Geometry and Mensuration

Perimeter of LMNOP=2πR/2 ​= πR

Perimeter of $LSRQP=$ Perimeter of $LSR+$ Perimeter of $RQP$ =πR/2​ + πR​/2 = πR

$\therefore$ Perimeter (LMNOP)/(LSRQP) ​= πR/πR​ = 1 : 1

Radius of the circle =2√2/2 ​​= √2​

$\therefore$ Volume to be removed $=$ ( Volume of the square - Volume of the circle ) ={[π√(2​)² × 1] − (2×2×1)} = 2π − 4

OX = OY = 10 m

d = 2r - 3 ∴ x × y = c × d

$\therefore$ 10 × 10 = 3(2r - 3) $\Rightarrow$ 6r - 9 = 100, $\Rightarrow$ 6r = 109

$\therefore$ r = 109/6​ m

Let the radius of hemispherical bowl = r
∴ Volume of hemispherical bowl = (2/3)πr³

Let the height of cylindrical vessel = h
r = h(1 + 50/100) = (2/3)r

Volume of cylindrical vessel = πr²(2r/3) = (2/3)πr³

Hence, volume of the beverage in the cylindrical vessel = (2/3)πr³ / (2/3)πr³ × 100% = 100%

In △ABD and △EBC,

BD/AB = BC/BE = 10.4/6x = 6/8

x = 1.8 m.
Where, x = height of the man.

$\therefore$ angles are 80∘, 60∘ and 40∘.

Major diameter of ellipse = 2a = 2(2b)

2a = 4b $\therefore$ a = 2b

Let the radius of circle = r

Now, $\pi$ ab = $2\pi r²$

or $\pi$ (2b)b = $2\pi r²$

∴ r = b

$\therefore$ Radius of circle = 1/2​× minor diameter of ellipse

= 50% of minor diameter of an ellipse

( a ) Side of the square = √36 = 6 cm

Perimeter of the square = 4 ( 6 ) = 24 cm

( b ) Perimeter of the triangle = 9 + 9 + 9 = 27 cm

( c ) Area of the rectangle = 40

lb = 40

b = 40/10 = 4

Perimeter = 2 ( l + b ) = 2 ( 4 + 10 ) = 28 cm

( d ) Perimeter of the circle = 2πr = 2 ( 3.14 ) ( 4 ) = 25.12 cm
Clearly , Perimeter of the rectangle is maximum .