Area — Part 2 MCQs with Explanations

$$\eqalign{ & \Rightarrow 2\pi {R_1} = 132 \cr & \Rightarrow {R_1} = \frac{{132 \times 7}}{{2 \times 22}} \cr & \Rightarrow {R_1} = 21\,m \cr & \Rightarrow 2\pi {R_2} = 176 \cr & \Rightarrow {R_2} = \frac{{176 \times 7}}{{2 \times 22}} \cr & \Rightarrow {R_2} = 28\,m \cr & \therefore {\text{Required difference :}} \cr & = \pi \left( {R_2^2 - R_2^2} \right) \cr & = \pi \left( {{R_2} + {R_1}} \right)\left( {{R_2} - {R_1}} \right) \cr & = \left( {\frac{{22}}{7} \times 49 \times 7} \right){m^2} \cr & = 1078\,{m^2} \cr} $$

Let the radii of the two circle be r and 3r respectively. Then, required ratio : $$ = \frac{{\pi {r^2}}}{{\pi {{\left( {3r} \right)}^2}}} = \frac{{\pi {r^2}}}{{9\pi {r^2}}} = 1:9$$

Let the radius of the ground be R metres. Then, $$\eqalign{ & \Rightarrow 2\pi R = 88 \cr & \Rightarrow R = \left( {\frac{{88 \times 7}}{{2 \times 22}}} \right) \cr & \Rightarrow R = 14\,m \cr} $$ Area of land strip : $$\eqalign{ & = \pi \left[ {{{\left( {14} \right)}^2} - {{\left( {11} \right)}^2}} \right]{m^2} \cr & = \left( {\frac{{22}}{7} \times 25 \times 3} \right){m^2} \cr & = \left( {\frac{{1650}}{7}} \right){m^2} \cr} $$ ∴ Cost of levelling : $$\eqalign{ & = {\text{Rs}}{\text{.}}\left( {\frac{{1650}}{7} \times 7} \right) \cr & = {\text{Rs}}{\text{. 1650}} \cr} $$

$$2\pi r = 100$$ So, length of the arc : $$\eqalign{ & = \frac{{2\pi r\theta }}{{360}} \cr & = \left( {\frac{{100 \times 20}}{{360}}} \right){\text{units}} \cr & = \left( {\frac{{50}}{9}} \right){\text{ units}} \cr & = 5.55{\text{ units}} \cr} $$

$$\eqalign{ & {\text{Radius}} = \frac{{{\text{Area}}}}{{{\text{Semi - perimeter}}}} \cr & {\text{Radius}} = \left( {{\text{Area}} \times \frac{2}{{{\text{Area}}}}} \right) \cr & {\text{Radius}} = 2 \cr} $$

Let length of each side of square = 2π According to the question, $$\eqalign{ & \frac{{\pi {r^2}}}{2} + \frac{{\pi {r^2}}}{2} + 42 = {\text{Area of square}} \cr & \Rightarrow \pi {r^2} + 42 = 4{r^2} \cr & \Rightarrow 4{r^2} - \pi {r^2} = 42 \cr & \Rightarrow {r^2}\left( {4 - \frac{{22}}{7}} \right) = 42 \cr & \Rightarrow {r^2}\left( {\frac{{28 - 22}}{7}} \right) = 42 \cr & \Rightarrow \frac{{6{r^2}}}{7} = 42 \cr & \Rightarrow {r^2} = \frac{{42 \times 7}}{6} \cr & \Rightarrow {r^2} = 7 \times 7 \cr & \Rightarrow r = 7 \cr & \therefore 2r = 14\,cm \cr} $$

Given base of triangle and its height is 15 cm and 12 cm respectively Area of first triangle : $$\eqalign{ & = \frac{1}{2} \times {\text{Base}} \times {\text{Height}} \cr & = \frac{1}{2} \times 15 \times 12 \cr & = 90{\text{ sq}}{\text{. cm}} \cr} $$ According to the question, Let height of triangle be h cm Area of new triangle = 180 sq. cm Base = 20 sq. cm $$\eqalign{ & \Rightarrow 180 = \frac{1}{2} \times 20 \times {\text{Height}} \cr & \Rightarrow {\text{h}} = \frac{{2 \times 180}}{{20}} \cr & \Rightarrow {\text{h}} = 18{\text{ cm}} \cr} $$

Let breadth = x cm Then, length = 3x cm $$\eqalign{ & \Rightarrow {x^2} + {\left( {3x} \right)^2} = {\left( {8\sqrt {10} } \right)^2} \cr & \Rightarrow 10{x^2} = 640 \cr & \Rightarrow {x^2} = 64 \cr & \Rightarrow x = 8 \cr} $$ So, length = 24 cm and breadth = 8 cm ∴ Perimeter = [2(24 + 8)] cm = 64 cm

Area of the sheet = (30 × 15) cm 2 = 450 cm 2 Area used for typing = [(30 - 5) × (15 - 2.5)] cm 2 = 312.5 cm 2 ∴ Required percentage : $$\eqalign{ & = \left( {\frac{{312.5}}{{450}} \times 100} \right)\% \cr & = 69.4\% \approx 70\% \cr} $$

PQ = $$\sqrt {100} $$ + $$\sqrt {16} $$ + $$\sqrt {49} $$ = (10 + 4 + 7) = 21 Side of middle square = $$\sqrt {16} $$ = 4 Reduction in PQ = (21 - 19) = 2 New side of middle square = (4 - 2) = 2 ∴ Reduction in area of middle square = (4 2 - 2 2 ) = 12

Since (20) 2 + (21) 2 = (29) 2 So, it is a right-angled triangle with base = 20 m and height = 21 m ∴ Area = $$\left( {\frac{1}{2} \times 20 \times 21} \right)$$ m 2 = 210 m 2

$$\eqalign{ & d = 4\sqrt 2 \,cm \cr & \Rightarrow {\text{Area }} = \frac{1}{2}d_1^2 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{2} \times {\left( {4\sqrt 2 } \right)^2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{ = 16}}\,{\text{c}}{{\text{m}}^2} \cr} $$ Area of new square = (2 × 16) cm 2 = 32 cm 2 $$\eqalign{ & \therefore \frac{1}{2}d_2^2 = 32 \cr & \Rightarrow d_2^2 = 64 \cr & \Rightarrow {d_2} = 8\,cm \cr} $$

Perimeter = 64 m ⇒ 2(l + b) = 64 ∴ Area of 4 walls = 2(l + b) × h = (64 × 4) m 2 = 256 m 2

Let ABC be the equilateral triangle and AD be the altitude on base BC In an equilateral triangle, the altitude and the median coincide. So, BC = DC = $$\left( {\frac{{2\sqrt 3 }}{2}} \right)$$ cm = $$\sqrt 3 $$ cm Let the length of the altitude AD be x cm Then, in right angled ΔADB, AB 2 = AD 2 + BD 2 ⇒ $${\left( {2\sqrt 3 } \right)^2}$$ = x 2 + $${\left( {\sqrt 3 } \right)^2}$$ ⇒ x 2 = (12 - 3) ⇒ x 2 = 9 ⇒ x = 3 cm

$$\eqalign{ & \Leftrightarrow \frac{1}{2}{d_1} \times 2{d_1} = 25 \cr & \Leftrightarrow d_1^2 = 25 \cr & \Leftrightarrow {d_1} = 5 \cr} $$ ∴ Sum of lengths of diagonals = (5 + 10) cm = 15 cm

Circumference of the plot : $$\eqalign{ & = \left( {\frac{{3300}}{{15}}} \right)m \cr & = 220\,m \cr & \Rightarrow 2\pi R = 220 \cr & \Rightarrow R = \frac{{220 \times 7}}{{2 \times 22}} \cr & \Rightarrow R = 35\,m \cr} $$ Area of the plot : $$\eqalign{ & = \left( {\frac{{22}}{7} \times 35 \times 35} \right){m^2} \cr & = 3850\,{m^2} \cr} $$ ∴ Cost of flooring : $$\eqalign{ & = {\text{Rs}}{\text{.}}\left( {3850 \times 100} \right) \cr & = {\text{Rs}}{\text{. 385000}} \cr} $$

$$\eqalign{ & \Rightarrow \frac{{2\pi {R_1}}}{{2\pi {R_2}}} = \frac{2}{3} \cr & \Rightarrow \frac{{{R_1}}}{{{R_2}}} = \frac{2}{3} \cr & \Rightarrow \frac{{\pi R_1^2}}{{\pi R_2^2}} = {\left( {\frac{2}{3}} \right)^2} \cr & \Rightarrow \frac{{\pi R_1^2}}{{\pi R_2^2}} = \frac{4}{9} \text{ or } 4:9 \cr} $$

Let the radii of the outer and inner circle be R and r respectively Then, $$\eqalign{ & 2\pi R - 2\pi r = 44 \cr & \Rightarrow 2\pi \left( {R - r} \right) = 44 \cr & \Rightarrow \left( {R - r} \right) = \frac{{44 \times 7}}{{22 \times 2}} \cr & \Rightarrow \left( {R - r} \right) = 7 \cr} $$

Side of the square : $$\eqalign{ & = \frac{{120}}{4}cm \cr & = 30\,cm \cr} $$ Radius of the required circle : $$\eqalign{ & = \left( {\frac{1}{2} \times 30} \right)cm \cr & = 15\,cm \cr & = \pi \times {r^2} \cr & = \left[ {\frac{{22}}{7} \times {{\left( {15} \right)}^2}} \right]c{m^2} \cr} $$

Required area : $$\eqalign{ & = \frac{1}{2} \times {\text{Base}} \times {\text{Height}} \cr & = \left( {\frac{1}{2} \times 2r \times r} \right) \cr & = {r^2} \cr} $$

Given: Radius of a circle = 3 cm Area of circle : $$\eqalign{ & = \pi {r^2} \cr & = \pi \times {3^2} \cr & = 9\pi {\text{ sq}}{\text{.}}\,{\text{cm}} \cr} $$

Let each equal side of isosceles triangle be x cm Perimeter of an isosceles triangle = 36 cm $$\eqalign{ & \therefore x + x + 14 = 36 \cr & \Rightarrow 2x = 36 - 14 \cr & \Rightarrow x = \frac{{22}}{2} \cr & \Rightarrow x = 11\,cm \cr} $$ BD = DC = 7cm From ΔABD By using Pythagoras theorem : $$\eqalign{ & AD = \sqrt {A{B^2} - B{D^2}} \cr & \,\,\,\,\,\,\,\,\,\,\,\, = \sqrt {{{11}^2} - {7^2}} \cr & \,\,\,\,\,\,\,\,\,\,\,\, = \sqrt {121 - 49} \cr & \,\,\,\,\,\,\,\,\,\,\,\, = \sqrt {72} \cr & \,\,\,\,\,\,\,\,\,\,\,\, = 3 \times 2\sqrt 2 \cr & \,\,\,\,\,\,\,\,\,\,\,\, = 6\sqrt 2 \,cm \cr} $$ ∴ Area of ΔABC $$\eqalign{ & = \frac{1}{2} \times BC \times AD \cr & = \frac{1}{2} \times 14 \times 6\sqrt 2 \cr & = 42\sqrt 2 {\text{ sq}}{\text{. cm}} \cr} $$

Area of the floor : $$\eqalign{ & = \left( {\frac{{510}}{{8.50}}} \right){m^2} \cr & = 60\,{m^2} \cr} $$ ∴ Breadth of the room : $$\eqalign{ & = \left( {\frac{{60}}{8}} \right)m \cr & = 7.5\,m \cr} $$

Let l and b be the length and breadth of the rectangle respectively. Then, $$\eqalign{ & \Rightarrow \sqrt {{l^2} + {b^2}} = 15 \cr & \Rightarrow \left( {{l^2} + {b^2}} \right) = {\left( {15} \right)^2} \cr & \Rightarrow {l^2} + {b^2} = 225 \cr} $$ And, $$\eqalign{ & \Rightarrow l + b = 3 \cr & \Rightarrow {\left( {l - b} \right)^2} = 9 \cr & \Rightarrow {l^2} + {b^2} - 2lb = 9 \cr & \Rightarrow 225 - 2lb = 9 \cr & \Rightarrow 2lb = 216 \cr & \Rightarrow lb = 108 \cr} $$ Hence, area of the field $$ = lb = 108\,{m^2}$$

Area of the carpet : = [(5 - 0.20) × (8 - 0.20)] m 2 = (4.8 × 7.8) m 2 = 37.44 m 2 ∴ Cost of carpeting : = Rs. (37.44 × 18) = Rs. 673.92

Let the breadth of the plot be x metres Then, length of the plot = (3x) metres x × 3x = 7803 ⇒ 3x 2 = 7803 ⇒ x 2 = 2601 ⇒ x = $$\sqrt {2601} $$ ⇒ x = 51 m

Number of tiles required : $$\eqalign{ & = \frac{{{\text{Area of hall}}}}{{{\text{Area of each tile}}}} \cr & = \left( {\frac{{60 \times 40}}{{0.4 \times 0.4}}} \right) \cr & = 15000 \cr} $$ ∴ Total cost of tiles : = Rs. (15000 × 5) = Rs. 75000

$$\eqalign{ & {\text{Required ratio}} \cr & = \frac{{{a^2}}}{{{{\left( {\sqrt 2 a} \right)}^2}}} \cr & = \frac{{{a^2}}}{{2{a^2}}} \cr & = \frac{1}{2} \cr & = 1:2 \cr} $$

Let the breadth = x metres and length = (2x) metres Area of 4 walls = [2(2x + x)× 4] m 2 = (24x) m 2 ∴ 24x = 120 ⇒ x = 5 So, length = 10 m, and breadth = 5 m Area of the floor = (10 × 5) m 2 = 50 m 2

Let the sides of the triangle be 5x, 5x and 4x cm respectively Then, 5x + 5x = 4x + 14 ⇒ 14 x = 14 ⇒ x = 1 So, a = 5 cm, b = 5 cm, c = 4 cm $$\eqalign{ & s = \frac{{{\text{a + b + c}}}}{2} \cr & \,\,\,\,\,\,\,\, = \left( {\frac{{14}}{2}} \right)cm \cr & \,\,\,\,\,\,\,\, = 7cm \cr} $$ (s - a) = 2 cm, (s - b) = 2 cm, (s - c) = 3 cm ∴ Area of the triangle : $$\eqalign{ & = \sqrt {7 \times 2 \times 2 \times 3} \cr & = 2\sqrt {21} \,c{m^2} \cr} $$

Let the length of sides of the two triangles be a 1 and a 2 respectively and their altitudes be h 1 and h 2 respectively. Then, $$\eqalign{ & \Leftrightarrow \frac{{\frac{{\sqrt 3 }}{4}a_1^2}}{{\frac{{\sqrt 3 }}{4}a_2^2}} = \frac{{25}}{{36}} \cr & \Rightarrow {\left( {\frac{{{a_1}}}{{{a_2}}}} \right)^2} = {\left( {\frac{5}{6}} \right)^2} \cr & \Rightarrow \frac{{{a_1}}}{{{a_2}}} = \frac{5}{6} \cr} $$ And $$\eqalign{ & \Leftrightarrow \frac{{\frac{1}{2} \times {a_1} \times {h_1}}}{{\frac{1}{2} \times {a_2} \times {h_2}}} = \frac{{25}}{{36}} \cr & \Rightarrow \frac{5}{6} \times \frac{{{h_1}}}{{{h_2}}} = \frac{{25}}{{36}} \cr & \Rightarrow \frac{{{h_1}}}{{{h_2}}} = \frac{{25}}{{36}} \times \frac{6}{5} \cr & \Rightarrow \frac{{{h_1}}}{{{h_2}}} = \frac{5}{6} \cr} $$

$$\eqalign{ & \frac{1}{2} \times 6 \times d = 24 \cr & \Rightarrow d = \frac{{24}}{3} \cr & \Rightarrow d = 8\,cm \cr & OA = 4\,cm\,{\text{and}}\,OB = 3\,cm \cr & \therefore AB = \sqrt {{{\left( {OA} \right)}^2} + {{\left( {OB} \right)}^2}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = \sqrt {{4^2} + {3^2}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = 5\,cm \cr} $$

$$\eqalign{ & \pi {R^2} = 7 \times \left( {2\pi R} \right) \cr & \Rightarrow R = 14 \cr & \therefore {\text{ Circumference :}} \cr & {\text{= }}\left( {2 \times \frac{{22}}{7} \times 14} \right)\text{units} \cr & = 88\,\text{units} \cr} $$

Since weight of the disc is proportional to its area, we have : $$\eqalign{ & \pi \left( {{R^2} - {r^2}} \right) = \frac{{24}}{{25}}\pi {R^2} \cr & \Rightarrow {R^2} - {r^2} = \frac{{24}}{{25}}{R^2} \cr & \Rightarrow {r^2} = \frac{1}{{25}}{R^2} \cr & \Rightarrow {R^2} = 25{r^2} \cr & \Rightarrow R = 5r \cr} $$

Radius of the required circle : $$\eqalign{ & = \left( {\frac{1}{2} \times 14} \right)cm \cr & = 7\,cm \cr} $$ Area of the circle : $$\eqalign{ & = \left( {\frac{{22}}{7} \times 7 \times 7} \right)\,c{m^2} \cr & = 154\,c{m^2} \cr} $$

Let original radius be R cm Then, original circumference = $$2\pi r$$ cm New radius : $$\eqalign{ & = \left( {175\% {\text{ of }}R} \right)cm \cr & = \left( {\frac{{175}}{{100}} \times R} \right)cm \cr & = \frac{{7R}}{4}\,cm \cr} $$ New circumference : $$\eqalign{ & = \left( {2\pi \times \frac{{7R}}{4}} \right)cm \cr & = \frac{{7\pi R}}{2}\,cm \cr} $$ Increase in circumference : $$\eqalign{ & = \left( {\frac{{7\pi R}}{2} - 2\pi R} \right)cm \cr & = \frac{{3\pi R}}{2}\,cm \cr} $$ Increase % : $$\eqalign{ & = \left( {\frac{{3\pi R}}{2} \times \frac{1}{{2\pi R}} \times 100} \right)\% \cr & = 75\% \cr} $$

Given length and width of a square base plate of brass is x cm and 1 mm Volume of the plate of square base = Area of base × height $$\eqalign{ & = {x^2} \times \frac{1}{{10}} \cr & = \frac{{{x^2}}}{{10}}\,cu.cm. \cr} $$ According to the question, $$\eqalign{ & \Rightarrow \frac{{{x^2}}}{{10}} \times 8.4 = 4725 \cr & \Rightarrow {x^2} = \frac{{4725 \times 10}}{{8.4}} \cr & \Rightarrow {x^2} = 5625 \cr & \Rightarrow x = \sqrt {5625} = 75\,cm \cr} $$

Radius of circle = 7cm Given area of rectangle : = Area of circle = $${\frac{22}{7} \times 7 \times 7}$$ = 154 cm 2

Let the breadth of the rectangle be x metres Then, length of the rectangle = 2x metres ⇒ 2(2x + x) = 60 ⇒ 6x = 60 ⇒ x = 10 So, length = 20 m, breadth = 10 m ∴ Area = (20 × 10) m 2 = 200 m 2

Let the length of the field be x metres Then, breadth of the field = $$\frac{3x}{4}$$ metres $$\eqalign{ & x \times \frac{{3x}}{4} = 300 \cr & \Rightarrow {x^2} = 300 \times \frac{4}{3} \cr & \Rightarrow {x^2} = 400 \cr & \Rightarrow x = 20 \cr} $$ So, length = 20 m, breadth = 15 m ∴ Area of the garden : $$\eqalign{ & = \left[ {\left\{ {\left( {20 + 3} \right) \times \left( {15 + 3} \right)} \right\} - \left( {20 \times 15} \right)} \right]{m^2} \cr & = \left[ {\left( {23 \times 18} \right) - \left( {20 \times 15} \right)} \right]{m^2} \cr & = \left( {414 - 300} \right){m^2} \cr & = 114\,{m^2} \cr} $$