Area Questions and 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 :

  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}

• 16. The area of incircle of an equilateral triangle of side 42 cm is :

  1. \begin{aligned} 462 cm^2 \end{aligned}
  2. \begin{aligned} 452 cm^2 \end{aligned}
  3. \begin{aligned} 442 cm^2 \end{aligned}
  4. \begin{aligned} 432 cm^2 \end{aligned}

  Answer :

  Option A

  Explanation:

  \begin{aligned}
  \text{Radius of incircle} = \frac{a}{2\sqrt3} \\
  = \frac{42}{2\sqrt3} \\
  = 7\sqrt{3} \\
  \text{Area of incircle =} \\
  \frac{22}{7}*49*3 = 462 cm^2
  \end{aligned}

• 17. 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 is:

  1. 15 cm
  2. 20 cm
  3. 25 cm
  4. 30 cm

  Answer :

  Option 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}

• 18. Find the circumference of a circle, whose area is 24.64 meter sqaure

  1. 17.90m
  2. 17.80m
  3. 17.60m
  4. 17.40m
  5. Only assumption I is implicit
  6. Only assumption II is implicit
  7. Either I or II is implicit
  8. Neither I nor II is implicit
  9. Both I and II are implicit

  Answer :

  Option E

  Explanation:

  \begin{aligned}
  \text{Area of Square =} \pi*r^2\\
  => \pi*r^2 = 24.64 \\
  => r^2 = \frac{24.64}{22}*7 \\
  => r^2 = 7.84 \\
  => r = \sqrt{7.84} \\
  => r = 2.8 \\
  \text{Circumference =}2\pi*r\\
  = 2*\frac{22}{7}*2.8 \\
  = 17.60m
  
  \end{aligned}

• 19. A rectangular field is to be fenced on three sides leaving a side of 20 feet uncovered.If the area of the field is 680 sq.ft, how many feet of fencing will be required ?

  1. 88 feet
  2. 86 feet
  3. 84 feet
  4. 82 feet
  5. Only assumption I is implicit
  6. Only assumption II is implicit
  7. Either I or II is implicit
  8. Neither I nor II is implicit
  9. Both I and II are implicit

  Answer :

  Option F

  Explanation:

  We are given with length and area, so we can find the breadth.
  as Length * Breadth = Area
  => 20 * Breadth = 680
  => Breadth = 34 feet
  
  Area to be fenced = 2B + L = 2*34 + 20
  = 88 feet

• 20. The percentage increase in the area of a rectangle, if each of its sides is increased by 20% is:

  1. 32%
  2. 34%
  3. 42%
  4. 44%
  5. art is governed by external rules and conditions.
  6. art is for the sake of art and life.
  7. art is for the sake of art alone.
  8. artist realises his dreams through his artistic creation.
  9. artist should use his art for the sake of society.

  Answer :

  Option

  Explanation:

  Let original length = x metres and original breadth = y metres.
  
  \begin{aligned}
  \text{Original area } = \text{xy } m^2 \\
  \text{New Length }= \frac{120}{100}x = \frac{6}{5}x \\
  \text{New Breadth }= \frac{120}{100}y = \frac{6}{5}y \\
  =>\text{New Area }= \frac{6}{5}x * \frac{6}{5}y \\
  =>\text{New Area }= \frac{36}{25}xy \\
  
  \text{Area Difference} = \frac{36}{25}xy - xy \\
  = \frac{11}{25}xy \\
  
  Increase \% = \frac{Differnce}{Actual}*100 \\
  = \frac{11xy}{25}*\frac{1}{xy}*100 = 44\%
  
  \end{aligned}
  
  

• 21. 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 is.

  1. a
  2. a/2
  3. 2a
  4. None of above

  Answer :

  Option 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}