# Area Calculation - Online Quiz

Following quiz provides Multiple Choice Questions (MCQs) related to **Area Calculation**. You will have to read all the given answers and click over the correct answer. If you are not sure about the answer then you can check the answer using **Show Answer** button. You can use **Next Quiz** button to check new set of questions in the quiz.

Q 1 - The aggregate territory of 64 little square of a chessboard is 400 cm^{2}. There is 3cm wide fringe around the chess board. What is the Length of every side of the chessboard?

### Answer : D

### Explanation

Length of each side of the chess board excluding path=√400=20cm. Length of each side = (20+3+3) = 26cm

Q 2 - The edge of a rectangle and a square are 160m each. The range of the rectangle is not exactly that of the square by 100sq. meters. The length of the rectangle is:

### Answer : C

### Explanation

Each side of the square= 160/4m = 40m 2(L+b) =160 ⇒ (L+b) = 80 (40)^{2}-Lb= 100 ⇒ lb= (1600-100) = 1500 (L-b)^{2}= (L+b)^{2}- 4Lb= (80)^{2}-4*1500= (6400-6000) =400⇒L-b=20 ∴L+b= 80, L-b= 20 ⇒ 2L=100 ⇒ L= 50 m

Q 3 - The length of a rectangular plot is expanded by 25%. To keep its region unaltered, the width of the plot ought to be:

### Answer : D

### Explanation

Let the length be x meter and breadth be y mtr. Then, its area = (xy) m^{2}New length = (125/100*x) m = (5x/4) m. let the new breadth be z meters. Then, xy = 5x/4*z ⇒z= 4/5 y Decrease in width = (y-4/5y) = y/5 mtr. Decrease % in width = (y/5*1/y*100) % = 20%

Q 4 - A Verandah 40m long and 15 m wide is to be cleared with stones every measuring 6dm by 5dm. The quantity of stones required is:

### Answer : B

### Explanation

Area of the verandah= (40*15)m^{2}= 600m^{2}Area of one stone= (6/10*5/10) m^{2}= 3/10m^{2}No. of stones = (600*10/3) = 2000

Q 5 - The largest possible square is increased in a circle of unit radius. The area of the square is:

### Answer : A

### Explanation

Diagonal of the square = 2* Radius =(2*1)= 2 units Area of the square= [1/2* (2)^{2}] sq. units= 2 sq. unit

Q 6 - If an area enclosed by a circle or a square or an equilateral triangle is the same, then the maximum perimeter is possessed by:

### Answer : A

### Explanation

πR^{2}= s^{2}= √3/4 a^{2}=A ⇒ R= √ (A/π), s =√A and a =√4A/ (√3) Perimeter of circle = 2πR= 2π√ (A/π) =2√A*√π= 2*√3.14*√A= (2*1.77*√A) = (3.54* √A) Perimeter of square = 4√A Perimeter of triangle = 3a = 3√ (4A/√3= 6/3 (ki power1/4). √A =3 (ki power3/4).2√A = (27)⅟4 .2√A>4√A ∴ Perimeter of triangle is maximum.

Q 7 - The areas of two similar triangles are 12 cm^{2} and 48 cm^{2} .If the height of the smaller one is 2.1 cm, then the corresponding height of the bigger one is:

### Answer : B

### Explanation

The areas of two similar triangles are in the ratio of the square of the corresponding sides. ∴12/48= (2.1)^{2}/h^{2}⇒h^{2}=4* (2.1)^{2}⇒h= (2* 2.1) = 4.2 cm

Q 8 - A typist uses a paper 30cm*15 cm. He leaves an edge of 2.5cm at the top and the base and 1.25 cm on either side. What rate of paper range around accessible for writing?

### Answer : D

### Explanation

Total area = (30*15) cm^{2}Area used = [(30-1.25*2)* (15-2.5*2)] = (27.5*10) cm^{2}= 275cm^{2}Percentage of area used = (275/450*100) % = 61.1%

Q 9 - One side and one inclining of a rhombus measure 5cm and 8cm individually. What is the territory of the rhombus?

### Answer : B

### Explanation

Let the diagonals AC and BD of rhombus. ABCD intersect at O. Then ∠AOB =90° AB= 5cm and AO= 1/2 * AC= (1/2 *8) = 4cm BO =√ (ab)^{2}-(OA)^{2}= √ (5)^{2}-(4)^{2}=√25-16= √9 =3cm BD =2* BO = (2*3) cm =6 Area =1/2*AC*BD= (1/2*8*6) =24cm^{2}

Q 10 - A round greenery enclosure has an outline of 440 m. There is a 7m wide fringe inside the patio nursery along its outskirts. The territory of the fringe is:

### Answer : D

### Explanation

2πR =440 ⇒ 2*22/7*R= 440 ⇒R= (440* 7/44)=70 m Outer radius = 70m, inner radius = (70-7) =63 m Required area = π [(70)^{2}-(63)^{2}] m^{2}= 22/7 *(70+63) (70-63) m^{2}= (22*133) m^{2}, = 2926m^{2}