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area of theplot = 110 * 65 = 7150 sq.m
area of the plot excluding the path = (110-5)* (65-5) = 6300 sq.m
area of the path = 7150 - 6300 =850 sq.m
cost of gravelling the path = 850 x (80/100) = 680 Rs

let length = x and breadth = y then  

2(x+y) = 46         =>   x+y = 23  

x²+y² = 17² = 289  

now (x+y)² = 23² 

=> x²+y²+2xy= 529 

=> 289+ 2xy = 529

=> xy = 120  

area = xy = 120 sq.cm

let the inner and outer radii be r and R meters
then, 2πr = 352/7

=> r = (352/7) * (7/22) * (1/2) = 8m
2πR = 528/7

=> R= (528/7) * (7/22) * (1/2)= 12m
width of the ring = R-r = 12-8 = 4m

let original radius = r and  new radius = (50/100) r = r/2

original area = πr2 and new area = πr/22

decrease in area =  3πr2/4*1πr2 *100 = 75%

let each side of the square be a , then area = a2 

As given that The side is increased by 25%, then 

New side = 125a/100 = 5a/4 

New area = 5a42  

Increased area= 25a216-a2 

Increase %=9a2/16a2*100  % = 56.25%

let ABC be the isosceles triangle, the AD be the altitude 

Let AB = AC = x then BC= 32-2x       [because parameter = 2 (side) + Base] 

since in an isoceles triange the altitude bisects the base so 

BD = DC = 16-x 

In a triangle ADC, AC2=AD2+DC2 

x2=82+16-x2 ⇒x=10 

BC = 32-2x = 32-20 = 12 cm 

Hence, required area = 12*BC*AD= 12*12*10 = 60 sq cm

let x and y be the sides of the rectangle then 

correct area =105100x×96100y 

Error% = 504500xy-xy=4500xy %

4500xy×1xy×100=45=0.8

let length = 2x and breadth = x then
(2x-5) (x+5) = (2x * x)+75
5x-25 = 75 => x=20
length of the rectangle = 40 cm