The area of the sector of a circle, whose radius is 12 meters and whose angle at the centre is 42, is?

? ?d-d = 105 cm
? (? -1 ) d = 105 cm
? [(22/7)-1] x d = 105 cm
? d = (7/15) x 105 cm = 49 cm

? 2?R - 2?r = (176-132)
? 2?(R-r) = 44
? R-r = ( 44 x 7 )/ (2 x 22)
= 7 m

Area left ungrazed
= [(63 x 63) - (4 x 1/4 x 22/7 x (63/2)²] m²
= (63 x 63 - (99 x 63)/2 ) m²
= 63 x (63 - 99/2) m²
= 850.5 m²

Length of each side of hexagoan = radius of circle
? its perimeter = 6r

Original circumference = 2?r
New circumference = (150 /100) x 2 ?r
= 3?r

2?R = 3?r
? R = 3r/2
Original area = ?r²
New area = ?R²
= ?9r² / 4 = 9?r²/4

Increase in area = (9?r²/4 ) - (?r²)
= (5/4) ?r²

Req. increase per cent = [{(5/4) ?r²} / {?r²}] x 100
= 125 %

Arc length = 2?r (?° / 360°)
= 2 x (22/7) x 21 x (72° / 360°) cm
= 26.4 cm

Area of sector = ( arc length x radius ) / 2 cm²
= (3.5 x 5 ) / 2
= 8.75 cm²

Let length = a metres and breadth = b metres
Then, 2(a+b)= 46
? (a+b) =23 and ab =120
? Diagonal = ?a² + b²
?(23)²-2 x 120
?289 = 17 m

Speed = 12 x (5/18) m/sec
=10/3 m/sec

there4; perimeter = (10/3) x 15 x 60 m=3000 m
? 2( a + 4a) = 3000 m
? a = 300 m
So, length = 1200 m and breadth = 300 m

? Area = (1200 x 300 ) m² = 360000m²

Area of a triangle formed by joining the mid point of the sides of the triangle is 1/4^th of area of the original triangle .
So area of ? DEF = 36/ 4 = 9 m²