MCQ: Area and Perimeter- 3 - SSC CGL MCQ

Let circumference = 100 cm .
Then, ∵ 2πr = 100
⇒ r = 100/2π
=50/π
⇒ New circumference
= 105 cm
Then, 2πR = 105
⇒ R = 105 / (2π)
&rArr Original area = [ π x (50/π) x (50/π) ]
= 2500/π cm²
⇒ New Area = [π x (105/2π) x (105/2π)]
= 11025 / (4π) cm²
⇒ Increase in area = [11025/(4π)] - 2500/π cm²
= 1025 / 4π cm²
Required increase percent [1025/(4π)] x 2500/π x 100 = 41/4%
= 10.25%

∵ 22/7 x r² = 462
⇒ r² = (462 x 7) /22 = 147
⇒ r = 7√3 cm
∴ Height of the triangle = 3r = 21√3 cm
Now, ∵ a² = a²/4 + (3r)²
⇒ 3a²/4 = (21√3)²
⇒ a² = (1323 x 4)/3
⇒ a = 21x 2 = 42 cm
∴ Perimeter = 3a = 3 x 42
=126 cm 

Area of the room =(544 x 374) cm²
size of largest square tile = H.C.F. of 544 & 374
= 34 cm
Area of 1 tile = (34 x 34) cm²
∴ Least number of tiles required
= (544 x 374) / (34 x 34) = 176

Let there be n sides of the polygon. Then it has n vertices.
The total number of straight lines obtained by joining n vertices by talking 2 at a time is ⁿC₂
These ⁿC₂ lines also include n sides of polygon.
Therefore, the number of diagonals formed is ⁿC₂ - n.
Thus, ⁿC₂ - n = 44
⇒ [n(n - 1)/2] - n = 44
⇒ ( n² - 3n) / 2 = 44
⇒ n² - 3n = 88
⇒ n² - 3n - 88 = 0
⇒(n - 11) (n + 8) = 0
∴ n = 11 

Area of equilateral triangle = √3a²/4 = x ......(i)
And perimeter = 3a = y
⇒ a = y/3 ....(ii)
Now, Putting the value of a from Eq. (ii) in Eq. (i). we get
√3 (y/3)²/4 = x
⇒ x = √3 x y²/36
⇒ x = y²/3√3x = y²/12√3
12√3 x = y²
On squaring both sides, we get
y⁴ = 432x² 

Given that, l = 2b [Here l = length and b = breadth]
Decrease in length = Half of the 10 cm = 10/2 = 5 cm
Increase in breadth = Half of the 10 cm = 10/2 = 5 cm
Increase in the area = (70 + 5) = 75 sq cm
According to the question,
(l - 5) (b + 5) = lb + 75
⇒ (2b - 5) (b + 5) = 2b² + 75 [since l = 2b]
⇒ 5b - 25 = 75
⇒ 5b = 100
∴ b = 100/ 5 = 20
∴ l = 2b = 2 x 20 = 40 cm

Area of square plate = (Side)²
= (2d)²
= 4d²
Area of circular plate = π (d/2)²
= πd²/4
∵ Number of square plates
= [(4d²)/4] / [(πd²)/4]
= (4 x 4)/π
≈ 5
Since, nearest integer value is 5.

∵ 1/2 x 2 √ x h
= √3/4 x (2 √3)²
∴ h = 3 cm.

Angle swept in 30 min= 180°
Area swept = [(22/7) x 7 x 7] x [180°/360°] cm²
= 77 cm²

Area of park = 100 x 100 = 10000 m²
Area of circular lawn = Area of park - area of park excluding circular lawn
= 10000 - 8614
= 1386
Now again area of circular lawn = (22/7) x r² = 1386 m²
⇒ r² = (1386 x 7) / 22
= 63 x 7
= 3 x 3 x 7 x 7
∴ r = 21 m

Ratio of similar triangle
= Ratio of the square of corresponding sides
= (3x)² / (4x)² = 9x² / 16x²
= 9/16 = 9 : 16

Let the radius of circle is 'r' and a side of a square is 'a',
then given condition
2πr = 4a
⇒ a = πr/2
∴ Area of square = (πr/2)² = π² /4r² = 9.86r²/4 = 2.46r²
and area of circle = πr² = 3.14;r²
and let the side of equilateral triangle is x.
Then, given condition,
3x = 2πr
⇒ x = 2πr/3
∴ Area of equilateral triangle = √3/4 x ²
= √3/4 x 4π²r²/9
= π²/3√3r²
= 1.89r²
Hence, Area of circle > Area of square > Area of equilateral triangle.

We know that, the radius of a circle inscribed in a equilateral triangle = a/[2√3]
Where, a be the length of the side of an equilateral triangle.
Given that, area of a circle inscribed in an equilateral tringle = 154 cm²
∴ π(a/2√3)² = 154
⇒ (a/2√3)² = 154 x (7/22) = (7)a²
⇒ a = 42√3 cm
Perimeter of an equilateral triangle = 3a
= 3(14√3)
= 42√3 cm

Area of square = (Side)² = 20²
= 400 sq cm
∴ Area of rectangle
= 1.8 x 400 = 720 sq cm
Let length and breadth of rectangle be 5k and k respectively.
Then, according to the question,
5k x k = 720
⇒ 5k² = 720
⇒ k² = 720/5 = 144
∴ k = √144 = 12 cm
Perimeter of rectangle = 2(5k + k) = 12k
= 12 x12 = 144 cm

Let one diagonal be k.
Then, other diagonal = (60k/100) = 3k/5 cm
Area of rhombus =(1/2) x k x (3k/5) = (3/10)
= 3/10 (square of longer diagonal)
Hence, area of rhombus is 3/10 times.