Area Questions and Answers

• 8. One side of rectangular field is 15 meter and one of its diagonals is 17 meter. Then find the area of the field.

  1. \begin{aligned} 120m^2 \end{aligned}
  2. \begin{aligned} 130m^2 \end{aligned}
  3. \begin{aligned} 140m^2 \end{aligned}
  4. \begin{aligned} 150m^2 \end{aligned}

  Answer :

  Option A

  Explanation:

  \begin{aligned}
  \text{We know }h^2 = b^2+h^2 \\
  =>\text{Other side }= \sqrt{(17)^2-(15)^2} \\
  = \sqrt{289-225} = \sqrt{64} \\
  = 8 meter \\
  Area = Length \times Breadth \\
  = 15\times8 m^2 = 120 m^2
  
  \end{aligned}

• 9. 50 square stone slabs of equal size were needed to cover a floor area of 72 sq.m. Find the length of each stone slab.

  1. 110 cm
  2. 116 cm
  3. 118 cm
  4. 120 cm

  Answer :

  Option D

  Explanation:

  Area of each slab =
  \begin{aligned}
  \frac{72}{50}m^2 = 1.44 m^2\\
  \text{Length of each slab =}\sqrt{1.44} \\
  = 1.2m = 120 cm
  
  \end{aligned}

• 10. 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 is.

  1. 200 m
  2. 150 m
  3. 148 m
  4. 140 m

  Answer :

  Option 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}

• 11. The perimeters of 5 squares are 24 cm, 32 cm, 40 cm, 76 cm and 80 cm respectively. The perimeter of another square equal in area to the sum of the area of these square is:

  1. 124 cm
  2. 120 cm
  3. 64 cm
  4. 56 cm

  Answer :

  Option A

  Explanation:

  Clearly first we need to find the areas of the given squares, for that we need its side.
  
  Side of sqaure = Perimeter/4
  
  So sides are,
  
  \begin{aligned}
  \left(\frac{24}{4}\right),\left(\frac{32}{4}\right),\left(\frac{40}{4}\right),\left(\frac{76}{4}\right),\left(\frac{80}{4}\right) \\
  = 6,8,10,19,20 \\
  \text{Area of new square will be }\\
  = [6^2+8^2+10^2+19^2+20^2] \\
  = 36+64+100+361+400 \\
  = 961 cm^2 \\
  \text{Side of new Sqaure =}\sqrt{961} \\
  = 31 cm \\
  \text{Required perimeter =}(4\times31) \\
  = 124 cm
  
  \end{aligned}

• 12. A towel, when bleached, was found to have lost 20% of its length and 10% of its breadth. The percentage of decrease in area is ?

  1. 25%
  2. 26%
  3. 27%
  4. 28%

  Answer :

  Option D

  Explanation:

  Let original length = x
  and original width = y
  Decrease in area will be
  \begin{aligned}
  = xy-\left( \frac{80x}{100}\times\frac{90y}{100}\right) \\
  = \left(xy- \frac{18}{25}xy\right) \\
  = \frac{7}{25}xy \\
  \text{Decrease = }\left(\frac{7xy}{25xy} \times100\right) \% \\
  = 28\%
  \end{aligned}

• 13. The height of an equilateral triangle is 10 cm. find its area.

  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}

  Answer :

  Option 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}

• 14. If the circumference of a circle increases from 4pi to 8 pi, what change occurs in the area ?

  1. Area is quadrupled
  2. Area is tripled
  3. Area is doubles
  4. Area become half

  Answer :

  Option A

  Explanation:

  \begin{aligned}
  2\pi R1 = 4 \pi \\
  => R1 = 2 \\
  2\pi R2 = 8 \pi \\
  => R2 = 4 \\
  \text{Original Area =} 4\pi * 2^2 \\
  = 16 \pi \\
  \text{New Area =} 4\pi * 4^2 \\
  = 64 \pi
  \end{aligned}
  
  So the area quadruples.