Area Questions Answers

• 15. What are the least number of square tiles required to pave the floor of a room 15 m 17 cm long and 9 m 2 cm broad ?

  1. 714
  2. 814
  3. 850
  4. 866

  Answer And Explanation

  Answer: Option B

  Explanation:

  In this type of questions, first we need to calculate the area of tiles. With we can get by obtaining the length of largest tile.
  Length of largest tile can be obtained from HCF of length and breadth.
  So lets solve this,
  
  Length of largest tile = HCF of (1517 cm and 902 cm)
  = 41 cm
  
  Required number of tiles =
  \begin{aligned}
  \frac{\text{Area of floor}}{\text{Area of tile}} \\
  = \left(\frac{1517\times902}{41 \times 41}\right)\\
  = 814
  
  \end{aligned}

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• 16. The area of a square is 69696 cm square. What will be its diagonal ?

  1. 373.196 cm
  2. 373.110 cm
  3. 373.290 cm
  4. 373.296 cm

  Answer And Explanation

  Answer: Option 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}

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• 17. If the ratio of the areas of two squares is 225:256, then the ratio of their perimeters is :

  1. 15:12
  2. 15:14
  3. 15:16
  4. 15:22

  Answer And Explanation

  Answer: Option 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}

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• 18. 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 cm.

  1. 9 cm
  2. 8 cm
  3. 7 cm
  4. 6 cm

  Answer And Explanation

  Answer: Option 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}

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• 19. 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 is :

  1. 22 cm
  2. 20 cm
  3. 18 cm
  4. 10 cm

  Answer And Explanation

  Answer: Option 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}

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• 20. 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 side.

  1. 12 cm
  2. 14 cm
  3. 16 cm
  4. 18 cm

  Answer And Explanation

  Answer: Option 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
  

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• 21. 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 And Explanation

  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}

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