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31. The diameter of a circle is equal to the perimeter of a square whose area is 3136 cm² . What is the circumference of the circle ?

A. 352 cm

B. 704 cm

C. 39424 cm

D. 1024 cm

Discuss

Solution:

Area of square = 3136 cm²
Side of squared =

\sqrt{3136}

= 56 cm
Perimeter of square :
= 4a
= (4 × 56) cm
= 224 cm
= diameter of circle
∴ Circumference of circle :

\begin{aligned} = π d \\ = \frac{22}{7} \times 224 \\ = 704 cm \end{aligned}

32. The area of a rectangular field is 2100 sq. meters. If the field is 60 metres long, what is its perimeter ?

A. 180 m

B. 210 m

C. 240 m

D. Cannot be determined

Discuss

Solution:

\begin{aligned} Breadth = \frac{Area}{Length} \\ = (\frac{2100}{60}) m \\ = 35 m \\ ∴ Perimeter = 2 (60 + 35) m \\ = 190 m \end{aligned}

33. How many metres of carpet 63 cm wide will be required to cover the floor of a room 14 metres by 9 metres ?

A. 185 metres

B. 200 metres

C. 210 metres

D. 220 metres

Discuss

Solution:

Area of the floor = (14 × 9) m² = 126 m²
∴ Length of the carpet :

\begin{aligned} = (\frac{126}{63} \times 100) m \\ = 200 metres \end{aligned}

34. If the length and breadth of a rectangular field are increased, the area increases by 50%. If the increase in length was 20 %, by what percentage was the breadth increased ?

A. 20%

B. 25%

C. 30%

D. Data inadequate

Discuss

Solution:

Let the original length and breadth of the rectangle be l and b respectively
Then, original area = lb
New length = 120% of l =

\frac{6 l}{5}

New area = 150% of lb =

\frac{3 l b}{2}

New breadth :

\begin{aligned} = (\frac{3 l b}{2} \times \frac{5}{6 l}) \\ = \frac{5 b}{4} \end{aligned}

Increase in breadth :

\begin{aligned} = (\frac{5 b}{4} - b) \\ = \frac{b}{4} \end{aligned}

∴ Increase % :

\begin{aligned} = (\frac{b}{4} \times \frac{1}{b} \times 100) % \\ = 25 % \end{aligned}

35. Total area of 64 small squares of a chessboard is 400 sq. cm. There is 3 cm wide border around the chess board. What is the length of the side of the chess board ?

A. 17 cm

B. 20 cm

C. 23 cm

D. 26 cm

Discuss

Solution:

Area of each square :

\begin{aligned} = (\frac{400}{64}) c m^{2} \\ = 6.25 c m^{2} \end{aligned}

Side of each small square :

\begin{aligned} = \sqrt{6.25} c m \\ = 2.5 c m \end{aligned}

Since there are 8 squares along each side of the chess board, we have :
Side = [(8 × 2.5) + 6] cm
= 26 cm

36. Area of a square natural lake is 50 sq. kms. A driver wishing to cross the lake diagonally, will have to swim a distance of :

A. 10 miles

B. 12 miles

C. 15 miles

D. None of these

Discuss

Solution:

Let the length of the diagonal be x km
Then,

\begin{aligned} \Rightarrow \frac{1}{2} x^{2} = 50 \\ \Rightarrow x^{2} = 100 \\ \Rightarrow x = 10 km \\ \Rightarrow x = (\frac{10}{1.6}) miles \\ \Rightarrow x = 6.25 miles \\ [∵ 1 mile = 1.609 km] \end{aligned}

37. A rectangular plank $\sqrt{2}$ metre wide is placed symmetrically on the diagonal of a square of side 8 metres as shown in the figure. The area of the plank is :

A. $7 \sqrt{2}$ sq. m

B. 14 sq. m

C. 98 sq. m

D. ( $16 \sqrt{2}$ - 3) sq. m

Discuss

Solution:

\begin{aligned} Let A P = A Q = x metres \\ Then, \\ \Rightarrow x^{2} + x^{2} = {(\sqrt{2})}^{2} \\ \Rightarrow 2 x^{2} = 2 \\ \Rightarrow x^{2} = 1 \\ \Rightarrow x = 1 m \\ S o, △ P A Q is isosceles . \\ ∴ P T = Q T = (\frac{\sqrt{2}}{2}) m = (\frac{1}{\sqrt{2}}) m \\ In △ P T A, we have : ∠ P T A = 90^{\circ} \\ ∴ A T^{2} = A P^{2} - P T^{2} \\ = 1^{2} - {(\frac{1}{\sqrt{2}})}^{2} \\ = 1 - \frac{1}{2} = \frac{1}{2} \\ O r, A T = (\frac{1}{\sqrt{2}}) m \\ Similarly, \\ C X = (\frac{1}{\sqrt{2}}) m \\ ∴ P S = Q R = X T \\ = A C - 2 \times A T \\ = [8 \sqrt{2} - (2 \times \frac{1}{\sqrt{2}})] m \\ = (8 \sqrt{2} - \frac{2}{\sqrt{2}}) m \\ = \frac{14}{\sqrt{2}} m \\ Area of the plank : \\ = (\frac{14}{\sqrt{2}} \times \sqrt{2}) m^{2} \\ = 14 m^{2} \end{aligned}

38. The area of a triangle whose sides are of lengths 3 cm, 4 cm and 5 cm is :

A. 8 cm²

B. 6 cm²

C. 10 cm²

D. None of these

Discuss

Solution:

Since 3² + 4² = 5²,
So it is a right-angled triangle with Base = 3 cm and Height = 4 cm
∴ Area :

\begin{aligned} = (\frac{1}{2} \times 3 \times 4) c m^{2} \\ = 6 c m^{2} \end{aligned}

39. The perimeter of a triangle is 30 cm and its area is 30 cm². If the largest side measures 13 cm, then what is the length of the smallest side of the triangle ?

A. 3 cm

B. 4 cm

C. 5 cm

D. 6 cm

Discuss

Solution:

Let the smallest side be x cm.
Then, other sides are 13 cm and (17 - x) cm
Let a = 13, b = x and c = (17 - x)
So, s = 15

\begin{aligned} Area = \sqrt{s (s - a) (s - b) (s - c)} \\ = \sqrt{15 \times 2 \times (15 - x) (x - 2)} \\ = \sqrt{30 (15 - x) (x - 2)} \\ ∴ 30 (15 - x) (x - 2) = {(30)}^{2} \\ \Rightarrow (15 - x) (x - 2) = 30 \\ \Rightarrow x^{2} - 17 x + 60 = 0 \\ \Rightarrow (x - 12) (x - 5) = 0 \\ \Rightarrow x = 12 or x = 5 \\ Smallest side  =  5 cm \end{aligned}

40. If a parallelogram with area P, a rectangle with area R and a triangle with area T are all constructed on the same base and all have the same altitude, then which of the following statements is false ?

A. P = R

B. P + T = 2R

C. P = 2T

D. T = $\frac{1}{2}$ R

Discuss

Solution:

Let each have base = b and height = h
Then,
P = b × h, R = b × h, T =

\frac{1}{2}

× b × h
So, P = R, P = 2T and T =

\frac{1}{2}

R, are all correct statements.
∴ Option B is false

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