MCQ: Area - 2 - SSC CGL MCQ

Let the length of the rectangle = 4 units
And breadth of the rectangle = 3 units
Then diagonal of the rectangle = √(4² + 3²) = 5 units
According to the question, the length of a rectangle is increased by 25% and the breadth is reduced by
33.33%
New length = 125% of 4 units = 5 units
New breadth = 66.66% of 3 units = 2 units
In the new rectangle, New diagonal= √(5² + 2²) = √29 = approximately 5.38 units

Hence, option A is correct.

Area of square = 225 cm²
Side of square = √225 = 15 cm
Perimeter of square = 15 × 4 = 60 cm
So, perimeter of rectangle = 2 × 60 = 120 cm
2 (2 (x + 3) + 2 (x + 1)) = 120
2x + 6 + 2x + 2 = 60
4x + 8 = 60
4x = 52
x = 13
So, length of rectangle = 2 (13 + 3) = 32 cm
And, breadth of rectangle = 2 (13 + 1) = 28 cm
Therefore, area of rectangle = 32 × 28 = 896 cm²
Hence, option B is correct.

According to the question,
Let the radius of the circle be r cm.
So, the length of the rectangle = 1.25 r cm and the breadth of the rectangle = 0.90r cm
Given that,
πr² – 1.25 r x 0.90 r = 394.94
3.14r² – 1.125r² = 394.94

r² = 196
r = 14
length = 1.25 × 14 = 17.5 cm
breadth = 0.90 × 14 = 12.6 cm
Perimeter of rectangle = 2(17.5 + 12.6) = 60.2 cm
Hence, option C is correct.

Dimensions of rectangle inside the track = 47m (50 – 2 x 1.5) and 32m (35 – 2 x 1.5)
Area of track = area of park – area of rectangle inside the track
= (50 × 35 – 47 × 32) = (1750 – 1504) = 246m²
Cost of laying bricks at the rate of Rs 125 per 10m² 

Hence, option B is correct

Let the side of square be ‘s’ cm
So, length of rectangle = 1.4s cm
And, breadth of rectangle = 0.8s cm
According to the question,
2(1.4s + 0.8s) + 4s = 210
4.4s + 4s = 210
8.4s = 210

So, length of rectangle = 1.4 × 25 = 35 cm
And, breadth of rectangle = 0.8 × 25 = 20 cm
So, diagonal of rectangle

So option B is the correct answer.

Length = L; Width = B; Area = 1250;
L = 1250/B

Assume that the side CD in the figure is fenced by his home wall.
So, the total 100 meter wire is used to fence sides AC, AB & BD
AC + AB + BD = 100
AC = BD = B
AB = L
(2 x B) + L = 100

On solving, we get, B = 25 and L = 50 respectively
Therefore, L = 50 and B = 25 meters
Hence, option B is correct.

We know that,
Area of an equilateral triangle of side 

We know that,
Height of an equilateral triangle = √3a/2

⇒ Height of the equilateral triangle 
∴ Height of the equilateral triangle = 6√3 m
Hence, option D is correct.

Let the length and breadth of the field is 3x and x respectively

Therefore, the length of the field = 3x = 66 m
And the breadth of the field = x = 22 m
Required cost = 2 x (66 + 22) x 11 = Rs. 1936
Hence, option D is correct

The side of the square = 1000/4 = 25 cm

The area of the square = 25 × 25 = 625 sq. m
The area of the rectangle = 3x625 / 5 = 375 sq. cm
Let the length of the rectangle = 3x and breadth of the rectangle = 166.67% of 3x = 5x then perimeter =
2(l + b)
And area = 375 = 3x x 5x
x = 5 cm
Perimeter = 2(3x + 5x) = 16x = 80 cm
Hence, option B is correct.

The total surface area of the cuboid = 2(l x b + b × h + h x l) = 2(15 × 20 + 20 × 18 + 18 × 15) = 2 × (300 +
360 + 270) = 2 x 930 = 1860 sq. m
The cuboid is made of metal of negligible thickness then outer surface area = inner surface area
The required answer = 2 x 1860 x 5 = 18600 paisa = Rs. 186
Hence, option C is correct.

The boundary meets perpendicularly the opposite side it means it the perpendicular distance between the opposite sides.
The area of a parallelogram = base × perpendicular distance between the opposite sides = 15 × 16 = 240 sq. m
The total money required = 240 × 15 = Rs. 3600
Hence, option B is correct.

Let Radius = r then Circumference = 2πr cm and height = (2πr – 15) cm
Curved surface area = 2πrh = 2πr x (2πr – 15) = 154
By solving, r = 3.5 cm
Circumference of the base = 2πr cm 
Height = 22 – 15 = 7 cm

Hence, option B is correct.

Let Height of cylinder = h'
Volume of cylinder = πr²h

Hence, option B is correct.

For a triangle, Sum of two sides is always greater than third side.
Let the third side of the triangle be C
If sum of two side is 25
∴ 0 < C < 25 holds true for range of C.
C cannot be zero and cannot be more than 25
Perimeter of triangle
= 25 < (perimeter of triangle) < 50
Cost of fencing at Rs 15 per meter will lie between 375 – 750
Hence, option D Rs 800 cannot be the cost of fencing

Area of Circle = A = πr²

Now, breadth of rectangle is half the radius of circle.
Breadth of Rectangle = 14/2 = 7
Perimeter of Rectangle = 2 (L + B)

4L = 3L + 21
L = 21
Side of Square = 2 x 21 = 42
Area of Square = 42 x 42 = 1764
Hence, option A is correct