Solution: (d)

Circumference of the circle = 39.6 m

or, 2pr = 39.6 m

or, r = \(\displaystyle \frac{{39.6\times 7}}{{22\times 2}}=6.3m\)

 Area of circle = \(\displaystyle \pi {{r}^{2}}=\frac{{22}}{7}\times 6.3\times 6.3=124.74{{m}^{2}}\)

 Area of the rectangle =124.74

 Length of the rectangle = \(\displaystyle \frac{{124.74}}{{4.5}}=27.72m\)

Solution: (c)

One side of square = \(\displaystyle \frac{{perimeter}}{4}=\frac{{44}}{4}=11cm\)

Perimeter of rectangle = \(\displaystyle 4\times perimeterofsquare=4\times 44=176cm\)

Width of rectangle = \(\displaystyle \frac{{perimeterofrec\tan gle}}{2}-length\)

\(\displaystyle \frac{{176}}{2}-51=88-51=37cm\)

Therefore, Required difference = width – side = 37 – 11 = 26 cm.

Solution: (d)

The area of the shaded region area of square – Area of the circle

Required answer = \(\displaystyle {{(28)}^{2}}-\frac{{22}}{7}\times 14\times 14=784-616=168{{m}^{2}}\)

Solution: (d)

Area of field = \(\displaystyle 3584{{m}^{2}}\)

Let the length and breadth be 7x and 2x

Then \(\displaystyle 7x\times 2x\)= \(\displaystyle 3584{{m}^{2}}\)

\(\displaystyle 14{{x}^{2}}=3584{{m}^{2}}\)

\(\displaystyle {{x}^{2}}=256\)

x = 16 m

Length = 7x = 112 m, Breadth = 2x = \(\displaystyle 16\times 2=32m\)

Perimeter = 2(l + b) = 2(112 + 32) = 288 m

Solution: (b)

\(\displaystyle {{x}^{2}}+{{y}^{2}}={{(9\sqrt{2})}^{2}}\)

\(\displaystyle 2{{x}^{2}}=81\times 9\)   

x= 9