Area Calculation - Online Quiz


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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.

Questions and Answers

Q 1 - The aggregate territory of 64 little square of a chessboard is 400 cm2. There is 3cm wide fringe around the chess board. What is the Length of every side of the chessboard?

A - 17 cm

B - 20 cm

C - 23 cm

D - 26 cm

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:

A - 30 m

B - 40 m

C - 50 m

D - 60 m

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:

A - kept unaltered

B - expanded by 25%

C - expanded by 20%

D - Diminished by 20%

Answer : D

Explanation

Let the length be x meter and breadth be y mtr.
Then, its area = (xy) m2
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:

A - 1000

B - 2000

C - 3000

D - none of these

Answer : B

Explanation

Area of the verandah= (40*15)m2= 600m2
Area of one stone= (6/10*5/10) m2= 3/10m2
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:

A - 2 sq. unit

B - π sq. unit

C - 2√2π sq. unit

D - 4√2π sq. unit

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:

A - triangle

B - square

C - equilateral triangle

D - both triangle and square

Answer : A

Explanation

πR2 = s2= √3/4 a2 =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 cm2 and 48 cm2 .If the height of the smaller one is 2.1 cm, then the corresponding height of the bigger one is:

A - 0.525 cm

B - 4.2 cm

C - 4.41 cm

D - 8.4 cm

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/h2 ⇒h2=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?

A - 65%

B - 70%

C - 80%

D - 61.1%

Answer : D

Explanation

Total area = (30*15) cm2
Area used = [(30-1.25*2)* (15-2.5*2)]
= (27.5*10) cm2 = 275cm2
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?

A - 20cm2

B - 24 cm2

C - 26cm2

D - 40cm2

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) =24cm2

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:

A - 2918m2

B - 2921 m2

C - 2924 m2

D - 2926 m2

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] m2= 22/7 *(70+63) (70-63) m2
= (22*133) m2, = 2926m2


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