Area Questions Answers

Area Questions Answers (MCQ) listings with explanations includes questions of Area of Rectangle, Square, Triangle, Rhombus, Trapezium, Circle etc which are important for many competitive exams.

11) The area of a square is 69696 cm square. What will be its diagona

  • 1) 373.196 cm
  • 2) 373.110 cm
  • 3) 373.290 cm
  • 4) 373.296 cm
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Ans.   D
Explanation :
If area is given then we can easily find side of a square as, \begin{aligned}Side = \sqrt{69696} \= 264 cm \\text{we know diagonal =}\sqrt{2}\times side \= \sqrt{2}\times 264 \= 1.414 \times 264 \= 373.296 cm \end{aligned}

12) If the ratio of the areas of two squares is 225:256, then the ratio of their perimeters i

  • 1) 15:12
  • 2) 15:14
  • 3) 15:16
  • 4) 15:22
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Ans.   C
Explanation :
\begin{aligned} \frac{a^2}{b^2} = \frac{225}{256} \\frac{15}{16} \<=> \frac{4a}{4b} = \frac{4*15}{4*16} \= \frac{15}{16} = 15:16 \end{aligned}

13) The difference of the areas of two squares drawn on two line segments in 32 sq. cm. Find the length of the greater line segment if one is longer than the other by 2

  • 1) 9 cm
  • 2) 8 cm
  • 3) 7 cm
  • 4) 6 cm
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Ans.   C
Explanation :
Let the lengths of the line segments be x and x+2 cm then, \begin{aligned} {(x+2)}^2 - x^2 = 32 \x^2 + 4x + 4 - x^2 = 32 \4x = 28 \x = 7 cm \end{aligned}

14) The base of a triangle is 15 cm and height is 12 cm. The height of another triangle of double the area having the base 20 cm i

  • 1) 22 cm
  • 2) 20 cm
  • 3) 18 cm
  • 4) 10 cm
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Ans.   C
Explanation :
\begin{aligned} \text{Area of triangle, A1 = }\frac{1}{2}*base*height \= \frac{1}{2}*15*12 = 90 cm^2 \\text{Area of second triangle =} 2*A1 \= 180 cm^2 \\frac{1}{2}*20*height = 180 \=> height = 18 cm \end{aligned}

15) The sides of a triangle are in the ratio of \begin{aligned}\frac{1}{2}:\frac{1}{3}:\frac{1}{4}\end{aligned}. If the perimeter is 52 cm, then find the length of the smallest si

  • 1) 12 cm
  • 2) 14 cm
  • 3) 16 cm
  • 4) 18 cm
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Ans.   A
Explanation :
\begin{aligned} \text{Ratio of sides =}\frac{1}{2}:\frac{1}{3}:\frac{1}{4} \=6:4:3\Perimeter = 52 cm \\text{So sides are =} \\left( 52*\frac{6}{13}\right)cm,\left( 52*\frac{4}{13}\right)cm, \left( 52*\frac{3}{13}\right)cm \end{aligned} a = 24 cm, b = 16 cm and c = 12 cm Length of the smallest side = 12 cm

16) The height of an equilateral triangle is 10 cm. find its ar

  • 1) \begin{aligned} \frac{120}{\sqrt{3}} cm^2 \end{aligned}
  • 2) \begin{aligned} \frac{110}{\sqrt{3}} cm^2 \end{aligned}
  • 3) \begin{aligned} \frac{100}{\sqrt{3}} cm^2 \end{aligned}
  • 4) \begin{aligned} \frac{90}{\sqrt{3}} cm^2 \end{aligned}
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Ans.   C
Explanation :
Let each side be a cm, then \begin{aligned} \left(\frac{a}{2}\right)^2+{10}^2 = a^2 \<=>\left(a^2-\frac{a^2}{4}\right) = 100 \<=> \frac{3a^2}{4} = 100 \a^2 = \frac{400}{3} \Area = \frac{\sqrt{3}}{4}*a^2 \= \left(\frac{\sqrt{3}}{4}*\frac{400}{3}\right)cm^2 \= \frac{100}{\sqrt{3}}cm^2 \end{aligned}

17) If the area of a square with the side a is equal to the area of a triangle with base a, then the altitude of the triangle

  • 1) a
  • 2) a/2
  • 3) 2a
  • 4) None of above
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Ans.   C
Explanation :
\begin{aligned} \text{We know area of square =}a^2 \\text{Area of triangle =}\frac{1}{2}*a*h \=> \frac{1}{2}*a*h = a^2 \=> h = 2a \end{aligned}

18) What will be the ratio between the area of a rectangle and the area of a triangle with one of the sides of the rectangle as base and a vertex on the opposite side of the rectangl

  • 1) 1:1
  • 2) 1:2
  • 3) 2:3
  • 4) 2:1
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Ans.   D
Explanation :
As far as questions of Area or Volume and Surface area are concerned, it is all about formulas and very little logic. So its a sincere advice to get all formulas remembered before solving these questions. Lets solve this, \begin{aligned} \text{Area of rectangle =}l*b\\text{Area of triangle =}\frac{1}{2}l*b\\text{Ratio =}l*b:\frac{1}{2}l*b \= 1:\frac{1}{2} \= 2:1 \end{aligned} One little thing which should be taken care in this type of question is, be sure you are calculating ration in the given order of the question. If it is ratio of triangle and rectangle then we have to write triangle formula first. cheers :)

19) A triangle and a parallelogram are constructed on the same base such that their areas are equal. If the altitude of the parallelogram is 100 m , then the altitude of the triangle

  • 1) 200 m
  • 2) 150 m
  • 3) 148 m
  • 4) 140 m
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Ans.   A
Explanation :
Let the triangle and parallelogram have common base b, let the Altitude of triangle is h1 and of parallelogram is h2(which is equal to 100 m), then \begin{aligned} \text{Area of triangle =}\frac{1}{2}*b*h1\\text{Area of rectangle =}b*h2\\text{As per question }\\frac{1}{2}*b*h1 = b*h2 \ \frac{1}{2}*b*h1 = b*100 \ h1 = 100*2 = 200 m \ \end{aligned}

20) The area of rhombus is 150 cm square. The length of one of the its diagonals is 10 cm. The length of the other diagonal

  • 1) 15 cm
  • 2) 20 cm
  • 3) 25 cm
  • 4) 30 cm
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Ans.   D
Explanation :
We know the product of diagonals is 1/2*(product of diagonals) Let one diagonal be d1 and d2 So as per question \begin{aligned} \frac{1}{2}*d1*d2 = 150 \\frac{1}{2}*10*d2 = 150 \d2 = \frac{150}{5} = 30 \\end{aligned}
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