be 10.1 cm. The radius of this circle must be ?

Solution:

The largest chord of a circle is its diameter. So,

Radius = Diameter/2

= 10.1/2

= 5.05 cm.

Example 15:

Consider the following statements

I. The tangent of a circle is a line that meets the circle in one and only one point.

II. The tangent of a circle at the end point of the diameter is perpendicular to the diameter.

Which of the above statements is/are correct?

Solution:

By definition of tangent,

A tangent to a circle is straight line that touches the circle at a single point. Also, tangent at the end points of a diameter of a circle is perpendicular to the diameter.

So, both statements are correct.