The length of each side of an equilateral triangle having an area of 4√3 cm² is?
Answer: D
Use the equilateral-triangle area formula and solve for the side.
\[ A = \frac{\sqrt{3}}{4}a^{2} = 4\sqrt{3} \]
\[ a^{2} = \frac{4\sqrt{3} \times 4}{\sqrt{3}} = 16 \]
\[ a = 4 \text{ cm} \]
The \( \sqrt{3} \) cancels from both sides, which is why the answer comes out a whole number.
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The length and breadth of a rectangle is increased by 10% and 25% respectively. What is the increase in the area?
Answer: B
Both dimensions grow, so multiply the two factors.
\[ \frac{A'}{A} = (1 + 0.10)(1 + 0.25) = 1.10 \times 1.25 = 1.375 \]
\[ \text{increase} = 1.375 - 1 = 0.375 = 37.5\% \]
Adding the percentages gives 35%; the extra 2.5% is the cross term \( 0.10 \times 0.25 \), the part of the rectangle that gains in both directions at once.
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The perimeters of two squares are 40 cm and 32 cm. Find the perimeter of a third square whose area is equal to the difference of the areas of the two squares .
Answer: A
We know perimeter of square = 4(side)
So Side of first square = 40/4 = 10 cm
Side of second square = 32/4 = 8 cm
Area of third Square = 10*10 - 8*8
= 36 cm
So side of third square = 6 (because area of square = side*side)
Perimeter=4*Side=4*6=24 cm
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In a Circle of Radius 5 cm if AB and AC are two equal chords of length 6cm each, then the length of chord BC is
Answer: D
Place the centre at the origin with \( A = (0,\,5) \). By symmetry, equal chords \( AB \) and \( AC \) put B and C at the same height, so \( BC \) is horizontal.
For \( B = (x,\,y) \) on the circle with \( AB = 6 \):
\[ x^{2} + y^{2} = 25, \qquad x^{2} + (y - 5)^{2} = 36 \]
Subtracting:
\[ -10y + 25 = 11 \quad\Longrightarrow\quad y = 1.4 \]
\[ x^{2} = 25 - 1.96 = 23.04 \quad\Longrightarrow\quad x = 4.8 \]
So \( B = (4.8,\,1.4) \) and \( C = (-4.8,\,1.4) \):
\[ BC = 2 \times 4.8 = 9.6 = \frac{48}{5} \text{ cm} \]
None of the listed lengths is \( \tfrac{48}{5} \), so the answer is none of these. The \( \tfrac{24}{5} \) on offer is only half of \( BC \) - the distance from B to the axis of symmetry.
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How many paying stones, each measuring 2 1/2 m * 2 m are required to pave a rectangular court yard 30 m long and 16 1/2 m board?
Answer: A
Convert the mixed numbers to decimals and divide the areas.
\[ \text{stone} = 2.5 \times 2 = 5 \text{ m}^{2} \]
\[ \text{courtyard} = 30 \times 16.5 = 495 \text{ m}^{2} \]
\[ n = \frac{495}{5} = 99 \]
The division is exact, so the stones tile the courtyard with no cutting - which is what makes the volume comparison legitimate here.
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The breadth of a rectangular field is 60% of its length. If the perimeter of the field is 800 m,what is the area of the field?
Answer: B
Express the breadth in terms of the length and use the perimeter.
\[ b = 0.6l \quad\Longrightarrow\quad 2(l + 0.6l) = 800 \]
\[ 2(1.6l) = 800 \quad\Longrightarrow\quad l = 250 \text{ m} \]
\[ b = 0.6 \times 250 = 150 \text{ m} \]
\[ A = 250 \times 150 = 37{,}500 \text{ m}^{2} \]
Reading the 60% as applying to the length rather than the breadth reverses the two and halves the area to 18,750 - the first distractor.
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What is the perimeter of a square field whose diagonal is 8√2?
Answer: B
From \( d = a\sqrt{2} \):
\[ a = \frac{8\sqrt{2}}{\sqrt{2}} = 8 \]
\[ \text{perimeter} = 4a = 32 \text{ m} \]
The \( \sqrt{2} \) in the given diagonal is a deliberate signal that the side is a whole number.
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The area of a sector of a circle of radius 5 cm formed by an arc of length 3.5 cm is?
Answer: C
A sector is a triangle-like wedge whose "base" is its arc and whose "height" is the radius:
\[ A = \frac{1}{2} \times r \times \text{arc length} \]
\[ = \frac{1}{2} \times 5 \times 3.5 = 8.75 \text{ cm}^{2} \]
This form needs no angle at all. It follows from \( A = \tfrac{\theta}{2\pi}\pi r^{2} \) and \( \text{arc} = r\theta \), which combine to give exactly \( \tfrac{1}{2}r \times \text{arc} \).
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A paper is in a square form whose one side is 20 cm. Two semi circles are drawn on its opposites as diameters. If these semi circles are cut down what is the area of the remaining paper?
Answer: A
Each semicircle sits on a side of the square as its diameter, so each has diameter 20 cm and radius 10 cm.
\[ A_{\text{square}} = 20^{2} = 400 \text{ cm}^{2} \]
\[ A_{\text{one semicircle}} = \frac{1}{2}\pi (10)^{2} = 50\pi \text{ cm}^{2} \]
Two of them together make a full circle of radius 10:
\[ A_{\text{cut away}} = 100\pi \text{ cm}^{2} \]
\[ A_{\text{remaining}} = (400 - 100\pi) \text{ cm}^{2} \approx 85.8 \text{ cm}^{2} \]
The semicircles are on opposite sides, so they do not overlap and their areas simply add.
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The difference between the length and the breadth of a rectangle is 23 m. If its perimeter is 206 m, then its area is:
Answer: D
The perimeter gives the sum of the sides and the difference is given directly.
\[ 2(l + b) = 206 \quad\Longrightarrow\quad l + b = 103 \]
\[ l - b = 23 \]
Adding and subtracting the two:
\[ 2l = 126 \Rightarrow l = 63, \qquad 2b = 80 \Rightarrow b = 40 \]
\[ A = 63 \times 40 = 2520 \text{ m}^{2} \]
Halving the perimeter before pairing it with the difference is the step that turns two awkward numbers into a simple pair of simultaneous equations.
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