MCQ Questions on Mensuration 2D for Competitive Exams

1

A₁ and A₂ are two regular polygons. The sum of all the interior angles of A₁ is 1080°. Each interior angle of A₂ exceeds its exterior angle by 132°. The sum of the number of sides A₁ and A₂ is:

A) 22 B) 24 C) 23 D) 21

Discuss

Answer & Solution

Answer: Option C

Solution:

A₁ and A₂ are 2 regular polygon
Sum of internal angle of A₁ = 1080°
(n - 2) × 180° = 1080°
(n - 2) = 6
n = 6 + 2 = 8
Internal angle of each A₂ is grater than exterior angle by 132°
I + E = 180°
I - E = 132°
2I = 312°
I = 156°
E = 24°
n $$ = \frac{{{{360}^ \circ }}}{{{\text{Each Angle}}}} = \frac{{{{360}^ \circ }}}{{{{24}^ \circ }}} = 15$$
Sum of side of A₁ and A₂ = 15 + 8 = 23

2

The sum of the interior angles of a regular polygon A is 1260 degrees and each interior angle of a regular polygon B is $$128\frac{4}{7}$$ degrees. The sum of the number of sides of polygons A and B is:

A) 17 B) 18 C) 19 D) 16

Discuss

Answer & Solution

Answer: Option D

Solution:

Sum of internal angle = (n - 2) × 180°
Polygon A :
1260° = (n - 2) × 180°
7 = (n - 2)
n = 7 + 2 = 9
Polygon B :
$$\eqalign{ & {180^ \circ } - \frac{{{{360}^ \circ }}}{n} = 128\frac{{{4^ \circ }}}{7} \cr & {180^ \circ } - \frac{{{{360}^ \circ }}}{n} = \frac{{{{900}^ \circ }}}{7} \cr & \frac{{{{360}^ \circ }}}{n} = \frac{{{{1260}^ \circ } - {{900}^ \circ }}}{7} \cr & \frac{{{{360}^ \circ }}}{n} = \frac{{{{360}^ \circ }}}{7} \cr & n = 7 \cr} $$
Sum of Side of polygon A and B = A + B = 9 + 7 = 16

3

If the given figure, in a right angle triangle ABC, AB = 12 cm and AC = 15 cm. A square is inscribed in the triangle. One of the vertices of square coincides with the vertex of triangle. What is the maximum possible area (in cm²) of the square?
Mensuration 2D mcq question image

A) $$\frac{{1296}}{{49}}$$ B) $$25$$ C) $$\frac{{1225}}{{36}}$$ D) $$\frac{{1225}}{{64}}$$

Discuss

Answer & Solution

Answer: Option A

Solution:

Then BC = 9 cm
Side (a) square $$ = \frac{{12 \times 9}}{{21}} = \frac{{36}}{7}$$
Area of largest square that can be formed inside the circle $$ = {\left( {\frac{{36}}{7}} \right)^2} = \frac{{1296}}{{49}}$$

4

In the given figure, four identical semicircles are drawn in quadrant. XA = 7 cm. What is the area (in cm²) of shaded region?
Mensuration 2D mcq question image

A) 70 B) 140 C) 77 D) 84

Discuss

Answer & Solution

Answer: Option D

Solution:

$$\eqalign{ & {\text{Area of quadrant XDTB}} \cr & = \frac{1}{4}\pi \times 14 \times 14 \cr & = \frac{1}{4} \times \frac{{22}}{7} \times 14 \times 14 \cr & = 154{\text{ c}}{{\text{m}}^2} \cr & {\text{Area of four semi - circle}} \cr & = 4 \times \frac{1}{2}\left[ {\pi \times \frac{7}{2} \times \frac{7}{2}} \right] \cr & = 2 \times \frac{{22}}{7} \times \frac{7}{2} \times \frac{7}{2} \cr & = 77{\text{ c}}{{\text{m}}^2} \cr} $$
Mensuration 2D mcq question image

$$\eqalign{ & {\text{Area of shaded region = 2}} \times {\text{Area of quadrilateral MONX}} \cr & = 2 \times \left[ {\frac{1}{4}\pi \times 3.5 \times 3.5 - \frac{1}{2} \times 3.5 \times 3.5} \right] \cr & = {\left( {3.5} \right)^2}\left[ {\frac{{11}}{7} - 1} \right] \cr & = \frac{4}{7} \times \frac{7}{2} \times \frac{7}{2} \cr & = 7{\text{ c}}{{\text{m}}^2} \cr} $$

5

The diameter of the front wheel of an engine is 2x cm and that of rear wheel is 2y cm to cover the same distance, find the number of times the rear wheel will revolve when the front wheel revolves 'n' times,

A) $$\frac{n}{{xy}}{\text{ times}}$$ B) $$\frac{{yn}}{x}{\text{ times}}$$ C) $$\frac{{nx}}{y}{\text{ times}}$$ D) $$\frac{{xy}}{n}{\text{ times}}$$

Discuss

Answer & Solution

Answer: Option C

Solution:

Circumference of front wheel × number of its revolutions = circumference of rear wheel × number of its revolutions
2πx × n = 2πy × m (let 'm' is the revolution of rear wheel)
$$m = \frac{{nx}}{y}$$

6

In the given figure, PQR is a quadrant whose radius is 7 cm. A circle is inscribed in the quadrant as shown in the figure. What is the area (in cm²) of the circle?
Mensuration 2D mcq question image

A) 385 - 221√2 B) 308 - 154√2 C) 154 - 77√2 D) 462 - 308√2

Discuss

Answer & Solution

Answer: Option D

Solution:

In ΔABQ
AQ² + AB² = BQ²
R² + R² = (7 - R)²
2R² = 49 + R² - 14R
R² + 14R - 49 = 0
∴ R = $$ = \frac{{ - 14 \pm \sqrt {196 + 196} }}{2}$$
R = -7 ± 7$$\sqrt 2 $$
∴ R = 7$$\sqrt 2 $$ - 7
∴ Area of circle = πR²
= $$\frac{{22}}{7}$$ × (7$$\sqrt 2 $$ - 7)(7$$\sqrt 2 $$ - 7)
= 154(2 + 1 - 2$$\sqrt 2 $$ )
= 462 - 308$$\sqrt 2 $$

7

If the length of each side of an equilateral triangle is increased by 2 units, the area is found to be increased by 3 + √3 square unit. The length of each side of the triangle is

A) √3 units B) 3 units C) 3√3 units D) 3√2 units

Discuss

Answer & Solution

Answer: Option A

Solution:

Let each side of the triangle be a units
$$\eqalign{ & \Rightarrow \frac{{\sqrt 3 }}{4}\left\{ {{{\left( {a + 2} \right)}^2} - {a^2}} \right\} = 3 + \sqrt 3 \cr & \frac{1}{4}\left( {{a^2} + 4 + 4a - {a^2}} \right) = 1 + \sqrt 3 \cr & \frac{1}{2}\left( {4 + 4a} \right) = 1 + \sqrt 3 \cr & 1 + a = 1 + \sqrt 3 \cr & a = \sqrt 3 {\text{ units}} \cr} $$

8

The perimeter of a rectangle and an equilateral triangle are same. Also one of the sides of the rectangle is equal to the side of the triangle. The ratio of the area of the rectangle and the triangle is

A) √3 : 1 B) 1 : √3 C) 2 : √3 D) 4 : √3

Discuss

Answer & Solution

Answer: Option C

Solution:

2($$l$$ + b) = 3a
(a = side of equilateral triangle)
Let (b = a)
⇒ 2($$l$$ + a) = 3a
⇒ 2$$l$$ + 2a = 3a
⇒2$$l$$ = a
⇒ $$l$$ = $$\frac{{\text{a}}}{2}$$
Required Ratio
$$\eqalign{ & = \frac{{l \times b}}{{\frac{{\sqrt 3 }}{4}{a^2}}} \cr & = \frac{{\frac{a}{2} \times a}}{{\frac{{\sqrt 3 }}{4}{a^2}}} \cr & = \frac{{{a^2}}}{2} \times \frac{4}{{\sqrt 3 {a^2}}} \cr & = \frac{2}{{\sqrt 3 }} \cr & = {\bf{2:}}\sqrt {\bf{3}} \cr} $$

9

A parallelogram has sides 60 m and 40 m and one of its diagonals is 80 m long. Its area is

A) $$500\sqrt {15} {\text{ }}{{\text{m}}^2}$$ B) $$600\sqrt {15} {\text{ }}{{\text{m}}^2}$$ C) $$400\sqrt {15} {\text{ }}{{\text{m}}^2}$$ D) $$450\sqrt {15} {\text{ }}{{\text{m}}^2}$$

Discuss

Answer & Solution

Answer: Option B

Solution:

$$\eqalign{ & S\left( {\Delta ABD} \right) = \frac{{60 + 80 + 40}}{2} = 90 \cr & {\text{ar }}\Delta ABD = \sqrt {90\left( {90 - 80} \right)\left( {90 - 60} \right)\left( {90 - 40} \right)} \cr & = \sqrt {90 \times 10 \times 30 \times 50} \cr & = 300\sqrt {15} {\text{ }}{{\text{m}}^2} \cr & {\text{ar }}\square ABCD = 2 \times {\text{ar }}\Delta ABD = 600\sqrt {15} {\text{ }}{{\text{m}}^2} \cr} $$

10

The area of a regular hexagon of side 2√3 cm is:

A) 18√3 cm² B) 12√3 cm² C) 36√3 cm² D) 27√3 cm²

Discuss

Answer & Solution

Answer: Option A

Solution:

A regular hexagon consists of 6 equilateral triangle
Area of regular hexagon
$$\eqalign{ & = 6 \times \frac{{\sqrt 3 }}{4} \times {\left( {{\text{side}}} \right)^3} \cr & = 6 \times \frac{{\sqrt 3 }}{4} \times {a^2} \cr & = 6 \times \frac{{\sqrt 3 }}{4} \times {\left( {2\sqrt 3 } \right)^2} \cr & = 6 \times \frac{{\sqrt 3 }}{4} \times 12 \cr & = 18\sqrt 3 {\text{ c}}{{\text{m}}^2} \cr} $$