Mensuration 2D — Part 2 MCQs with Explanations

Let width = x m Area of Park = 60 × 40 = 2400 m 2 Area of lawn = 2109 m 2 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. x 2 - 100x + 291 = 0 x 2 - (97 + 3)x + 291 = 0 x 2 - 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 units 2

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

Area of circle = A Radius of circle = r Circumference of circle = C πr 2 = 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

Let AB = AC = a cm BD = DC = b cm ∴ Altitude of isosceles triangle is also median In right ΔADC AD 2 = a 2 - b 2 64 = a 2 - b 2 . . . . . . (i) Perimeter = 64 a + a + 2b = 64 2a + 2b = 64 a + b = 32 . . . . . (ii) On dividing $$ = \frac{{{a^2} - {b^2}}}{{a + b}} = \frac{{64}}{{32}} = 2$$ ∴ a 2 - b 2 = (a + b)(a - b) a - b = 2 ∴ a + b = 32 On solving a = 17, b = 15 Area of ΔABC = $$\frac{1}{2}$$ × AD × BC = $$\frac{1}{2}$$ × 8 × 30 = 120 cm

Let radius of circle = r ∴ OA = 8 - r and OP = 8 + r Now ∠OAP is right angle triangle and use pythagoras theorem in ΔOAP ⇒ (8 + r) 2 = (8 - r) 2 + 4 2 64 + r 2 + 16r = 64 + r 2 - 16r + 16 32r = 16 r = $$\frac{1}{2}$$cm And area of circle = πr 2 $$\eqalign{ & = \frac{{22}}{7} \times \frac{1}{2} \times \frac{1}{2} \cr & = \frac{{11}}{{14}}{\text{ c}}{{\text{m}}^2} \cr} $$

Diagonal of a square = (a + b) ∴ side of square = $$\frac{{\left( {a + b} \right)}}{{\sqrt 2 }}$$ ∴ Area of square = (side) 2 $$\eqalign{ & = {\left( {\frac{{a + b}}{{\sqrt 2 }}} \right)^2} \cr & = \frac{1}{2}\left( {{a^2} + {b^2} + 2ab} \right) \cr & = \frac{1}{2}\left( {{a^2} + {b^2}} \right) + ab \cr} $$

$$\eqalign{ & {\text{In }}\sin \,{\text{rule}} \cr & \frac{{\sin {{75}^ \circ }}}{{PQ}} = \frac{{\sin {{60}^ \circ }}}{{PO}} \cr & \frac{{\sin \left( {{{30}^ \circ } + {{45}^ \circ }} \right)}}{8} = \frac{{\sin {{60}^ \circ }}}{{PO}} \cr & \frac{{\frac{1}{2} \times \frac{1}{{\sqrt 2 }} + \frac{{\sqrt 3 }}{2} \times \frac{1}{{\sqrt 2 }}}}{8} = \frac{{\sqrt 3 }}{{2PO}} \cr & \frac{{\sqrt 3 + 1}}{{2\sqrt 2 \times 8}} = \frac{{\sqrt 3 }}{{2PO}} \cr & PO = \frac{{8\sqrt 6 }}{{\left( {\sqrt 3 + 1} \right)}} \cr & {\text{After rationaltion}} \cr & PO = \frac{{8\sqrt 6 }}{{\left( {\sqrt 3 + 1} \right)}} = 4\sqrt 6 \left( {\sqrt 3 - 1} \right) \cr & {\text{Then in }}\Delta POD, \cr & {\text{Area}} = \frac{1}{2}PQ \times PO\sin {45^ \circ } \cr & = \frac{1}{2} \times 8 \times 4\sqrt 6 \left( {\sqrt 3 - 1} \right) \times \frac{1}{{\sqrt 2 }} \cr & = 2\sqrt 2 \times 4\sqrt 6 \left( {\sqrt 3 - 1} \right) \cr & = 16\left( {3 - \sqrt 3 } \right) \cr} $$

\[\Delta OAD \sim \Delta ABE\left( \begin{gathered} \because AE\,||\,DC \hfill \\ \therefore \angle E = \angle D \hfill \\ \angle B = \angle A \hfill \\ \end{gathered} \right)\] $$\eqalign{ & \frac{{{\text{ar}}{\text{. }}\Delta OAD}}{{{\text{ar}}{\text{. }}\Delta ABE}} = \frac{1}{4} \cr & \Delta OBC \sim \Delta ABE \cr & \therefore \frac{{{\text{ar}}{\text{. }}\Delta OBC}}{{{\text{ar}}{\text{. }}\Delta ABE}} = \frac{9}{4} \cr & \because 4 = 72{\text{ c}}{{\text{m}}^2} \cr & \therefore 1 = 18{\text{ c}}{{\text{m}}^2} \cr & 9 = 162{\text{ c}}{{\text{m}}^2} \cr & \therefore {\text{ ar}}{\text{. of }}\square ADEC \cr & \Rightarrow 162 - \left( {72 + 18} \right) = 72{\text{ c}}{{\text{m}}^2} \cr & \therefore {\text{Area of trapezium }}ADCB \cr & = 72 + 72 \cr & = 144{\text{ c}}{{\text{m}}^2} \cr} $$

Let sides of triangle are, a, b and c respectively ∴ largest side given = 17 cm = Perimeter = a + b + c = 40 cm (given) Area = 60 cm 2 (given) In such questions take the help of triplets which form right angle triangle So, here we have a side 17 cm ⇒ by triplet we get sides 8 and 15 ⇒ check the sides Perimeter = 8 + 15 + 17 = 40 Area = $$\frac{1}{2}$$ × 8 × 15 ⇒ 60 Hence sides are 15, 8 Smaller side = 8 cm

It is clear from the figure Area of shaded region = Area of sector of OFG and OAD + Area of circle BCHE - area of sector OHE and OBC $$\eqalign{ & = 2 \times \frac{\theta }{{360}} \times \pi {r^2} + \pi {r^2} - 2 \times \frac{\theta }{{360}} \times \pi {r^2} \cr & = 2 \times \frac{{45}}{{360}} \times 3.14 \times 24 \times 24 + 3.14 \times {8^2} - 2 \times \frac{{45}}{{360}} \times 3.14 \times {8^2} \cr & = \frac{1}{4} \times 3.14 \times 24 \times 24 + 3.14 \times 64 - \frac{1}{4} \times 3.14 \times 64 \cr & = 3.14 \times \left( {144 + 64 - 16} \right) \cr & = 3.14 \times 192 \cr & = 602.88{\text{ Answer}} \cr} $$

$$\eqalign{ & {\text{According to the question,}} \cr & {\text{Area}} \times {\text{Rate}} = {\text{Expenditure}} \cr & \pi {r^2} \times \frac{1}{2} = {\text{Rs}}{\text{. }}7700 \cr & {r^2} = \frac{{7700 \times 7}}{{22}} \times 2 \cr & \boxed{r = 70} \cr & {\text{Perimeter}} = 2\pi r \cr & = 2 \times \frac{{22}}{7} \times 70 \cr & = 440{\text{ meter}} \cr & {\text{Expenditure}} = 440 \times 1.20 = {\text{Rs}}{\text{. }}528 \cr} $$

Let the side of equilateral triangle = 'a' and the side of square = 'b' In circle radius of equilateral $$\Delta = \frac{a}{{2\sqrt 3 }}$$ $$\eqalign{ & \therefore {\text{Diagonal of square}} = 2 \times \frac{a}{{2\sqrt 3 }} = \frac{a}{{\sqrt 3 }} \cr & {\text{Now, }}b = \frac{{{\text{Diagonal}}}}{{\sqrt 2 }} = \frac{a}{{\frac{{\sqrt 3 }}{{\sqrt 2 }}}} = \frac{a}{{\sqrt 6 }} \cr & {\text{Required ratio}} = \frac{{\frac{{\sqrt 3 }}{4}{a^2}}}{{{{\left( {\frac{a}{{\sqrt 6 }}} \right)}^2}}} \cr & = \frac{{\sqrt 3 }}{4}{a^2} \times \frac{6}{{{a^2}}} \cr & = \frac{{3\sqrt 3 }}{2} \Rightarrow 3\sqrt 3 :2 \cr} $$

Let radius of circle = R Side of square = a Side of equilateral Δ = b According to question, $$\eqalign{ & 2\pi R = 4a = 3b \cr & \therefore a = \frac{{\pi R}}{2} \cr & b = \frac{2}{3}\pi R \cr} $$ Ratio of their areas \[\begin{array}{*{20}{c}} {\pi {R^2}}&:&{{a^2}}&:&{\frac{{\sqrt 3 }}{4}{b^2}} \\ {\pi {R^2}}&:&{{{\left( {\frac{{\pi R}}{2}} \right)}^2}}&:&{\frac{{\sqrt 3 }}{4}{{\left( {\frac{2}{3}\pi R} \right)}^2}} \\ 1&:&{\frac{\pi }{4}}&:&{\frac{{\sqrt 3 }}{9}\pi } \\ {\text{C}}&:&{\text{S}}&:&{\text{T}} \end{array}\] Here, we can see that C > S > T Quicker Approach: When perimeter of two or more figures are same then the figure which has more vertex is greater in the area. Since, here, circle has infinite vertex. Therefore, C > S > T

$$\eqalign{ & {\text{Area of half }}\Delta = \frac{1}{2}\left( {\frac{{\sqrt 3 }}{4}{a^2}} \right) \cr & = \frac{1}{2} \times \frac{{\sqrt 3 }}{4} \times 6 \times 6 \cr & = \frac{{9\sqrt 3 }}{2} \cr & {\text{Area of shaded region}} \cr & \Rightarrow \frac{{9\sqrt 3 }}{2} - \frac{{90}}{{360}}\pi {r^2} \cr & = \frac{{9\sqrt 3 }}{2} - \frac{1}{4} \times \frac{{22}}{7} \times 1 \times 1 \cr & = \frac{{9\sqrt 3 }}{2} - \frac{{11}}{{2 \times 7}} \cr & = \frac{1}{2}\left( {9\sqrt 3 - \frac{{11}}{7}} \right){\text{c}}{{\text{m}}^2} \cr} $$

In this condition always square area is greater than any other quadrilateral So, option (C) is correct. Alternate Solution:- Let perimeter = 16 cm Perimeter of square = 4a = 16 a = 4 Area (A) of square = 16 m 2 Perimeter of rectangle = 2($$l$$ + b) 16 = 2($$l$$ + b) $$l$$ + b = 8 Area (B) of rectangle = $$\left( {\frac{{4,\,4}}{{6,\,2}}} \right)$$ = 6 × 2 = 12 cm 2

a√2 = 9√2 a = 9 a + b + c = 36 Circle of maximum radius is only exist in equilateral Δ 3a = 36 a = 12 $$\eqalign{ & S = \frac{{a + b + c}}{2}, \cr & r = \frac{{\frac{{\sqrt 3 }}{4} \times {a^2}}}{{\frac{{12 + 12 + 12}}{2}}} \cr & \Rightarrow r = \frac{{\sqrt 3 }}{4} \times \frac{{12 \times 12 \times 2}}{{36}} \cr & \Rightarrow r = 2\sqrt 3 \cr} $$ ∴ Area of circle = πr 2 = 12π

Perimeter of rectangle 2(5x + 7x) = 168 5x + 7x = 84 12x = 84 x = 7 Sides of rectangle = 35 cm, 49 cm $$\eqalign{ & {\text{AC}} = \sqrt {{{35}^2} + {{49}^2}} \cr & {\text{AC}} = \sqrt {1225 + 2401} \cr & {\text{AC}} = \sqrt {3626} \cr} $$

$$\eqalign{ & {\text{Area of square}} = {\left( {14} \right)^2} = 196 \cr & {\text{Area of semicircle}} = \frac{\pi }{2}{\left( 7 \right)^2} = \frac{{49}}{2}\pi \cr & {\text{Side of square LMNO}} = \frac{{{\text{diagonal}}}}{{\sqrt 2 }} = \frac{7}{{\sqrt 2 }} \cr & {\text{Area of square LMNO}} = {\left( {\frac{7}{{\sqrt 2 }}} \right)^2} = \frac{{49}}{2} \cr & {\text{Shaded portion}} = 196 - \frac{{49}}{2}\pi - \frac{{49}}{2} \cr & = 196 - \frac{{49}}{2} \times \frac{{22}}{7} - \frac{{49}}{2} \cr & = 119 - \frac{{49}}{2} \cr & = \frac{{189}}{2} \cr & = 94.5 \cr} $$

$$\eqalign{ & {\text{Area}} = \frac{{270}}{{360}}\pi {r^2} \cr & = \frac{3}{4} \times \frac{{22}}{7} \times 20 \times 20 \cr & = \frac{{6600}}{7}\,{\text{c}}{{\text{m}}^2} \cr} $$

$$\eqalign{ & {\text{The area of largest }}\Delta = \frac{1}{2} \times b \times h \cr & = \frac{1}{2} \times 12 \times 6 \cr & = 36{\text{ }}{{\text{m}}^2} \cr} $$

$$\eqalign{ & {\text{Area of }}\Delta {\text{SQU}} = \frac{1}{2}{\text{ area of hexagon}} \cr & = \frac{1}{2} \times \left[ {6 \times \frac{{\sqrt 3 }}{4}{{\left( {12} \right)}^2}} \right] \cr & = \frac{1}{2} \times 216\sqrt 3 \cr & = 108\sqrt 3 {\text{ c}}{{\text{m}}^2} \cr} $$

$$\eqalign{ & \Delta DHC \sim \Delta DFG \cr & {\text{So, }}\frac{{DC}}{{DG}} = \frac{{CH}}{{FG}} \cr & \Rightarrow \frac{8}{{28}} = \frac{{CH}}{{20}} \cr & \Rightarrow CH = \frac{{40}}{7} \cr & {\text{So, area of }}\Delta DCH \cr & = \frac{1}{2} \times DC \times CH \cr & = \frac{1}{2} \times 8 \times \frac{{40}}{7} \cr & = \frac{{160}}{7}{\text{ c}}{{\text{m}}^2} \cr} $$

Distance between two parallel line = $$\frac{2}{{11}}$$ (sum of both parallel line) = $$\frac{2}{{11}}$$ × (7x + 4x) = 2x Area = $$\frac{1}{2}$$ (sum of parallel sides) × (distance between them) ⇒ $$\frac{1}{2}$$ (7x + 4x) × 2x = 176 ⇒ 11x 2 = 176 ⇒ x 2 = 16 ⇒ x = 4 AB = 7 × 4 = 28 cm CD = 4 × 4 = 16 cm CM = 2 × 4 = 8 cm AM = AN + NM ⇒ AM = AN + 16 ⇒ AM = 6 + 16 ⇒ AM = 22 $$\eqalign{ & \left( {{\text{AN}} = {\text{BM}} = \frac{{12}}{2} = 6} \right) \cr & {\text{A}}{{\text{C}}^2} = {\text{C}}{{\text{M}}^2} + {\text{A}}{{\text{M}}^2} \cr & \Rightarrow {\text{A}}{{\text{C}}^2} = {8^2} + {22^2} \cr & \Rightarrow {\text{AC}} = \sqrt {64 + 484} \cr & \Rightarrow {\text{AC}} = \sqrt {548} \cr & \Rightarrow {\text{AC}} = 2\sqrt {137} \cr} $$

(∴ Rhombus is a parallelogram ∴ area of Rhombus = base × height) Area of Rhombus = base × height = 6.5 × 10 = 65 cm 2 Also area of Rhombus = $$\frac{1}{2}$$ × d 1 × d 2 ⇒ $$\frac{1}{2}$$ × 26 × d 2 = 65 ⇒ 13 × d 2 = 65 ⇒ d 2 = 5 cm

Breadth = 12 m Length = 24 m Cost = 100 Rs./m 2 r = 6 R = 12 π × 144 + π × 36 + 288 = 3.14 × 144 + 3.14 × 36 + 288 = 853.2 ∴ Total cost = 853.2 × 100 = Rs. 85320

$$\eqalign{ & {\text{Let EN }} \bot \,{\text{AD}} \cr & {\text{Area of }}\Delta {\text{AED}} = \frac{1}{2} \times {\text{EN}} \times {\text{AD}} \cr & {\text{Area of trapezium ABCD}} \cr & = \frac{1}{2}\left( {{\text{AD}} + {\text{BC}}} \right) \times {\text{EN}} \cr & \frac{{{\text{ar}}\left( {{\text{ABCD}}} \right)}}{{{\text{ar}}\left( {{\text{AED}}} \right)}} \cr & = \frac{{\frac{1}{2}\left( {{\text{AD}} + {\text{BC}}} \right) \times {\text{EN}}}}{{\frac{1}{2} \times {\text{EN}} \times {\text{AD}}}} \cr & = \frac{{{\text{AD}} + {\text{BC}}}}{{{\text{AD}}}} \cr} $$

Length of perpendicular drawn from the right angle to hypotenuse, $$\eqalign{ & {\text{P}} = \frac{{{\text{a}} \times {\text{b}}}}{{\text{H}}} \cr & {{\text{P}}^2} = \frac{{{{\text{a}}^2}{{\text{b}}^2}}}{{{{\text{H}}^2}}} \cr & {{\text{P}}^2} = \frac{{{{\text{a}}^2}{{\text{b}}^2}}}{{{{\text{a}}^2} + {{\text{b}}^2}}}\,\,\,\,\left[ {\because {{\text{H}}^2} = {{\text{a}}^2} + {{\text{b}}^2}} \right] \cr} $$

BD = length of diagonal = speed × time $$\eqalign{ & = \frac{{52}}{{60}} \times 15 \cr & = 13{\text{ m}} \cr & {\text{BD}} = \sqrt {{l^2} + {b^2}} \cr & \Rightarrow {l^2} + {b^2} = {13^2} \cr & \Rightarrow {l^2} + {b^2} = 169 \cr & {\text{Again, }}l + b = \frac{{68}}{{60}} \times 15 = 17 \cr & {\left( {l + b} \right)^2} = {l^2} + {b^2} + 2lb \cr & {17^2} = 169 + 2lb \cr & lb = \frac{{120}}{2} = 60{\text{ }}{{\text{m}}^2} \cr} $$

Area of shaded region = Area of semicircle - Area of triangle $$\eqalign{ & = \frac{{\pi {{\left( a \right)}^2}}}{2} - \frac{1}{2} \times a \times 2a \cr & = \frac{{\pi {a^2}}}{2} - {a^2} \cr & = {a^2}\left( {\frac{\pi }{2} - 1} \right){\text{ sq}}{\text{. units}} \cr} $$

Let the length of interior square = $$x$$ The length of exterior square = ($$x$$ + 9) According to question $$\eqalign{ & {\left( {x + 9} \right)^2} - {x^2} = 105.75 \cr & \left( {x + 9 + x} \right)\left( {x + 9 - x} \right) = \frac{{10575}}{{100}} \cr & \left( {2x + 9} \right) \times 9 = \frac{{423}}{4} \cr & 2x + 9 = \frac{{47}}{4} \cr & 2x = \frac{{47}}{4} - 9 \cr & 2x = \frac{{11}}{4} \cr & x = \frac{{11}}{8} \cr & {\text{Total cost of fencing}} \cr & = \frac{{11}}{8} \times 4 \times 100 \cr & = {\text{Rs}}{\text{. }}550 \cr} $$

Area of Hexagon $$ = 6 \times \frac{{\sqrt 3 }}{4}{\left( 6 \right)^2} = 54\sqrt 3 {\text{ c}}{{\text{m}}^2}$$ Area of one part of Hexagon $$ = \frac{{54}}{6}\sqrt 3 = 9\sqrt 3 $$ Required area $$ = 54\sqrt 3 + 2 \times 9\sqrt 3 = 72\sqrt 3 {\text{ c}}{{\text{m}}^2}$$ Alternate Solution: Counting the equilateral triangle figure = 8 Area of equilateral triangle is equal to $$ = 8 \times \frac{{\sqrt 3 }}{4} \times 6 \times 6 = 72\sqrt 3 {\text{ c}}{{\text{m}}^2}$$

Each interior angle of polygon is given by $$\eqalign{ & = \frac{{\left( {x - 2} \right)}}{x} \times 180 \cr & {\text{Sides is }}a,\,2a \cr & \frac{{\left( {\frac{{a - 2}}{a}} \right) \times 180}}{{\left( {\frac{{2a - 2}}{{2a}}} \right) \times 180}} = \frac{3}{4} \cr & \frac{{\left( {a - 2} \right)}}{a} \times \frac{{2a}}{{2\left( {a - 1} \right)}} = \frac{3}{4} \cr & \frac{{\left( {a - 2} \right)}}{{\left( {a - 1} \right)}} = \frac{3}{4} \cr & 4a - 8 = 3a - 3 \cr & a = 5 \cr & {\text{So, sides }}5,\,10 \cr} $$

Diagonal of square ABCD = 8√2 Diagonal of square EBGF = 6√2 Height of triangle EDG = 8√2 - 3√2 = 5√2 Area of triangle EDG = $$\frac{1}{2}$$ × 6√2 × 5√2 = 30 Area of triangle EFG = (3√2) 2 = 18 Area of shaded region = 30 - 18 = 12