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.

Questions and Answers

Q 1 - A rectangular patio 3.78m long and 5.25 m wide is to be cleared precisely with square tiles, the greater part of the same size. The base No. such tiles are:

A - 430

B - 440

C - 450

D - 460

Answer : C

Explanation

L=378 m and b= 525 cm
Maximum length of the square tile= H.C.F (378cm, 525cm) =21cm
No. of tiles =378*525/21*21= (18*25) = 450

Q 2 - The border of the rectangle is 60 m. on the off chance that its length is twice its broadness, and then its region is:

A - 160 m2

B - 180 m2

C - 200 m2

D - 220 m2

Answer : C

Explanation

Let the breadth be x mtr. Then, length = 2x mtr.
2(2x+x) = 60 ⇒ 6x=60 ⇒x= 60
∴ L=20m and b= 10m ⇒area = (20*10) =200 m2

Q 3 - If every side of a square is expanded by half, the proportion of the territory of the subsequent square to the range of the given square is:

A - 5:4

B - 9:4

C - 4:5

D - 4:9

Answer : B

Explanation

Let the original side be x mtr. Then, area =x2 sq. mtr.
New side = (150% of x) m = (150/100* x) m = (3x/2) m
New area = (3x/2*3x/2) m2= 9x2/4 m2
Required ratio of areas = 9x2/4: x2= 9:4

Q 4 - The length of a rectangular plot is twice its expansiveness. On the off chance that the length of its corner to corner is 9√5m, the border of the perimeter of the rectangle is:

A - 27 m

B - 54 m

C - 81 m

D - none of these

Answer : B

Explanation

let breadth = x meter, Then, length = 2x meter
Diagonal = √ (2x) 2+ x2 =√5x2= √5x meter
∴ √5x= 9√5 ⇒9m ⇒Perimeter = 2(18+9) m =54m

Q 5 - The area of the largest circle that can be drawn insight rectangle with sides 8m by 7m is:

A - 352m2/7

B - 77/2m2

C - 32m2

D - 49m2

Answer : B

Explanation

Radius of the circle= 7/2 m
Area of the circle = (22/7 *7/2 *7/2) m2 =77/2m2

Q 6 - A circle and a rectangle have same perimeter.The side of the rectangle is 18cm and 26 cm. what is the area of the circle?

A - 88 cm2

B - 154 cm2

C - 616 cm2

D - 1250 cm2

Answer : C

Explanation

Circumference of the circle= Perimeter of the rectangle
= 2 (18+26) cm= 88cm
2*22/7* R= 88 ⇒ R= (88*7/44) =14 cm
Area of the circle = πr2 = (22/7 *14*14) cm2 =616 cm2

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 - If the side of a rhombus is 20cm and its shorter corner to corner is three-fourth of its more extended askew, then the range of the rhombus is:

A - 375 cm2

B - 380 cm2

C - 384 cm2

D - 395 cm2

Answer : C

Explanation

Let the longer diagonal be x cm , then shorter diagonal = (3/4)x cm
∴ AC= x cm and BD=(3/4 )x cm
AO =1/2*AC =x/2 cm, BO= 1/2 BD= (3/8) X cm and AB= 20 cm
In right ∆ AOB, we have AO2 +BO2 = AB2
(x/2) 2+ (3x/8) 2= (20) 2⇒x2/4+9x2/64=400 ⇒16x2+9x2=25600 ⇒25x2= 25600 ⇒x2=1024
⇒ x=√1024= 32 cm
∴ AO =32/2=16cm, BO= (3/8*32) cm=12cm
∴ AC=2*AO= 32 cm, BD= 2*BO= 24cm
Area of the rhombus = (1/2*32*24) cm2 =384 cm2

Q 10 - If the range of a circle is expanded by 6%, then its region is expansion by:

A - 0.36%

B - 3.6%

C - 12.36%

D - 36%

Answer : C

Explanation

Let initial radius= R. Then, area = πR2
New radius =106% of R= (106/100* R) = 53R/50
New area = π* (53R/50)2
Increase in area = π*{(53R/50)2 -R2}=π(53R/50+R)( 53R/50 -R)
= (π*103R/50*3R/50) = (309/2500) πR2
Increased % = (309/2500 *πR2*1/πR2*100) %= 309/25%= 12.36%
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