MCQ Questions on Mensuration 2D for Competitive Exams

Let AB = AC = a cm
BD = DC = b cm
∴ Altitude of isosceles triangle is also median
In right ΔADC
AD² = a² - b²
64 = a² - b² . . . . . . (i)
Perimeter = 64
a + a + 2b = 64
2a + 2b = 64
a + b = 32 . . . . . (ii)
On dividing $$ = \frac{{{a^2} - {b^2}}}{{a + b}} = \frac{{64}}{{32}} = 2$$
∴ a² - b² = (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 + 4²
64 + r² + 16r = 64 + r² - 16r + 16
32r = 16
r = $$\frac{1}{2}$$cm
And area of circle = πr²
$$\eqalign{ & = \frac{{22}}{7} \times \frac{1}{2} \times \frac{1}{2} \cr & = \frac{{11}}{{14}}{\text{ c}}{{\text{m}}^2} \cr} $$

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 cm² (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