# 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 proportion of the length and expansiveness of a plot is 3:2. On the off chance that the expansiveness is 40m not exactly the length of its slanting is:

### Answer : C

### Explanation

Let the length be 3x meter. Then, breadth = 2x meters Then, 3x-2x= 40 ⇒ x=40 ∴length = (3*40) = 120m and b= (2*40) =80m ∴ Perimeter = 2(120+80) = 400m

Q 2 - The length of a rectangle is expanded by 10% and its expansiveness is diminished by 10%. At that point, the range of the new rectangle is:

### Answer : C

### Explanation

Let length be L unit and breadth be b unit. Area = Lb sq.units New length = (110/100*L) = 11L/10, new breadth = (90/100*b)= 9b/10 New area = (11L/10 *9b/10) Sq. units= (99/100 *Lb) Area decreased = (Lb-99/100 Lb) sq.units = Lb/100 sq. units Percent decreased = (Lb/100*1/lb*100) %= 1%

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:

### Answer : C

### Explanation

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

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 perimeter of a square circumscribed about a circle of radius r is:

### Answer : C

### Explanation

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

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 - The stature of an equilateral triangle is √6cm. Its range is:

### Answer : B

### Explanation

Let the base be A cm , Then, 1/2 *a* √6= (√3/4) a^{2}⇒ a= 1/2 *√6* 4/√3= 2√2 Area of the triangle = {(√3/4* (2√2)^{2}} cm^{2}= (√3/4*8) cm^{2}= 2√3cm^{2}

Q 9 - Each side of an equilateral triangle is equivalent to the span of a circle whose region is 154 cm^{2}. The range of the equilateral triangle is:

### Answer : B

### Explanation

Let each side of the equilateral triangle be a cm and radius of the circle is r cm. Then , a =r πr^{2}=154 ⇒22/7 *r^{2}=154⇒ r^{2}= (154*7/22) = 49 ⇒r= 7 ∴ a = 7cm ⇒ area of ∆ = √ 3/4 *(7)^{2}cm^{2}=49√3/4 cm^{2}

Q 10 - The distinction between the radii of the smaller circle and the greater circle is 7cm and the contrast between the territories of the two circles is 1078 cm^{2}. Span of the littler circle is:

### Answer : B

### Explanation

Let the radii of the inner and outer circle be r and R cm. Then, R-r= 7 and π (R^{2}-r^{2}) = 1078 ∴ Π (R^{2}-r^{2})/ (R-r) = 1078/7 ⇒R+r = (1078/7*7/22) =49 On solving (R-r) =7 and (R+r) = 49, we get r= 21 cm