Mensuration 2D — Part 1 MCQs with Explanations

A 1 and A 2 are 2 regular polygon Sum of internal angle of A 1 = 1080° (n - 2) × 180° = 1080° (n - 2) = 6 n = 6 + 2 = 8 Internal angle of each A 2 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 A 1 and A 2 = 15 + 8 = 23

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

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 $$ = {\left( {\frac{{36}}{7}} \right)^2} = \frac{{1296}}{{49}}$$

$$\eqalign{ & {\text{Area of quadrant XDTB}} \cr & = \frac{1}{4}\pi \times 14 \times 14 \cr & = \frac{1}{4} \times \frac{{22}}{7} \times 14 \times 14 \cr & = 154{\text{ c}}{{\text{m}}^2} \cr & {\text{Area of four semi - circle}} \cr & = 4 \times \frac{1}{2}\left[ {\pi \times \frac{7}{2} \times \frac{7}{2}} \right] \cr & = 2 \times \frac{{22}}{7} \times \frac{7}{2} \times \frac{7}{2} \cr & = 77{\text{ c}}{{\text{m}}^2} \cr} $$ $$\eqalign{ & {\text{Area of shaded region = 2}} \times {\text{Area of quadrilateral MONX}} \cr & = 2 \times \left[ {\frac{1}{4}\pi \times 3.5 \times 3.5 - \frac{1}{2} \times 3.5 \times 3.5} \right] \cr & = {\left( {3.5} \right)^2}\left[ {\frac{{11}}{7} - 1} \right] \cr & = \frac{4}{7} \times \frac{7}{2} \times \frac{7}{2} \cr & = 7{\text{ c}}{{\text{m}}^2} \cr} $$

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) $$m = \frac{{nx}}{y}$$

In ΔABQ AQ 2 + AB 2 = BQ 2 R 2 + R 2 = (7 - R) 2 2R 2 = 49 + R 2 - 14R R 2 + 14R - 49 = 0 ∴ R = $$ = \frac{{ - 14 \pm \sqrt {196 + 196} }}{2}$$ R = -7 ± 7$$\sqrt 2 $$ ∴ R = 7$$\sqrt 2 $$ - 7 ∴ Area of circle = πR 2 = $$\frac{{22}}{7}$$ × (7$$\sqrt 2 $$ - 7)(7$$\sqrt 2 $$ - 7) = 154(2 + 1 - 2$$\sqrt 2 $$ ) = 462 - 308$$\sqrt 2 $$

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

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 $$\eqalign{ & = \frac{{l \times b}}{{\frac{{\sqrt 3 }}{4}{a^2}}} \cr & = \frac{{\frac{a}{2} \times a}}{{\frac{{\sqrt 3 }}{4}{a^2}}} \cr & = \frac{{{a^2}}}{2} \times \frac{4}{{\sqrt 3 {a^2}}} \cr & = \frac{2}{{\sqrt 3 }} \cr & = {\bf{2:}}\sqrt {\bf{3}} \cr} $$

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

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

$$\eqalign{ & {\text{Side of square}} = \frac{{24}}{4} = 6 \cr & {\text{Then area of }}\Delta ABC = \frac{1}{2} \times 8 \times 6 = 24{\text{ c}}{{\text{m}}^2} \cr} $$

AB = BC = CD = DA BC = CX = XY = YZ = BZ AB = BE = AE In ΔABE ∠A = ∠B = ∠E = 60° In Pentagon BCXYZ Each internal angle of a pentagon = $$\frac{{\left( {n - 2} \right) \times 180}}{n}$$ ∠ZBC = $$\frac{{3 \times 180}}{5} \times {108^ \circ }$$ ∠EBZ + ∠ABE + ∠ABC + ∠ZBC = 360° ∠EBZ + 60° + 90° + 180° = 360° ∠EBZ = 102°

Length of rubber band = 3d + 2πr = 30 + 10π

Area of trapezium = $$\frac{1}{2}$$ × sum of parallel side × Distance between them 792 = $$\frac{1}{2}$$ (26 + 40) × Distance $$\frac{{792 \times 2}}{{66}}$$ = Distance ∴ Distance = 24

$$\eqalign{ & {\text{In }}\Delta PCB \cr & BC = \sqrt {{{10}^2} - {6^2}} = 8\,{\text{cm}} \cr & BC = AD = 8\,{\text{cm}} \cr & {\text{In }}\Delta ADP \cr & AP = \sqrt {{{13}^2} - {8^2}} \cr & AP = \sqrt {169 - 64} = \sqrt {105} \cr} $$

Area of verandah Area of verandah on outer boundary = 2x($$l$$ + b + 2x) = 2 × 3.5(25 + 15 + 2 × 3.5) = 329 m 2 Cost of flooring = 329 × 27.5 = Rs. 9047.50 (approx.)

Area of circle = 38.5 πr 2 = 38.5 r 2 = $$\frac{{385}}{{10}} \times \frac{7}{{22}}$$ r = $$\frac{7}{2}$$ cm MN = 2r = 7 cm $${\sqrt 2 }$$ a = 7 cm a = $$\frac{7}{{\sqrt 2 }}$$ cm Side of square EFGH = 2 × a EG = 7$${\sqrt 2 }$$ cm $${\sqrt 2 }$$ b = 7$${\sqrt 2 }$$ b = 7 cm Side of square ABCD = 7 × 2 = 14 cm Area = a 2 = 14 2 = 196 cm 2

Perimeter of road = 2(118 + 102) = 2 × (220) = 440 Required cost = 440 × $$\frac{5}{2}$$ = Rs. 1110

$$\eqalign{ & {\text{Side ratio}} = \frac{1}{3}:\frac{1}{5}:\frac{1}{6} = \boxed{10:6:5} \cr & 21\mu \to 147 \cr & 1\mu \to 7 \cr & {\text{Smallest side, }}5\mu \to 5 \times 7 = 35 \cr} $$

$$\eqalign{ & {\text{Area of the hexagon}} \cr & \Rightarrow 6 \times \frac{{\sqrt 3 }}{4} \times {a^2} \cr & \Rightarrow 6 \times \frac{{\sqrt 3 }}{4} \times 14 \times 14 \cr & \Rightarrow 294\sqrt 3 \cr} $$

$$\eqalign{ & {\text{Side of equilateral }}\Delta = 8{\text{ cm}} \cr & {\text{In circle radius}} = \frac{a}{{2\sqrt 3 }} = \frac{8}{{2\sqrt 3 }} = \frac{4}{{\sqrt 3 }} \cr & {\text{Circum radius}} = \frac{a}{{\sqrt 3 }} = \frac{8}{{\sqrt 3 }} \cr & {\text{Area bounded by both circle is}} \cr & = \pi \left\{ {{{\left( {\frac{8}{{\sqrt 3 }}} \right)}^2} - {{\left( {\frac{4}{{\sqrt 3 }}} \right)}^2}} \right\} \cr & = \frac{{22}}{7}\left( {\frac{{64}}{3} - \frac{{16}}{3}} \right) \cr & = \frac{{22}}{7}\left( {\frac{{48}}{3}} \right) \cr & = \frac{{22 \times 16}}{7} \cr & = 50\frac{2}{7}{\text{ c}}{{\text{m}}^2} \cr} $$

$$\eqalign{ & {\text{Area of trapezium}} = \frac{1}{2}\left( {AB + CD} \right) \times h \cr & = \frac{1}{2} \times 6 \times 1.44 \cr & = 4.32{\text{ sq}}{\text{. yards}} \cr & \left( {{\text{1 sq}}{\text{. yards}} = 1800} \right) \cr & \therefore 4.32 = 1800 \times \frac{{432}}{{100}} = {\text{Rs}}{\text{. }}7776 \cr} $$

$$\eqalign{ & = 3 \times 2\sqrt {Rr} \cr & = 3 \times 2\sqrt {4 \times 4} \cr & = 24 \cr & = 24 + 2\pi r \cr & = 24 + 2\pi \times 4 \cr & = 24 + 8\pi {\text{ cm}} \cr} $$

Distance covered in 30 seconds = 30 m/min × $$\frac{{30}}{{60}}$$ = 15 m This is the difference of distance of the boundary and the diameter. Let 'R' be the radius $$\eqalign{ & 2\pi R - 2R = 15 \cr & 2R\left( {\pi - 1} \right) = 15 \cr & 2R = \frac{{15}}{{\pi - 1}} \cr & 2R = \frac{{15}}{{\frac{{22}}{7} - 1}} \cr & 2R = \frac{{15 \times 7}}{{15}} \cr & 2R = 7 \cr & R = \frac{7}{2} \cr & R = 3.5{\text{ m}} \cr} $$

Perimeter of equilateral triangle = 100 + 100 + 100 = 300 Perimeter of New triangle = 120 + 150 + 130 = 400 % increase $$ = \frac{{100}}{{300}} \times 100 = 33\frac{1}{3}\% $$

In ΔABC & ΔDCQ ∠ABC = ∠DCQ ∠ACB = ∠DQC BC = CQ ΔABC ≅ ΔDCQ ar ΔABC = ar ΔDCQ

According to the figure, ⇒ Perimeter = r + r + $$l$$ ⇒ 75 cm = 2r + length of arc ⇒ 75 cm = 2r + $$\frac{{2\pi r}}{4}$$ ⇒ 75 cm = 2r + $$\frac{{22 \times r}}{{7 \times 2}}$$ ⇒ r = 21 cm ⇒ Its area $$ = \frac{1}{4}\left[ {\frac{{22}}{7} \times 21 \times 21} \right] = 346.5{\text{ c}}{{\text{m}}^2}$$

Area of parallelogram = BC × FC = 15 × 12 = 180 cm 2 Area of parallelogram DC × AE = 180 18 × AE = 180 AE = 10 cm ∴ Distance between bigger sides = 10 cm

Let P be the point inside the equilateral ΔABC Let, PD = √3, PE = 2√3, PF = 5√3 and AB = BC = AC = $$x$$ $$\eqalign{ & {\text{ar}}{\text{.}}\,\Delta {\text{ABC}} = {\text{ar}}{\text{.}}\,\Delta {\text{ABP}} + {\text{ar}}{\text{.}}\,\Delta {\text{ACP}} + {\text{ar}}{\text{.}}\,\Delta {\text{BCP}} \cr & \frac{{\sqrt 3 }}{4}{x^2} = \frac{1}{2} \times x \times \sqrt 3 + \frac{1}{2} \times x \times 2\sqrt 3 + \frac{1}{2} \times x \times 5\sqrt 3 \cr & \sqrt 3 x = 2\sqrt 3 + 4\sqrt 3 + 10\sqrt 3 \cr & x = 16 \cr} $$ ∴ Perimeter of triangle = 3$$x$$ = 3 × 16 = 48 cm

$$\eqalign{ & {\text{Area Grazed by the cow}} \cr & = \frac{{{A^ \circ }}}{{360}}\pi {\left( 7 \right)^2} + \frac{{{B^ \circ }}}{{360}}\pi {\left( 7 \right)^2} + \frac{{{C^ \circ }}}{{360}}\pi {\left( 7 \right)^2} \cr & = \pi {\left( 7 \right)^2}\left[ {\frac{{{A^ \circ } + {B^ \circ } + {C^ \circ }}}{{360}}} \right] \cr & = \pi {\left( 7 \right)^2} \times \frac{{180}}{{360}} \cr & = \frac{1}{2}\pi {\left( 7 \right)^2} \cr & = 77{\text{ }}{{\text{m}}^2} \cr & s = \frac{{26 + 30 + 28}}{2} = 42 \cr & {\text{Area of field}} = \sqrt {s\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right)} \cr & = \sqrt {42 \times 16 \times 14 \times 12} \cr & = 336{\text{ }}{{\text{m}}^2} \cr & \Rightarrow {\text{Remaining area}} = 336 - 77 = 259{\text{ }}{{\text{m}}^2} \cr} $$

Given, AB = 2 CD = 1 Let BC = x ∵ Area of middle circle = Average of areas of other two circle $$\eqalign{ & \frac{\pi }{4}{\left( {2 + x} \right)^2} = \frac{{\frac{\pi }{4}{{\left( {2 + x + 1} \right)}^2} + \frac{\pi }{4}{{\left( 2 \right)}^2}}}{2} \cr & 2\left[ {\frac{\pi }{4}{{\left( {2 + x} \right)}^2}} \right] = \frac{\pi }{4}{\left( {3 + x} \right)^2} + \frac{\pi }{4} \times 4 \cr & 2{\left( {2 + x} \right)^2} = {\left( {3 + x} \right)^2} + 4 \cr & 2\left( {4 + {x^2} + 4x} \right) = 9 + {x^2} + 6x + 4 \cr & 8 + 2{x^2} + 8x = 9 + {x^2} + 6x + 4 \cr & {x^2} + 2x - 5 = 0 \cr & x = \frac{{ - 2 \pm \sqrt {4 + 20} }}{2} \cr & x = \frac{{ - 2 \pm 2\sqrt 6 }}{2} \cr & x = - 1 \pm \sqrt 6 \cr & \therefore x = \sqrt 6 - 1 = {\text{BC}} \cr} $$

$$\eqalign{ & 2S = 50 + 70 + 80 \cr & S = 100 \cr & A = \sqrt {S\left( {S - a} \right)\left( {S - b} \right)\left( {S - c} \right)} \cr & A = \sqrt {100 \times 50 \times 30 \times 20} \cr & A = 1000\sqrt 3 \cr & \frac{{\sqrt 3 }}{4}{x^2} = 1000\sqrt 3 \cr & {x^2} = 4000 \cr & x = 63.2 \cr} $$

$$\eqalign{ & {\text{Length of side}} = \frac{2}{{\sqrt 3 }}\left( {{P_1} + {P_2} + {P_3}} \right) \cr & = \frac{2}{{\sqrt 3 }}\left( {6 + 7 + 8} \right) \cr & = \frac{2}{{\sqrt 3 }} \times 21 \cr & = \frac{{42}}{{\sqrt 3 }} \times \frac{{\sqrt 3 }}{{\sqrt 3 }} \cr & = \frac{{42\sqrt 3 }}{3} \cr & = 14\sqrt 3 {\text{ cm}} \cr} $$

AB = 56 cm AE = EF = FG = GB = 14 cm ∴ EM = MF = 7 cm PF = $$\frac{{56}}{2}$$ = 28 cm OF = 28 - r OM = 7 + r MF = 7 In ΔOMF, OM 2 = OF 2 + MF 2 (7 + r) 2 = (28 - r) 2 + 7 2 49 + r 2 + 14r = 784 + r 2 - 56r + 49 70r = 784 r = $$\frac{{784}}{{70}}$$ = 11.2 Area of circle = πr 2 = $$\frac{{22}}{7} \times \frac{{112}}{{10}} \times \frac{{112}}{{10}}$$ = 394.24 cm 2

Circumference of paper sheet = 352 $$\eqalign{ & 2\pi R = 352 \cr & R = \frac{{352}}{{2\pi }} = \frac{{352 \times 7}}{{2 \times 22}} = 56{\text{ cm}} \cr & r = \frac{R}{2} = \frac{{56}}{2} = 28{\text{ cm}} \cr} $$ ∴ Circumference of circular plate $$\eqalign{ & = 2\pi r \cr & = 2 \times \frac{{22}}{7} \times 28 \cr & = 176{\text{ cm}} \cr} $$

Let the length of square park = L Area = L 2 L 2 = 16x 2 + 8x + 1 Length of the park $$\eqalign{ & = \sqrt {16{x^2} + 8x + 1} \cr & = \sqrt {{{\left( {4x} \right)}^2} + 2 \times 4x + {1^2}} \cr & = \sqrt {{{\left( {4x + 1} \right)}^2}} \cr & = \left( {4x + 1} \right){\text{ units}} \cr} $$

$$\eqalign{ & \sqrt {{a^2} + {b^2}} ,\,\sqrt {{{\left( {2a} \right)}^2} + {b^2}} ,\,\sqrt {{a^2} + {{\left( {2b} \right)}^2}} \cr & {\text{Let }}a = b \cr & \sqrt 2 a,\,\sqrt 5 a,\,\sqrt 5 a \cr} $$ $$\eqalign{ & {\text{Height}}\left( h \right) = \sqrt {{l^2} - {b^2}} \cr & = \sqrt {{{\left( {\sqrt 5 a} \right)}^2} - {{\left( {\frac{{\sqrt 2 }}{2}a} \right)}^2}} \cr & = \sqrt {\frac{{9{a^2}}}{2}} \cr & = \frac{{3a}}{{\sqrt 2 }} \cr & {\text{Area}} = \frac{1}{2} \times \frac{{3a}}{{\sqrt 2 }} \times \sqrt 2 a \cr & = \frac{3}{2}{a^2} \cr & = \frac{3}{2}ab \cr} $$

Let the radius of swimming pool = R Outer radius of pool with concrete wall = (R + 4) According to question πR 2 × $$\frac{{11}}{{25}}$$ = π(R + 4) 2 - πR 2 R 2 × $$\frac{{11}}{{25}}$$ = R 2 + 16 + 8R - R 2 $$\frac{{11}}{{25}}$$ R 2 = 16 + 8R 11R 2 - 200R - 400 = 0 By option (D), (In such type of equation go through the option to save your valuable time) R = 20 11 × (20) 2 - 200 × 20 - 400 = 0 4400 - 4000 - 400 = 0 0 = 0 (satisfy) Therefore, radius of pool R = 20 m

$$\eqalign{ & {\text{ar }}\Delta EDC = \frac{1}{2} \times ED \times DC \times \sin \angle EDC \cr & = \frac{1}{2} \times 12 \times 12 \times \sin {120^ \circ } \cr & = 36\sqrt 3 {\text{ c}}{{\text{m}}^2} \cr} $$

AO = OC = 12 AB = 13 BO = 5 Area of rhombus 4 × Area of ΔAOB = 4 × $$\frac{1}{2}$$ × 12 × 5 = 120 cm 2