Question

help me in these question plzz tomorrow is my exam plzzz help me​

Answer 1

ANS:9


Let the diameter of the

circular park ( D ) = 24 m

Radius of the park ( R ) = D/2

R = 24 /2 = 12 m

If the width of the path inside the

park ( w ) = 3m

Radius of the inside circle (r ) = R - w

r = 12 - 3 = 9 m

Area of the path ( A ) = πR² - πr²

A = π(R + r ) (R - r )

= π ( R + r ) w

= ( 22 /7 ) ( 12 + 9 ) ( 3 )

= 22 × 3 × 3

A = 198m²

Cost of the paving the per square

meter = Rs 25

Cost of paving total area of the path

=25 × 198

= Rs4950

I hope this helps you.




ASN:10
METHOD:1

Area of remaining part = Area of rectangle-2 area of semicircle
= lb-2πr^2/2
= (10)(7) - 2((22/7) (3.5)^2/2)
= 70-38.5
= 31.5 cm^2


METHOD:2

As from the figure I made, it is clear that the area of the quadrilateral PQRS is:-
Area of rectangle= Length* breadth
                           = 10 cm* 7 cm
                           = 70 sq. cm.
Area of circle 1 - as can be clearly seen from the figure, the diameter is 7cm. So, radius will be= 3.5 cm
 Area of circle 1= π * r^2
                         = 22/7 * 3.5 * 3.5
                         = 38.48 cm^2
now, area of 1/2 of both circles is equal. SO, 1/2 of circle 1= 1/2 of circle 2
1/2+1/2= 1
therefore,
the area of both circles will be equal to area of 1 circle= 38.48 cm^2

Now,
 area of remaining figure= Area of PQRS- Area of circles= (70- 38.48) cm^2
= 31.52 cm^2

Hope it helps! Smash that mark as brainliest and comment for any doubt!