Quantitative Aptitude
Mensuration
2,507 Questions
Mensuration Questions
C
Correct answer
Explanation
For an ellipse x^2/a^2 + y^2/b^2 = 1, the maximum area of an inscribed rectangle is 2ab. Here a=5, b=4. Area = 2 * 5 * 4 = 40.
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R
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$\displaystyle \frac{R}{2}$
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$\displaystyle \sqrt{\frac{2}{3}}R$
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$\displaystyle \sqrt{\frac{3}{2}}R$
C
Correct answer
Explanation
For a cylinder of radius r and height h inscribed in a sphere of radius R, the relationship is r^2 + (h/2)^2 = R^2. Maximizing the volume V = pi * r^2 * h leads to the optimal radius r = sqrt(2/3) * R.
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$\displaystyle\frac{2r}{\sqrt{3}}$
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$\displaystyle\frac{\sqrt{2}r}{\sqrt{3}}$
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$r$
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$\sqrt{3}$r
B
Correct answer
Explanation
For a cylinder of radius x and height 2h inscribed in a sphere of radius r, x^2 + h^2 = r^2. Volume V = pi * x^2 * 2h = 2 * pi * (r^2 - h^2) * h = 2 * pi * (r^2*h - h^3). Setting dV/dh = 0 gives r^2 - 3h^2 = 0, so h = r/sqrt(3). Then x^2 = r^2 - r^2/3 = 2r^2/3, so x = r * sqrt(2/3).
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$\displaystyle \pi a^{2}\left ( 1+1/\sqrt{5} \right )$
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$\displaystyle \pi/2 a^{2}\left ( 1+\sqrt{5} \right )$
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$\displaystyle \pi a^{2}\left ( 1+\sqrt{5} \right )$
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$\displaystyle \pi a^{2}\left ( 2+\sqrt{5} \right )$
C
Correct answer
Explanation
Total surface area S = 2 * pi * r^2 + 2 * pi * r * h. For a cylinder in a sphere of radius a, r^2 + (h/2)^2 = a^2. Express S in terms of h, differentiate, and find the maximum. The result is pi * a^2 * (1 + sqrt(5)).
D
Correct answer
Explanation
Let sides be L and B. 2(L+B) = 36, so L+B = 18. Volume V = pi * r^2 * h. If revolved about L, V = pi * B^2 * L = pi * B^2 * (18-B). To maximize, dV/dB = pi * (36B - 3B^2) = 0, so 3B(12-B) = 0, B=12, L=6. Ratio L:B = 6:12 = 1:2. If revolved about B, L=12, B=6, ratio 2:1.
A
Correct answer
Explanation
Let r be the radius of the sphere and x be the edge of the cube. Surface area S = 4*pi*r^2 + 6*x^2. Volume V = (4/3)*pi*r^3 + x^3. Minimizing V subject to constant S leads to the condition where the derivative of V with respect to r (using x as a function of r) is zero, resulting in r = x/2, or r:x = 1:2.
D
Correct answer
Explanation
For a cone of height h and base radius R circumscribed about a sphere of radius r, the volume is V = (1/3)pi R^2 h. Using similar triangles, R^2 = r^2 * h / (h - 2r). Minimizing V leads to h = 4r.
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$\dfrac { r }{ 2 } $
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$\dfrac { r }{ 3 } $
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$\dfrac { 3r }{ 4 } $
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$\dfrac { 4r }{ 3 } $
D
Correct answer
Explanation
For a cone of maximum volume inscribed in a sphere of radius r, the height h is 4r/3 and the base radius is sqrt(8/9) * r. This is a standard optimization problem result.
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$\dfrac {V(a + d)^{2}}{a^{2}}$
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$\dfrac {V(a + d)}{a}$
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$V(a + d)$
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$a + d$
A
Correct answer
Explanation
For a hollow sphere with constant metallic volume, R^3 - r^3 is constant. Differentiating gives R^2 dR/dt = r^2 dr/dt, so with R = a + d, r = a, and dR/dt = V, the inner-radius rate is V(a + d)^2/a^2.
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$\displaystyle \frac{14\pi R^{\tfrac 52}}{15\sqrt{2}a\sqrt{g}}$
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$\displaystyle \frac{14\pi R^{\tfrac 52}}{\sqrt{2}a\sqrt{g}}$
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$\displaystyle \frac{14\pi R^{\tfrac 52}}{5\sqrt{2}a\sqrt{g}}$
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$\displaystyle \frac{4\pi R^{\tfrac 52}}{15\sqrt{2}a\sqrt{g}}$
A
Correct answer
Explanation
The time to empty a tank is given by t = (A_tank / a) * sqrt(2/g) * integral(sqrt(y) dy) from 0 to R. For a hemisphere, A_tank = pi * (R^2 - y^2). The integration leads to the standard formula t = (14 * pi * R^(5/2)) / (15 * a * sqrt(2g)).
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$4\pi$
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$5\pi$
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$16\pi$
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$25\pi$
D
Correct answer
Explanation
The equation (x+2)^2 + (y+3)^2 = 25 represents a circle with radius squared = 25, so r = 5. Area = pi * r^2 = 25 * pi.
C
Correct answer
Explanation
The greatest circle of a sphere is its cross-section through the center, which has the same radius as the sphere. Diameter = 20, so radius = 10.
A
Correct answer
Explanation
A great circle of a sphere is a circle that has the same radius as the sphere. If the radius is 16, the diameter of the great circle is 2 * 16 = 32.
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$0.3$%
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$0.6$%
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$0.9$%
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$\dfrac { 4 }{ 3 } \pi { \left( 0.3 \right) }^{ 3 }$
C
Correct answer
Explanation
Volume of a sphere V = (4/3) * pi * r^3. The relative error in volume is 3 times the relative error in radius. Given the error in radius is 0.3%, the error in volume is 3 * 0.3% = 0.9%.
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$\dfrac { 19\pi +6\sqrt { 3 } }{ 6 } $
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$\dfrac { 5\pi +6\sqrt { 3 } }{ 6 } $
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$\dfrac { \pi}{ 6 } $
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$\dfrac { 5\pi -6\sqrt { 3 } }{ 6 } $
D
Correct answer
Explanation
The area of intersection of two circles with radii r1, r2 and distance d is calculated using the formula for circular segments. For r1=1, r2=sqrt(3), d=2, the circles are orthogonal or intersect at specific angles. The calculation yields (5pi - 6sqrt(3))/6.