Answer: Option D. -> $$4\sqrt 3 $$
Radius of circum-circle :$$\eqalign{
& = \frac{a}{{\sqrt 3 }} \cr
& = \frac{{12}}{{\sqrt 3 }}\,cm \cr
& = 4\sqrt 3 \,cm \cr} $$

Answer: Option A. -> 1.967 cm2
Required area = (Area of an equilateral Δ of side 7 cm) - (3 × Area of sector with θ
= 60° and r = 3.5 cm)
$$ = \left[ {\left( {\frac{{\sqrt 3 }}{4} \times 7 \times 7} \right) - \left( {3 \times \frac{{22}}{7} \times 3.5 \times 3.5 \times \frac{{60}}{{360}}} \right)} \right]c{m^2}$$
$$\eqalign{
& = \left( {\frac{{49\sqrt 3 }}{4} - 11 \times 0.5 \times 3.5} \right)c{m^2} \cr
& = (21.217 - 19.25)c{m^2} \cr
& = 1.967\,c{m^2} \cr} $$

Answer: Option A. -> 6 m
Let the length and breadth of the rectangle be 3x and 2x respectively
Then,
Perimeter = 2 (3x + 2x) = 10x
And,
Area = (3x × 2x) = 6x2
$$\eqalign{
& \therefore \frac{{10x}}{{6{x^2}}} = \frac{5}{9} \cr
& \Rightarrow 30x = 90 \cr
& \Rightarrow x = 3 \cr} $$
So, breadth = (2 × 3) m = 6 m

Answer: Option A. -> 12.5% increase
Let the original length and breadth of the rectangle be $$l$$ and b respectively
New length :
$$ = 150\% {\text{ of }}l = \frac{{3l}}{2}$$
New breadth :
$$ = 75\% {\text{ of }}b = \frac{{3b}}{4}$$
Original area = $$lb$$
New area :$$\eqalign{
& = \left( {\frac{{3l}}{2} \times \frac{{3b}}{4}} \right) \cr
& = \frac{{9lb}}{8} \cr} $$
Increase in area :
$$\eqalign{
& = \left( {\frac{{9lb}}{8} - lb} \right) \cr
& = \frac{{lb}}{8} \cr} $$
∴ Increase % :
$$\eqalign{
& = \left( {\frac{{lb}}{8} \times \frac{1}{{lb}} \times 100} \right)\% \cr
& = 12.5\% \cr} $$

Answer: Option E. -> None of these
Area of square = $$\sqrt {1024} $$  cm = 32 cm
Length of rectangle = (2 × 32) cm = 64 cm
Breadth of rectangle = (32 - 12) cm = 20 cm
∴ Required ratio = 64 : 20 = 16 : 5

Answer: Option D. -> 432 sq.metre
Height of triangle = perimeter of square
Diagonal of square = $$8\sqrt 2 $$ m
∴ Length of each side of square :
$$ = \frac{{8\sqrt 2 }}{{\sqrt 2 }} = 8\,m$$
∴ Perimeter of square = 4 × 8 = 32 m = Height
Area of other square = 729
Side of square = $$\sqrt {729} $$  = 27 m = base of triangle
∴ Area of triangle :
$$\eqalign{
& = \frac{1}{2} \times {\text{Base}} \times {\text{Height}} \cr
& = \frac{1}{2} \times 27 \times 32 \cr
& = 432{\text{ sq}}{\text{.metre}} \cr} $$

Answer: Option A. -> 1 : 2
Let r1 and r2 be the radii of the in-circle and circum-circle of a square respectively and let each side of the square be a.
Then,
$$\eqalign{
& {r_1} = \frac{a}{2} \cr
& {r_2} = \frac{1}{2} \times {\text{diagonal of the sequence}} \cr
& {r_2} = \frac{1}{2} \times \sqrt 2 a \cr
& {r_2} = \frac{{\sqrt 2 a}}{2}cm \cr} $$
∴ Required ratio :
$$\eqalign{
& = \frac{{\pi \times {{\left( {\frac{a}{2}} \right)}^2}}}{{\pi \times {{\left( {\frac{{\sqrt 2 a}}{2}} \right)}^2}}} \cr
& = 1:2 \cr} $$

Answer: Option B. -> 2 : 1
$$\eqalign{
& \frac{l}{{2\left( {l + b} \right)}} = \frac{1}{3} \cr
& \Rightarrow 3l = 2l + 2b \cr
& \Rightarrow l = 2b \cr
& \Rightarrow \frac{1}{b} = \frac{2}{1} = 2:1 \cr} $$

Answer: Option C. -> 17 metres
Let the length of carpet be l metres and breadth the b metres
∴ Diagonal = $$\sqrt {{l^2} + {b^2}} $$
According to the question,
$$\eqalign{
& lb = 120{\text{ and }} {\text{2}}\left( {l + b} \right) = 46 \cr
& \Rightarrow \left( {l + b} \right) = 23 \cr} $$
On squaring both sides :
$$\eqalign{
& \Rightarrow {\left( {l + b} \right)^2} = {23^2} \cr
& \Rightarrow {l^2} + {b^2} + 2lb = 529 \cr
& \Rightarrow {l^2} + {b^2} + 2 \times 120 = 529 \cr
& \Rightarrow {l^2} + {b^2} = 529 - 240 \cr
& \Rightarrow {l^2} + {b^2} = 289 \cr
& \therefore \sqrt {{l^2} + {b^2}} = \sqrt {289} = 17 \cr} $$
Diagonal of the carpet = 17 metres

Answer: Option B. -> 125%
Let original length = $$l$$ metres and original breadth = b metres
Original area : $$ = \left( {lb} \right){m^2}$$
New length :
$$\eqalign{
& = \left( {\frac{{150l}}{{100}}} \right)m \cr
& = \left( {\frac{{3l}}{2}} \right)m \cr
& \text{New breadth :} \cr
& = \left( {\frac{{150b}}{{100}}} \right)m \cr
& = \left( {\frac{{3b}}{2}} \right)m \cr
& \text{New area :} \cr
& = \left( {\frac{{3l}}{2} \times \frac{{3b}}{2}} \right){m^2} \cr
& = \left( {\frac{{9lb}}{4}} \right){m^2} \cr
& \text{Increase} = 1 - \frac{9lb}{4} = \frac{5lb}{4} \cr
& \therefore \text{ Increase % :} \cr
& = \left( {\frac{{5lb}}{4} \times \frac{1}{{lb}} \times 100} \right)\% \cr
& = 125\% \cr} $$