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 - The edge of a story of a room is 18m. What is the region of four dividers of the room, if its stature is 3m?

A - 21m2

B - 42m2

C - 54m2

D - 108m2

Answer : C

Explanation

Perimeter= 2 (L+b)=18 and height =3m
Area of 4 walls = 2(l+b)*h= (18*3) = 54 sq.m

Q 2 - The region of a rectangle is 12sq. mtr. Also, its length is 3 times that of its expansiveness. What is the edge of the rectangle?

A - 14 m

B - 18 m

C - 24 m

D - none of these

Answer : D

Explanation

Let the breadth be x mtr. then its length = 3x
∴ 3x*x= 12 ⇒ x2= 4 ⇒x=2
∴ L= 6m, b=2m ⇒perimeter = 2(6+2) m = 16m

Q 3 - A corridor 20m long and 15m expansive is encompassed by a verandah of Uniform width of 2.5 m. The expense of ground surfaces the verandah at Rs.17.50 per sq. mtr. is:

A - 2500

B - 3000

C - 3500

D - 4000

Answer : C

Explanation

Area of verandah = [(25*20)-(20*15)]m2= 200 m2
  Cost of flooring = (200*35/2) = 3500 rs.

Q 4 - The covering of a room twice the length it is board at the rate of 50 p for every square mtr. costs rs. 12.25 And the expense of painting its dividers the rate of 9 p for every square mtr. is rs. 6.30. The tallness of the room is:

A - 10/3 m

B - 13/3 m

C - 7 m

D - none of these

Answer : A

Explanation

Let breadth = x mtr. Then, length = 2x mtr.
Area of the floor = 1225/50 m2= 49/2 m2
∴ X*2x= 49/2 ⇒x2 =49/4 ⇒x=7/2
∴ L = 7m, b= 7/2 m
Area of 4 walls = 630/9 m2 = 70 m2
2(7+ 7/2)* h= 70 ⇒ 2*21/2*h =70
⇒h =70/21 m = 10/3m

Q 5 - The perimeter of a square circumscribed about a circle of radius r is:

A - 2r

B - 4r

C - 8r

D - 21πr

Answer : C

Explanation

Each side of the square = 2r
∴ Perimeter of the square = (4* 2r) = 8r.

Q 6 - Each side of an equilateral triangle measures 8 cm. Its area is:

A - 64 cm2

B - 16√3cm2

C - 4√3cm2

D - 21.3 cm2

Answer : B

Explanation

Area =( √3/4*8*8 ) cm2 = 16√3 cm2

Q 7 - ∆ABC and ∆ DEF are similar triangles such that BC= 4cm, EF= 5cm and area (∆ ABC) = 64cm2. The area of ∆DEF is:

A - 80 cm2

B - 100 cm2

C - 256/5 cm2

D - None of these

Answer : B

Explanation

The areas of two similar triangles are in the ratio of the square of  the corresponding sides.
Ar (∆ABC)/ ar (∆Def) =BC2/Ef2 ⇒64/ ar (∆Def) = (4)2/ (5)2 = 16/25
⇒ar (∆DEF) = (25*64/16) = 100 cm2

Q 8 - The length of every side of an equilateral triangle is 24 cm. The region of its engraved circle is:

A - 18 πcm2

B - 24 πcm2

C - 36 πcm2

D - 48 πcm2

Answer : D

Explanation

Let the height of the triangle be h then,
1/2 * 24*h = √3/4 *24*24 ⇒h= 12 √3
∴ 3r = 12 √3 ⇒r = 4√3
Area of in circle = π* (4√3)2 = 48πcm2

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 - 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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