Related Questions

A square is inscribed in a circle of diameter 2a and another square is circumscribing circle. The difference between the areas of outer and inner squares is__
A a2
B 2a2
C 3a2
D 4a2
A square is inscribed in a circle of diameter 2 a sand another square is circumscribing circle. The difference between the areas of outer and inner squares is.
A a2
B 2a2
C 3a2
D 4a2
The side length of a square inscribed inscribed in a circle is 2. What is the area of the circle shown at the right/
A π
B 2π
C πr2
D 22
The perimeters of two squares are 40cm and 30cm. Find the perimeter of a third square whose area is equal to the difference of the areas of the two squares.
A 18cm
B 24cm
C 32cm
D 36cm
In the figure below, a square is inscribed in a circle with diameter d. What is the sum of the areas of the shaded regions in terms of d?
A d2(π4-12)
B d2(π4-14)
C d2(π2-12)
D d2(π-2)
The difference between the radii of bigger circle and smaller circle is 14 cm and the difference between their areas is 1056 cm2. Radius of the smaller circle is:
A 7cm
B 5cm
C 9cm
D 3cm
The difference between the radii of bigger circle and smaller circle is 14 cm and the difference between their areas is 1056 cm2 . Radius of the smaller circle is _
A 7 cm
B 5 cm
C 9 cm
D 3 cm
The difference between the radio of bigger difference between circle and smaller circle is 14 and the difference between their sareas is 1056 cm2 . Radius of the smaller circle is .
A 7 cm
B 5 cm
C 9 cm
D 3 cm
The strain energy stored in a hollow circular shaft of outer diameter (D) and inner diameter (d) subjected to shear stress is
A $$\frac{{{\tau ^2}}}{{2{\text{C}}}}\left( {\frac{{{{\text{D}}^2} - {{\text{d}}^2}}}{{\text{D}}}} \right) \times {\text{Volume of shaft}}$$
B $$\frac{{{\tau ^2}}}{{2{\text{C}}}}\left( {\frac{{{{\text{D}}^2} + {{\text{d}}^2}}}{{\text{D}}}} \right) \times {\text{Volume of shaft}}$$
C $$\frac{{{\tau ^2}}}{{4{\text{C}}}}\left( {\frac{{{{\text{D}}^2} - {{\text{d}}^2}}}{{\text{D}}}} \right) \times {\text{Volume of shaft}}$$
D $$\frac{{{\tau ^2}}}{{4{\text{C}}}}\left( {\frac{{{{\text{D}}^2} + {{\text{d}}^2}}}{{\text{D}}}} \right) \times {\text{Volume of shaft}}$$
The polar moment of inertia of a hollow shaft of outer diameter (D) and inner diameter (d) is
A $$\frac{\pi }{{16}}\left( {{{\text{D}}^3} - {{\text{d}}^3}} \right)$$
B $$\frac{\pi }{{16}}\left( {{{\text{D}}^4} - {{\text{d}}^4}} \right)$$
C $$\frac{\pi }{{32}}\left( {{{\text{D}}^4} - {{\text{d}}^4}} \right)$$
D $$\frac{\pi }{{64}}\left( {{{\text{D}}^4} - {{\text{d}}^4}} \right)$$

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