# Area Questions Answers

1) 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

**Ans.**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.

2) The area of incircle of an equilateral triangle of side 42 cm i

- 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}

**Ans.**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}

3) A farmer wishes to start a 100 sq. m. rectangular vegetable garden. Since he has only 30 meter barbed wire, he fences three sides of the garden letting his house compound wall act as the fourth side fencing. Then find the dimension of the garde

- 1) 10 m * 5 m
- 2) 15 m * 5 m
- 3) 20 m * 5 m
- 4) 25 m * 5 m

**Ans.**C

**Explanation :**

From the question, 2b+l = 30 => l = 30-2b \begin{aligned} Area = 100m^2 \=> l \times b = 100 \=> b(30-2b) = 100 \b^2 - 15b + 50 = 0 \=>(b-10)(b-5)=0 \\end{aligned} b = 10 or b = 5 when b = 10 then l = 10 when b = 5 then l = 20 Since the garden is rectangular so we will take value of breadth 5. So its dimensions are 20 m * 5 m

4) A courtyard is 25 meter long and 16 meter board is to be paved with bricks of dimensions 20 cm by 10 cm. The total number of bricks required is

- 1) 16000
- 2) 18000
- 3) 20000
- 4) 22000

**Ans.**C

**Explanation :**

\begin{aligned} \text{Number of bricks =}\frac{\text{Courtyard area}}{\text{1 brick area}} \= \left( \frac{2500 \times 1600}{20 \times 10} \right) \= 20000 \end{aligned}

5) The length of a rectangle is three times of its width. If the length of the diagonal is \begin{aligned}8\sqrt{10}\end{aligned}, then find the perimeter of the rectang

- 1) 60 cm
- 2) 62 cm
- 3) 64 cm
- 4) 66 cm

**Ans.**C

**Explanation :**

Let Breadth = x cm, then, Length = 3x cm \begin{aligned} x^2+{(3x)}^2 = {(8\sqrt{10})}^2 \=> 10x^2 = 640 \=> x = 8 \\end{aligned} So, length = 24 cm and breadth = 8 cm Perimeter = 2(l+b) = 2(24+8) = 64 cm

6) A towel, when bleached, was found to have lost 20% of its length and 10% of its breadth. The percentage of decrease in area i

- 1) 25%
- 2) 26%
- 3) 27%
- 4) 28%

**Ans.**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}

7) What will be the cost of gardening 1 meter boundary around a rectangular plot having perimeter of 340 meters at the rate of Rs. 10 per square mete

- 1) Rs. 3430
- 2) Rs. 3440
- 3) Rs. 3450
- 4) Rs. 3460

**Ans.**B

**Explanation :**

In this question, we are having perimeter. We know Perimeter = 2(l+b), right So, 2(l+b) = 340 As we have to make 1 meter boundary around this, so Area of boundary = ((l+2)+(b+2)-lb) = 2(l+b)+4 = 340+4 = 344 So required cost will be = 344 * 10 = 3440

8) 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 i

- 1) 124 cm
- 2) 120 cm
- 3) 64 cm
- 4) 56 cm

**Ans.**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}

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 sl

- 1) 110 cm
- 2) 116 cm
- 3) 118 cm
- 4) 120 cm

**Ans.**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) 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 broa

- 1) 714
- 2) 814
- 3) 850
- 4) 866

**Ans.**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}