Given that, The surface area of a closed rectangular box, which is inscribed in a sphere, is 846 sq cm

So, $2(lb+bh+hl)=846$. 

And $4(l+b+h)=144$

$(l+b+h)=36$

$\left(l+b+h\right)^2=l^2+b^2+h^2+2\left(lb+bh+hl\right)$

$1296=\left(l^2+b^2+h^2\right)+846$

$450=l^2+b^2+h^2$

We are told that this cuboid is inscribed in a sphere, the body diagonal of the cuboid equals the diameter of the sphere, this can be visualised as:

This is nothing but, $\sqrt{l^2+b^2+h^2}=2R$

$l^2+b^2+h^2=4R^2$

$450=4R^2$

$R^2=\frac{225}{2}$

$R=\frac{15}{\sqrt{2}}$

Volume of sphere will be $\dfrac{4}{3}\times\ \pi\ \times\ \left(\dfrac{15}{\sqrt{2}}\right)^3$

$\dfrac{4}{3}\pi\ \left(\dfrac{3375}{2\sqrt{\ 2}}\right)$

$\pi\ \times\ 1125\sqrt{\ 2}$

It is given that AB = 9 cm, BC = 6 cm.

It is also known that the areas of the figures ABP, APQ, and AQCD are in geometric progression.

Hence, the area of the ABP, APQ, and AQCD are k, 2k, and 4k respectively.

The ratio of BP, PQ, QC will be the ratio of the respective triangles. Hence, we can draw a line from point A to point C.

Let the area of triangle AQC be x, which implies the area of triangle ADC = ADQC - AQC = 4k -x, which is equal to the sum of the area of triangle APB, AQP, and ACQ, respectively.

Therefore, 4k-x = 3k+x 

=> x = k/2

Hence the ratio of BP: PQ: CQ = k:2k: k/2 =2:4:1

Since BC is the diameter of the circle, which implies angle BAC is 90 degrees. Let AB = a cm, which implies AC = b cm. Hence, $ BC = \sqrt{a^2 + b^2} $, which is diameter of the circle (2r).

Hence, $ 2r = \sqrt{a^2 + b^2} $

=> $ 4r^2 = a^2 + b^2 $

The area of the triangle is $ \frac{1}{2} \times a \times b $, which can be written as

=> $ \frac{a \cdot b}{2(a^2+b^2)} \times (a^2 + b^2) $

=> $ \frac{a \cdot b}{2(a^2+b^2)} \times 4r^2 $

=> $ \frac{a \cdot b}{a^2+b^2} \times 2r^2 $

The correct option is D

Let us assume the length of the rectangle is 'l' and breadth of the rectangle is 'b'.

The radius, l/2 and b in the above diagram form a right-angled triangle.

=> ${(\frac{l}{2})}^2 + b^2 = 2^2$

We know that the area of the rectangle is l*b, which can be obtained by considering 2 times the geometric mean of ${(\frac{l}{2})}^2$

Therefore, for the maximum area, the equality condition of AM-GM inequality should be satisfied

=> ${(\frac{l}{2})}^2$ = $b^2$ => l = 2b

=> l/b = 2/1

The sum of the interior angles of a polygon of 'n' sides is given by (2n−4)× 90, and the sum of the exterior angles of a polygon is 360 degrees.

So, the difference between them will be 120 * n

=> (2n−4)90 − 360 = 120n

=> 60n = 720 => n = 12.

We know that the number of diagonals of a regular polygon is nC2 - n = 12C2 - 12 = 66 - 12 = 54.

CD = $\sqrt{y^2 + 25}$

$11 + y + \sqrt{y^2 + 25} = 36$

$\sqrt{y^2 + 25} = 25 - y$

$y^2 + 25 = 25^2 + y^2 - 50y$

$2y = 24$

y = 12

$A = lb$

$l^2 + b^2 = 4r^2$

$P = 2(l + b)$

$\frac{P}{2} = l + b$

Squaring on both the sides, we get

$\frac{P^2}{4} = l^2 + b^2 + 2lb$

$\frac{P^2}{4} = 4r^2 + 2lb$

$\frac{P^2}{8} - 2r^2 = lb$

Area of a regular hexagon = $\frac{3\sqrt{3}}{2}x^2$

Area of an equilateral triangle = $\frac{\sqrt{3}}{4}(a)^2$ ; where $a =$ side of the triangle

Since the area of the two figures are equal, we can equate them as follows:

$\frac{3\sqrt{3}}{2}x^2 = \frac{\sqrt{3}}{4}(12)^2$

On simplifying: $x^2 = 24$

$x = 2\sqrt{6}$

All the sides of the rhombus are equal.

The area of a rhombus is 12 $cm^2$

Considering d1 to be the length of the longer diagonal, d2 to be the length of the shorter diagonal.

The area of a rhombus is $ \frac{1}{2} (d1)(d2) = 12 $.

So $d1 \cdot d2 = 24$.

The length of the side of a rhombus is given by  

$ \frac{\sqrt{d1^2 + d2^2}}{2} $.  

This is because the diagonals and a side form a right triangle with sides $d1/2$, $d2/2$, and the side length.

$\frac{\sqrt{d1^2 + d2^2}}{2} = 5$

Hence $\sqrt{d1^2 + d2^2} = 10$

So  

$d1^2 + d2^2 = 100$

Using $d1 \cdot d2 = 24$, we have $2 \cdot d1 \cdot d2 = 48$.

So two equations:

$d1^2 + d2^2 + 2 d1 d2 = 100 + 48 = 148$

$d1^2 + d2^2 - 2 d1 d2 = 100 - 48 = 52$

So:

$d1 + d2 = \sqrt{148}$ (1)

$d1 - d2 = \sqrt{52}$  (2)

(1) + (2)= 2*(d1) = 2*($\sqrt{37} + \sqrt{13}$)

d1 = $\sqrt{37} + \sqrt{13}$

or 

In a rhombus the area of a Rhombus is given by :

The diagonals perpendicularly bisect each other. Considering the length of the diagonal to be $2a, 2b$.

The area of a Rhombus is:

$ \left( \frac{1}{2} \right) (2a)(2b) = 12 $

So:

$ab = 6$

The length of each side is:

$\sqrt{a^2 + b^2} = 5$, so

$a^2 + b^2 = 25$

$(a + b)^2 = 37$, so

$a + b = \sqrt{37}$

$(a - b)^2 = 13$, so

$a - b = \sqrt{13}$

Now solving:

$2a = (\sqrt{37} + \sqrt{13})$

$2b = (\sqrt{37} - \sqrt{13})$

2a is longer diagonal which is equal to  ($\sqrt{37} - \sqrt{13}$)

Applying cosine rule in triangle ACD:

$100 + X^2 - 2 \times 10 \times X \cos 30 = 400$

$X^2 - 10X\sqrt{3} - 300 = 0$

Solving, we get $X = \frac{10\sqrt{3} + 10\sqrt{15}}{2}$

Hence, area = $10X\sin 30 = \frac{\left(10\sqrt{3} + 10\sqrt{15}\right)10}{2 \cdot 2}$

$= 25(\sqrt{3} + \sqrt{15})$

We can say 40m is the perimeter of the park  

so side of rhombus = 10

Now $\frac{1}{2} \times d_1 \times d_2 = 96$  

so we get $d_1 \times d_2 = 192$ (1)  

And as we know diagonals of a rhombus are perpendicular bisectors of each other:  

so $\frac{d_1^2}{4} + \frac{d_2^2}{4} = 100$  

so we get $d_1^2 + d_2^2 = 400$ (2)

Solving (1) and (2)  

We get $d_1 = 12$ and $d_2 = 16$.

Now the cost, in INR, of laying electric wires along its two diagonals, at the rate of ₹125 per m, is= (12+16)(125) =3500

Now, definitely, three sides should be fenced at Rs 100/ft, and one side should be fenced at Rs 200/ft.

In this question, we are going to assume that the L is greater than B.

Hence, the one side painted at Rs 200/ft should be B to minimise costs.

Hence, the total cost = 200B + 100B + 100L + 100L = 300B + 200L

Now, L x B = 60000

B = 60000/L

Hence, total cost = 300B + 200L = 18000000/L + 200L

To minimise this cost, we can use AM>=GM,

$\frac{\frac{18000000}{L} + 200L}{2} \ge \sqrt{\frac{18000000}{L} \times 200L}$

$\frac{18000000}{L} + 200L \ge 2\sqrt{18000000 \times 200}$

$\frac{18000000}{L} + 200L \ge 2 \times 60000$

Hence, minimum cost = Rs 120000.

Let the base radius be 3r.

Height of upper cone is 9 so, by symmetry radius of upper cone will be r.

Volume of frustum=$\fracπ3(9r^2⋅27−r^2⋅9)$

Volume of upper cone = $\fracπ3⋅r^2⋅9$

Difference= $\fracπ3⋅9⋅r^2⋅25 = 225 = \frac{\pi}{3} ⋅ r^2 = 1$

Volume of larger cone = $\fracπ3⋅9r^2⋅27=243$

Let the length of radius be ‘$r$’.

From the above diagram,

$x^2 +r^2 = 6^2$ ....(i)

$(10 - x)^2 + r^2 = 8^2$ ....(ii)

Subtracting (i) from (ii), we get: 

$x = 3.6$ => $r^2 = 36 - (3.6)^2$ = $23.04$ ==> $r^2 = 36 - (3.6)^2$ = $23.04$

Area of circle = $\pi r^2 = 23.04 \pi$ 

Area of rhombus = 1/2*d1*d2 = 1/2*12*16 = 96.

.’. Ratio of areas = 23.04$\pi$ / 96 = $\frac{6 \pi}{25}$

Total area = 5 parts

Let ABCD be the rectangle with length $2l$ and breadth $2b$ respectively.

Area of the circle i.e. area of painted region $= \pi b^{2}$.

Given, $4lb - \pi b^{2} = \frac{2}{3}\pi b^{2}$.

$\Rightarrow 4lb = \frac{5}{3}\pi b^{2}$.

$\Rightarrow l = \frac{5\pi}{12} b$.

Given, $4lb = 135$  

$\Rightarrow 4 \cdot \frac{5\pi}{12} b^{2} = 135$  

$\Rightarrow b = \frac{9}{\sqrt{\pi}}$.

$\Rightarrow l = \frac{15}{4}\sqrt{\pi}$.

Perimeter of rectangle $= 4(l + b)$  

$= 4\left( \frac{15}{4}\sqrt{\pi} + \frac{9}{\sqrt{\pi}} \right)$  

$= 3\sqrt{\pi}\left(5 + \frac{12}{\pi}\right)$.

Hence option A is correct.

Let the sides of the rectangle be "a" and "3a" m. Hence the perimeter of the rectangle is 8a.

Let the side of the equilateral triangle be "m" cm. Hence the perimeter of the equilateral triangle is "3m" cm. Now we know that 8a+3m=90......(1)

Moreover area of the equilateral triangle is  $\frac{\sqrt3}4m^2$ and area of the rectangle is $3a^2$

According to the relation given $(\frac{\sqrt3}4m^2)^2$= $3a^2$

$\frac3{16}m^4$= $3a^2$ or $a^2 = \frac{m^4}{16}$

a = $\frac{m^2}4$

Substituting this in (1) we get $2m^2+3m−90$ = 0

solving this we get m=6 (ignoring the negative value since side can't be negative)

Hence a=9 and the longer side of the rectangle will be 3a=27cm

Given, BC = DE = 4

CD = BE = 5

In triangle ADE, EAD=$45^{0}$

tan⁡ 45 = $\frac{DE}{AE}$ => AE = 4

Area of trapezium = Area of rectangle BCDE + Area of triangle AED

= 20 + 8 = 28