Area — Part 1 MCQs with Explanations

Perimeter = Distance covered in 8 min. Perimeter = $$\left( {\frac{{12000}}{{60}} \times 8} \right){\text{m}}$$ Perimeter = 1600 m Let length = 3x metres and breadth = 2x metres. Then, 2(3x + 2x) = 1600 or x = 160 Therefore Length = 480 m and Breadth = 320 m Therefore Area = (480 x 320) m 2 = 153600 m 2

100 cm is read as 102 cm. ∴ A 1 = (100 x 100) cm 2 and A 2 (102 x 102) cm 2 (A 2 - A 1 ) = [(102) 2 - (100) 2 ] = (102 + 100) x (102 - 100) = 404 cm 2 ∴ Percentage error $$\eqalign{ & = \left( {\frac{{404}}{{100 \times 100}} \times 100} \right)\% \cr & = 4.04\% \cr} $$

$$\eqalign{ & \frac{{2\left( {l + b} \right)}}{b} = \frac{5}{1} \cr & \Rightarrow 2l + 2b = 5b \cr & \Rightarrow 3b = 2l \cr & b = \frac{2}{3}l \cr & {\text{Then,}} \cr & {\text{Area = }}216\,c{m^2} \cr & \Rightarrow l \times b = 216 \cr & \Rightarrow l \times \frac{2}{3}l = 216 \cr & \Rightarrow {l^2} = 324 \cr & l = 18\,cm \cr} $$

Let original length = x metres and original breadth = y metres $$\eqalign{ & {\text{Original}}\,{\text{are}} = \left( {xy} \right){m^2} \cr & {\text{New}}\,{\text{length}} = \left( {\frac{{120}}{{100}}x} \right)m = \left( {\frac{6}{5}x} \right)m \cr & {\text{New}}\,{\text{breadth}} = \left( {\frac{{120}}{{100}}y} \right)m = \left( {\frac{6}{5}y} \right)m \cr & {\text{New}}\,{\text{area}} = \left( {\frac{6}{5}x \times \frac{6}{5}y} \right){m^2} = \left( {\frac{{36}}{{25}}xy} \right){m^2} \cr & {\text{The}}\,{\text{difference}}\,{\text{between}}\,{\text{the}}\,{\text{original}}\,{\text{area = }}xy \cr & {\text{and}}\,{\text{new}}\,{\text{area}}\,\frac{{36}}{{25}}xy\,{\text{is}} \cr & = \left( {\frac{{36}}{{25}}} \right)xy - xy \cr & = xy\left( {\frac{{36}}{{25}} - 1} \right) \cr & = xy\left( {\frac{{11}}{{25}}} \right)\,or\,\left( {\frac{{11}}{{25}}} \right)xy \cr & \therefore {\text{Increase}}\,\% \cr & = \left( {\frac{{11}}{{25}}xy \times \frac{1}{{xy}} \times 100} \right)\% \cr & = 44\% \cr} $$

Area of the park = (60 x 40) m 2 = 2400 m 2 Area of the lawn = 2109 m 2 ∴ Area of the crossroads = (2400 - 2109) m 2 = 291 m 2 Let the width of the road be x metres. Then, 60 x + 40 x - x 2 = 291 ⇒ x 2 - 100 x + 291 = 0 ⇒ ( x - 97)( x - 3) = 0 ⇒ x = 3 m

$$\eqalign{ & {\text{Outer}}\,{\text{Side}} \cr & = \sqrt {{{\left( {\frac{{15}}{2}} \right)}^2} - {{\left( {\frac{9}{2}} \right)}^2}ft} \cr & = \sqrt {\frac{{225}}{4} - \frac{{81}}{4}ft} \cr & = \sqrt {\frac{{144}}{4}ft} \cr & = 6ft \cr & \therefore {\text{Area}}\,{\text{of}}\,{\text{closet}} = \left( {6 \times 4.5} \right)sq.\,ft = 27\,sq.\,ft. \cr} $$

$$\eqalign{ & {\text{Let}}\,{\text{original}}\,{\text{length}} = x\,{\text{and}} \cr & {\text{original}}\,{\text{breadth}} = y \cr & {\text{Decrease}}\,{\text{in}}\,{\text{area}} \cr & = xy - \left( {\frac{{80}}{{100}}x \times \frac{{90}}{{100}}y} \right) \cr & = {xy - \frac{{18}}{{25}}xy} \cr & = \frac{7}{{25}}xy \cr & \therefore {\text{Decrease}}\,\% = \cr & \left( {\frac{7}{{25}}xy \times \frac{1}{{xy}} \times 100} \right)\% \cr & = 28\% \cr} $$

Let the side of the square(ABCD) be x metres. Then, AB + BC = 2x metres AC =$$\sqrt 2 $$ x = (1.41x) m Saving on 2x metres = (0.59x) m $$\eqalign{ & {\text{Saving}}\,\% = \left( {\frac{{0.59x}}{{2x}} \times 100} \right)\% \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 30\% \,\left( {{\text{approx}}} \right) \cr} $$

$$\eqalign{ & \sqrt {{l^2} + {b^2}} = \sqrt {41} \cr & {\text{Also}},\,lb = 20 \cr & {\left( {l + b} \right)^2} = \left( {{l^2} + {b^2}} \right) + 2/b = 41 + 40 = 81 \cr & \Rightarrow \left( {l + b} \right) = 9 \cr & \therefore {\text{Perimeter}} = 2\left( {l + b} \right) = 18cm \cr} $$

$$\eqalign{ & {\text{Length}}\,{\text{of}}\,{\text{largest}}\,{\text{tile}} = \cr & {\text{H}}{\text{.C}}{\text{.F}}{\text{.}}\,{\text{of}}\,1517\,cm\,{\text{and}}\,902\,cm = 41\,cm \cr & {\text{Area}}\,{\text{of}}\,{\text{each}}\,{\text{tile}} = \left( {41 \times 41} \right)c{m^2} \cr & \therefore {\text{Required}}\,{\text{number}}\,{\text{of}}\,{\text{tiles}} \cr & = {\frac{{1517 \times 902}}{{41 \times 41}}} \cr & = 814 \cr} $$

We have: (l - b) = 23 and 2(l + b) = 206 or (l + b) = 103 Solving the two equations, we get: l = 63 and b = 40 ∴ Area = (l x b) = (63 x 40) m 2 = 2520 m 2

$$\eqalign{ & {\text{Let}}\,{\text{original}}\,{\text{length}} = x\,{\text{and}}\, \cr & {\text{Original}}\,{\text{breadth}} = y \cr & {\text{Original}}\,{\text{area}} = xy \cr & {\text{New}}\,{\text{length}} = \frac{x}{2} \cr & {\text{New}}\,{\text{breadth}} = 3y \cr & {\text{New}}\,{\text{area}} = \left( {\frac{x}{2} \times 3y} \right) = \frac{3}{2}xy \cr & \therefore {\text{Increase}}\,\% = {\frac{1}{2}xy \times \frac{1}{{xy}} \times 100} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 50\% \cr} $$

$$\eqalign{ & {\text{Let}}\,{\text{breadth}} = x\,{\text{metres}} \cr & {\text{Then,}}\,{\text{length}} = \,\left( {x + 20} \right)\,{\text{metres}} \cr & {\text{Perimeter}} = {\frac{{5300}}{{26.50}}} m = 200m \cr & \therefore 2\left[ {\left( {x + 20} \right) + x} \right] = 200 \cr & \Rightarrow 2x + 20 = 100 \cr & \Rightarrow 2x = 80 \cr & \Rightarrow x = 40 \cr & {\text{Hence,}}\,{\text{length}} = x + 20 = 60m \cr} $$

We have: l = 20 ft and lb = 680 sq. ft. So, b = 34 ft. ∴ Length of fencing = (l + 2b) = (20 + 68) ft = 88 ft.

$$\eqalign{ & {\text{Area}}\,{\text{to}}\,{\text{be}}\,{\text{plastered}} \cr & = \left[ {2\left( {l + b} \right) \times h} \right] + \left( {l \times b} \right) \cr & = \left\{ {\left[ {2\left( {25 + 12} \right) \times 6} \right] + \left( {25 \times 12} \right)} \right\}{m^2} \cr & = \left( {444 + 300} \right){m^2} \cr & = 744{m^2} \cr & \therefore {\text{Cost}}\,{\text{of}}\,{\text{Plastering}} \cr & = Rs.\,\left( {744 \times \frac{{75}}{{100}}} \right) \cr & = Rs.\,558 \cr} $$

Area of the floor = (5.5 × 3.75)m 2 = 20.625m 2 ∴ Cost of paying = Rs. (800 × 20.625) = Rs. 16500

Let the area of the whole painting be x cm 2 Then, $$\eqalign{ & \Rightarrow \frac{1}{4}x + 100 = \frac{3}{4}x \cr & \Rightarrow \frac{1}{2}x = 100 \cr & \Rightarrow x = 200 \cr} $$ ∴ Area of painting = 200 cm 2 And, Height = 10 cm Length of painting : $$\eqalign{ & = \left( {\frac{{200}}{{10}}} \right)cm \cr & = 20\,\,cm \cr} $$

Let original length = x and original breadth = y Then, original area = xy New length : $$\eqalign{ & = \frac{{160x}}{{100}} \cr & = \frac{{8x}}{5} \cr} $$ Let the new breadth = z Then, $$\eqalign{ & \Rightarrow \frac{{8x}}{5} \times z = xy \cr & \Rightarrow z = \frac{{5y}}{8} \cr} $$ ∴ Decrease in breadth : $$\eqalign{ & = \left( {\frac{{3y}}{8} \times \frac{1}{y} \times 100} \right)\% \cr & = 37\frac{1}{2}\% \cr} $$

$$\frac{{{\text{Monthly income of family A}}}}{{{\text{Monthly income of family B}}}}$$ $$ = \frac{{{\text{Area of square A}}}}{{{\text{Area of square B}}}}$$ $$\eqalign{ & \Rightarrow \frac{{40000}}{x} = \frac{{{2^2}}}{{{3^2}}} \cr & \Rightarrow x = \left( {\frac{{40000 \times 9}}{4}} \right) \cr & \Rightarrow x = 90000 \cr} $$

$$\eqalign{ & {\text{Area :}} \cr & = \left( {45 \times 40} \right){m^2} \Leftrightarrow \frac{1}{2} \times {\left( {{\text{Diagonal}}} \right)^2} \cr & = 1800\,\,\,\, \Leftrightarrow \,\,\,\,{\text{Diagonal}} \cr & = 60\,\,m \cr} $$

Let the sides of the two squares be x metres and y metres respectively Then, $$ \Rightarrow 29{x^2} = 6{y^2} - 1.....(i)$$ And, $$\eqalign{ & \Rightarrow 9x - 4y = 1 \cr & \Rightarrow 4y = 9x - 1 \cr & \Rightarrow y = \frac{{9x - 1}}{4}.....(ii) \cr} $$ From (i) and (ii), we get : $$\eqalign{ & \Rightarrow 29{x^2} = 6{\left( {\frac{{9x - 1}}{4}} \right)^2} - 1 \cr & \Rightarrow 29{x^2} = 6\left( {\frac{{81{x^2} + 1 - 18x}}{{16}}} \right) - 1 \cr & \Rightarrow 243{x^2} + 3 - 54x - 8 = 232{x^2} \cr & \Rightarrow 11{x^2} - 54x - 5 = 0 \cr & \Rightarrow \left( {x - 5} \right)\left( {11x + 1} \right) = 0 \cr & \Rightarrow x = 5\,\,m \cr & \therefore y = \frac{{9x - 1}}{4} = \frac{{9 \times 5\,\,m - 1}}{4} = 11\,\,m \cr & \text{Required difference} \cr & = \left(11-5\right) m \cr & = 6\,m } $$

$$\eqalign{ & {\text{Area of }}\Delta {\text{ }}PQR \cr & = \frac{1}{2} \times QR \times PQ \cr & {\text{Given}},\,PQ = 3\,cm \cr & QR = \sqrt {{{\left( {PR} \right)}^2} - {{\left( {PQ} \right)}^2}} \cr & \,\,\,\,\,\,\,\,\,\, = \sqrt {{{\left( {25} \right)}^2} - {3^2}} \,cm \cr & \,\,\,\,\,\,\,\,\,\, = \sqrt {625 - 9} \,cm \cr & \,\,\,\,\,\,\,\,\,\, = \sqrt {616} \,cm \cr & \,\,\,\,\,\,\,\,\,\, = 2\sqrt {154} \,cm \cr & \therefore {\text{Area}}\,{\text{of}}\,\Delta \,PQR \cr & = \left( {\frac{1}{2} \times 2\sqrt {154} \times 3} \right)c{m^2} \cr & = 3\sqrt {154} {\mkern 1mu} {\mkern 1mu} c{m^2} \cr} $$

$$\eqalign{ & {\text{Area of the traingle :}} \cr & = \frac{1}{2}ab\sin \theta \cr & = \left( {\frac{1}{2} \times 10 \times 10 \times \sin {{45}^ \circ }} \right)c{m^2} \cr & = \left( {\frac{1}{2} \times 10 \times 10 \times \frac{1}{{\sqrt 2 }}} \right)c{m^2} \cr & = \left( {\frac{{50}}{{\sqrt 2 }} \times \frac{{\sqrt 2 }}{{\sqrt 2 }}} \right)c{m^2} \cr & = 25\sqrt 2 \,c{m^2} \cr} $$

Let the altitude of the triangle be h 1 and base of each be b. Then, $$\frac{1}{2}$$ × b 1 × h 1 = b × h 2 , where h 2 = 100 m ⇔ h 1 = 2 h 2 ⇔ h 1 = (2 × 100) m ⇔ h 1 = 200 m

$$\eqalign{ & \because \pi {R^2} = 24.64 \cr & \Leftrightarrow {R^2} = \left( {\frac{{24.64}}{{22}} \times 7} \right) \cr & \Leftrightarrow {R^2} = 7.84 \cr & \Leftrightarrow R = \sqrt {7.84} \cr & \Leftrightarrow R = 2.8\,m \cr & \therefore {\text{Circumference :}} \cr & = \left( {2 \times \frac{{22}}{7} \times 2.8} \right)m \cr & = 17.60\,m \cr} $$

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

Radius of the plot = 21 m Area of the path : $$\eqalign{ & = \pi \left[ {{{\left( {24.5} \right)}^2} - {{\left( {21} \right)}^2}} \right]{m^2} \cr & = \left[ {\pi \left( {24.5 + 21} \right)\left( {24.5 - 21} \right)} \right]{m^2} \cr & = \left( {\frac{{22}}{7} \times 45.5 \times 3.5} \right){m^2} \cr & = 500.5\,{m^2} \cr} $$ ∴ Cost of gravelling : = Rs. (500.5 × 4) = Rs. 2002

Required area = Area of the quadrant with radius 14 m : $$\eqalign{ & = \left( {\frac{{22}}{7} \times 14 \times 14 \times \frac{{90}}{{360}}} \right){m^2} \cr & = 154\,{m^2} \cr} $$

$$\eqalign{ & s = \frac{{13 + 14 + 15}}{2} = \frac{{42}}{2} = 21 \cr & \therefore \,\,\vartriangle = \sqrt {21 \times 8 \times 7 \times 6} \cr & \,\,\,\,\,\,\,\,\,\,\, = \sqrt {s\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right)} \cr & \,\,\,\,\,\,\,\,\,\,\, = \left( {2 \times 2 \times 3 \times 7} \right) \cr & \,\,\,\,\,\,\,\,\,\,\, = 84\,c{m^2} \cr} $$ Radius of circle : $$\eqalign{ & = \left( {\frac{{13 \times 14 \times 15}}{{4 \times 84}}} \right)c{m^2} \cr & = \left( {\frac{{65}}{8}} \right)cm \cr & = 8.125\,\,cm \cr} $$

Distance mode by the skater in 44 sec = Circumference of the circle = $$\left( {2 \times \frac{{22}}{7} \times 28} \right)$$ m = 176 m Speed of skater : = $$\left( {\frac{{176}}{{44}}} \right)$$ m/sec = 4 m/sec Perimeter of hexagon : = (6 × 48) m = 288 m ∴ Required difference : = $$\left( {\frac{{288}}{{4}}} \right)$$ sec = 72 sec

Area of square = 3136 cm 2 Side of squared = $$\sqrt {3136} $$ = 56 cm Perimeter of square : = 4a = (4 × 56) cm = 224 cm = diameter of circle ∴ Circumference of circle : $$\eqalign{ & = \pi d \cr & = \frac{{22}}{7} \times 224 \cr & = 704{\text{ cm}} \cr} $$

$$\eqalign{ & {\text{Breadth}} = \frac{{{\text{Area}}}}{{{\text{Length}}}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {\frac{{2100}}{{60}}} \right)m \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 35m \cr & \therefore {\text{Perimeter}} = 2\left( {60 + 35} \right)m \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 190\,m \cr} $$

Area of the floor = (14 × 9) m 2 = 126 m 2 ∴ Length of the carpet : $$\eqalign{ & = \left( {\frac{{126}}{{63}} \times 100} \right){\text{m}} \cr & = 200{\text{ metres}} \cr} $$

Let the original length and breadth of the rectangle be l and b respectively Then, original area = lb New length = 120% of l = $$\frac{6l}{5}$$ New area = 150% of lb = $$\frac{3lb}{2}$$ New breadth : $$\eqalign{ & = \left( {\frac{{3lb}}{2} \times \frac{5}{{6l}}} \right) \cr & = \frac{{5b}}{4} \cr} $$ Increase in breadth : $$\eqalign{ & = \left( {\frac{{5b}}{4} - b} \right) \cr & = \frac{b}{4} \cr} $$ ∴ Increase % : $$\eqalign{ & = \left( {\frac{b}{4} \times \frac{1}{b} \times 100} \right)\% \cr & = 25\% \cr} $$

Area of each square : $$\eqalign{ & = \left( {\frac{{400}}{{64}}} \right)c{m^2} \cr & = 6.25\,c{m^2} \cr} $$ Side of each small square : $$\eqalign{ & = \sqrt {6.25} \,cm \cr & = 2.5\,cm \cr} $$ Since there are 8 squares along each side of the chess board, we have : Side = [(8 × 2.5) + 6] cm = 26 cm

Let the length of the diagonal be x km Then, $$\eqalign{ & \Rightarrow \frac{1}{2}{x^2} = 50 \cr & \Rightarrow {x^2} = 100 \cr & \Rightarrow x = 10\,{\text{km}} \cr & \Rightarrow x = \left( {\frac{{10}}{{1.6}}} \right){\text{miles}} \cr & \Rightarrow x = 6.25{\text{ miles}} \cr & \left[ {\because 1{\text{ mile}} = {\text{1.609 km}}} \right] \cr} $$

$$\eqalign{ & {\text{ Let }}AP{\text{ }} = {\text{ }}AQ{\text{ }} = {\text{ }}x{\text{ metres}} \cr & {\text{Then,}} \cr & \Rightarrow {x^2} + {x^2} = {\left( {\sqrt 2 } \right)^2} \cr & \Rightarrow 2{x^2} = 2 \cr & \Rightarrow {x^2} = 1 \cr & \Rightarrow x = 1\,m \cr & So,\,\,\vartriangle PAQ{\text{ is isosceles}}{\text{.}} \cr & \therefore PT = QT = \left( {\frac{{\sqrt 2 }}{2}} \right)m = \left( {\frac{1}{{\sqrt 2 }}} \right)m \cr & {\text{In }}\vartriangle {\text{ }}PTA{\text{, we have : }}\angle PTA = {90^ \circ } \cr & \therefore A{T^2} = A{P^2} - P{T^2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {1^2} - {\left( {\frac{1}{{\sqrt 2 }}} \right)^2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1 - \frac{1}{2} = \frac{1}{2} \cr & Or,AT = \left( {\frac{1}{{\sqrt 2 }}} \right)m \cr & {\text{Similarly,}} \cr & CX = \left( {\frac{1}{{\sqrt 2 }}} \right)m \cr & \therefore PS = QR = XT \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = AC - 2 \times AT \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left[ {8\sqrt 2 - \left( {2 \times \frac{1}{{\sqrt 2 }}} \right)} \right]m \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {8\sqrt 2 - \frac{2}{{\sqrt 2 }}} \right)m \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{14}}{{\sqrt 2 }}m \cr & {\text{Area of the plank :}} \cr & = \left( {\frac{{14}}{{\sqrt 2 }} \times \sqrt 2 } \right){m^2} \cr & = 14\,{m^2} \cr} $$

Since 3 2 + 4 2 = 5 2 , So it is a right-angled triangle with Base = 3 cm and Height = 4 cm ∴ Area : $$\eqalign{ & = \left( {\frac{1}{2} \times 3 \times 4} \right)c{m^2} \cr & = 6\,c{m^2} \cr} $$

Let the smallest side be x cm. Then, other sides are 13 cm and (17 - x) cm Let a = 13, b = x and c = (17 - x) So, s = 15 $$\eqalign{ & {\text{Area}} = \sqrt {s\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right)} \cr & \,\,\,\,\,\,\,\,\,\,\, = \sqrt {15 \times 2 \times \left( {15 - x} \right)\left( {x - 2} \right)} \cr & \,\,\,\,\,\,\,\,\,\,\, = \sqrt {30\left( {15 - x} \right)\left( {x - 2} \right)} \cr & \therefore 30\left( {15 - x} \right)\left( {x - 2} \right) = {\left( {30} \right)^2} \cr & \Rightarrow \left( {15 - x} \right)\left( {x - 2} \right) = 30 \cr & \Rightarrow {x^2} - 17x + 60 = 0 \cr & \Rightarrow \left( {x - 12} \right)\left( {x - 5} \right) = 0 \cr & \Rightarrow x = 12{\text{ or }}x = 5 \cr & {\text{ Smallest side = 5 cm}} \cr} $$

Let each have base = b and height = h Then, P = b × h, R = b × h, T = $$\frac{1}{2}$$ × b × h So, P = R, P = 2T and T = $$\frac{1}{2}$$R, are all correct statements. ∴ Option B is false