MCQ Questions on Mensuration 2D for Competitive Exams

Let width = x m

Area of Park = 60 × 40 = 2400 m²
Area of lawn = 2109 m²
So, the area of mid way path = 2400 - 2109 = 291
Area of mid way path = x($$l$$ + b - x)
x(60 + 40 - x) = 291
x(100 - x) = 291
Now you can choose from option or solve in detail.
x² - 100x + 291 = 0
x² - (97 + 3)x + 291 = 0
x² - 97x - 3x + 291 = 0
x(x - 97) - 3(x - 97) = 0
(x - 3)(x - 97) = 0
If x - 3 = 0
x = 3 m

When we draw such figures as mentioned in the question the vertex of the old triangle are the mid points of the sides of new triangle and the sides of the old triangle are half of the opposite side.
∴ Required ratio = 2 : 1

$$\eqalign{ & \sqrt 2 \mu \to 6 \cr & \sqrt 1 \mu \to 3\sqrt 2 \cr & {\text{AC}} = \left( {1 + \sqrt 3 } \right)\mu \cr & \to 3\sqrt 2 \left( {1 + \sqrt 3 } \right) \cr & \to 3\left( {\sqrt 2 + \sqrt 6 } \right) \cr} $$

Let ∠ABC = 60°
∠OBC = 30°
∴ Diagonals of Rhombus are the angle bisectors
In right ΔBOC
$$\eqalign{ & \frac{{{\text{OB}}}}{{{\text{BC}}}} = \cos {30^ \circ } \cr & \frac{{{\text{OB}}}}{8} = \frac{{\sqrt 3 }}{2} \cr} $$
OB = $$4\sqrt 3 $$
∴ BD = 2 × OB
= 2 × $$4\sqrt 3 $$
= $$8\sqrt 3 $$  cm

Sum of diagonal of quadrilateral = L
Perimeter = 4P
$$\eqalign{ & {\left( {\frac{{{\text{Diagonal}}}}{2}} \right)^2} - {\left( {\frac{{{\text{Semi perimeter}}}}{2}} \right)^2} \cr & = \frac{{{{\text{L}}^2}}}{4} - {\left( {\frac{{2{\text{P}}}}{2}} \right)^2} \cr & = \frac{{{{\text{L}}^2}}}{4} - {{\text{P}}^2} \cr & = \frac{1}{4}\left( {{{\text{L}}^2} - 4{{\text{P}}^2}} \right) \cr} $$

$$\eqalign{ & {\text{Using Hero's formula}} \cr & {\text{S}} = \frac{{15 + 7 + 20}}{2} = 21{\text{ cm}} \cr & {\text{Area of }}\Delta {\text{ABC}} = \sqrt {21\left( {21 - 20} \right)\left( {21 - 7} \right)\left( {21 - 15} \right)} \cr & = \sqrt {21 \times 1 \times 14 \times 6} \cr & = 42{\text{ c}}{{\text{m}}^2} \cr & \Rightarrow {\text{Area of }}\square {\text{ABCD}} = 42 \times 2 = 84{\text{ c}}{{\text{m}}^2} \cr} $$

$$\eqalign{ & {\text{Area of }}\Delta {\text{BIC}} = \frac{{\sqrt 3 }}{4} \times 12 \times 12 = 36\sqrt 3 \cr & {\text{Area of }}\Delta {\text{BIA}} = 36\sqrt 3 \cr & {\text{Area of }}\Delta {\text{BGI}} = \frac{{36\sqrt 3 }}{2} = 18\sqrt 3 \cr & {\text{Area of shaded region}} \cr & = 36\sqrt 3 + 18\sqrt 3 \cr & = 54\sqrt 3 {\text{ c}}{{\text{m}}^2} \cr} $$

Area $$\square $$ ABED = 3 × ar ΔADC
= 3 × $$\frac{{\sqrt 3 }}{4}$$ (2)² (ADC is an equilateral triangle)
= 3√3 units²

Let the side of square = a cm
According to the question,
$$l$$ × b = 3a²
20 × $$\frac{3}{2}$$ = 3a²
a = 10 cm

Area of circle = A
Radius of circle = r
Circumference of circle = C
πr² = A . . . . . . (i)
2πr = C . . . . . . (ii)
From equation (i) ÷ equation (ii)
$$\eqalign{ & \frac{{\pi {{\text{r}}^2}}}{{2\pi {\text{r}}}} = \frac{{\text{A}}}{{\text{C}}} \cr & \frac{{\text{r}}}{2} = \frac{{\text{A}}}{{\text{C}}} \cr} $$
rC = 2A