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41. The circumferences of two circle are 132 metres and 176 metres respectively. What is the difference between the area of the larger circle and the smaller circle ?

A. 1048 m²

B. 1076 m²

C. 1078 m²

D. 1090 m²

Discuss

Solution:

\begin{aligned} \Rightarrow 2 π R_{1} = 132 \\ \Rightarrow R_{1} = \frac{132 \times 7}{2 \times 22} \\ \Rightarrow R_{1} = 21 m \\ \Rightarrow 2 π R_{2} = 176 \\ \Rightarrow R_{2} = \frac{176 \times 7}{2 \times 22} \\ \Rightarrow R_{2} = 28 m \\ ∴ Required difference : \\ = π (R_{2}^{2} - R_{2}^{2}) \\ = π (R_{2} + R_{1}) (R_{2} - R_{1}) \\ = (\frac{22}{7} \times 49 \times 7) m^{2} \\ = 1078 m^{2} \end{aligned}

42. The ratio of the radii of two circles is 1 : 3. Then the ratio of their areas is :

A. 1 : 3

B. 1 : 6

C. 1 : 9

D. None of these

Discuss

Solution:

Let the radii of the two circle be r and 3r respectively.
Then, required ratio :

= \frac{π r^{2}}{π {(3 r)}^{2}} = \frac{π r^{2}}{9 π r^{2}} = 1 : 9

43. The circumference of a circular ground is 88 metres. A strip of land, 3 metres wide, inside and along the circumference of the ground is to be levelled. What is the budgeted expenditure if the levelling costs Rs. 7 per square metre ?

A. Rs. 1050

B. Rs. 1125

C. Rs. 1325

D. Rs. 1650

Discuss

Solution:

Let the radius of the ground be R metres.
Then,

\begin{aligned} \Rightarrow 2 π R = 88 \\ \Rightarrow R = (\frac{88 \times 7}{2 \times 22}) \\ \Rightarrow R = 14 m \end{aligned}

Area of land strip :

\begin{aligned} = π [{(14)}^{2} - {(11)}^{2}] m^{2} \\ = (\frac{22}{7} \times 25 \times 3) m^{2} \\ = (\frac{1650}{7}) m^{2} \end{aligned}

∴ Cost of levelling :

\begin{aligned} = Rs . (\frac{1650}{7} \times 7) \\ = Rs . 1650 \end{aligned}

44. If the circumference of a circle is 100 units, then what will be the length of the arc described by an angle of 20 degrees ?

A. 5.55 units

B. 4.86 units

C. 5.85 units

D. None of these

Discuss

Solution:

2 π r = 100

So, length of the arc :

\begin{aligned} = \frac{2 π r θ}{360} \\ = (\frac{100 \times 20}{360}) units \\ = (\frac{50}{9}) units \\ = 5.55 units \end{aligned}

45. If in a triangle, the area is numerically equal to the perimeter, then the radius of the inscribed circle of the triangle is :

A. 1

B. 1.5

C. 2

D. 3

Discuss

Solution:

\begin{aligned} Radius = \frac{Area}{Semi - perimeter} \\ Radius = (Area \times \frac{2}{Area}) \\ Radius = 2 \end{aligned}

46. Two equal circle are drawn in square in such a way that a side of the square forms diameter of each circle. If the remaining area of the square is 42 cm², how much will the diameter of the circle measure ?

A. 3.5 cm

B. 4 cm

C. 14 cm

D. 7.5 cm

Discuss

Solution:

Let length of each side of square = 2π
According to the question,

\begin{aligned} \frac{π r^{2}}{2} + \frac{π r^{2}}{2} + 42 = Area of square \\ \Rightarrow π r^{2} + 42 = 4 r^{2} \\ \Rightarrow 4 r^{2} - π r^{2} = 42 \\ \Rightarrow r^{2} (4 - \frac{22}{7}) = 42 \\ \Rightarrow r^{2} (\frac{28 - 22}{7}) = 42 \\ \Rightarrow \frac{6 r^{2}}{7} = 42 \\ \Rightarrow r^{2} = \frac{42 \times 7}{6} \\ \Rightarrow r^{2} = 7 \times 7 \\ \Rightarrow r = 7 \\ ∴ 2 r = 14 c m \end{aligned}

47. The base of triangle is 15 cm and height is 12 cm the height of another triangle of double the area having base 20 cm is :

A. 22 cm

B. 20 cm

C. 18 cm

D. 10 cm

Discuss

Solution:

Given base of triangle and its height is 15 cm and 12 cm respectively
Area of first triangle :

\begin{aligned} = \frac{1}{2} \times Base \times Height \\ = \frac{1}{2} \times 15 \times 12 \\ = 90 sq . cm \end{aligned}

According to the question,
Let height of triangle be h cm
Area of new triangle = 180 sq. cm
Base = 20 sq. cm

\begin{aligned} \Rightarrow 180 = \frac{1}{2} \times 20 \times Height \\ \Rightarrow h = \frac{2 \times 180}{20} \\ \Rightarrow h = 18 cm \end{aligned}

48. The difference between the length and breadth of a rectangle is 23 m. If its perimeter is 206 m, then its area is :

A. 1520 m²

B. 2420 m²

C. 2480 m²

D. 2520 m²

Discuss

Solution:

We have :
(l - b) = 23 and 2(l + b) = 206 or (l + b) = 103
Solving the two equations, we get :
l = 63 and b = 40
∴ Area = (l × b) = (63 × 40)m² = 2520 m²

49. The length of a rectangle is three times of its width. If the length of the diagonal is $8 \sqrt{10}$ cm, then the perimeter of the rectangle is :

A. $15 \sqrt{10}$ cm

B. $16 \sqrt{10}$ cm

C. $24 \sqrt{10}$ cm

D. 64 cm

Discuss

Solution:

Let breadth = x cm
Then, length = 3x cm

\begin{aligned} \Rightarrow x^{2} + {(3 x)}^{2} = {(8 \sqrt{10})}^{2} \\ \Rightarrow 10 x^{2} = 640 \\ \Rightarrow x^{2} = 64 \\ \Rightarrow x = 8 \end{aligned}

So, length = 24 cm and breadth = 8 cm
∴ Perimeter = [2(24 + 8)] cm = 64 cm

50. A typist uses a paper 30 cm by 15 cm. He leaves a margin of 2.5 cm at the top and bottom and 1.25 cm on either side. What percentage of paper area is approximately available for typing ?

A. 60%

B. 65%

C. 70%

D. 80%

Discuss

Solution:

Area of the sheet = (30 × 15) cm² = 450 cm²
Area used for typing
= [(30 - 5) × (15 - 2.5)] cm²
= 312.5 cm²
∴ Required percentage :

\begin{aligned} = (\frac{312.5}{450} \times 100) % \\ = 69.4 % \approx 70 % \end{aligned}

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