Area - Study Mode

[#121] A triangle and a parallelogram are constructed on the same base such that their areas are equal. If the altitude of the parallelogram is 100 m, then the altitude of the triangle is :
Correct Answer

(D) 200 m

Explanation

Solution: Let the altitude of the triangle be h 1 and base of each be b. Then, $$frac{1}{2}$$ × b 1 × h 1 = b × h 2 , where h 2 = 100 m ⇔ h 1 = 2 h 2 ⇔ h 1 = (2 × 100) m ⇔ h 1 = 200 m

[#122] The circumference of a circle, whose area is 24.64 m 2 , is :
Correct Answer

(C) 17.60 m

Explanation

Solution: $$eqalign{
& x08ecause pi {R^2} = 24.64 cr
& Leftrightarrow {R^2} = left( {frac{{24.64}}{{22}} imes 7}
ight) cr
& Leftrightarrow {R^2} = 7.84 cr
& Leftrightarrow R = sqrt {7.84} cr
& Leftrightarrow R = 2.8,m cr
& herefore { ext{Circumference :}} cr
& = left( {2 imes frac{{22}}{7} imes 2.8}
ight)m cr
& = 17.60,m cr} $$

[#123] The ratio of the radii of two circle is 3 : 2. What is the ratio of their circumferences ?
Correct Answer

(B) 3 : 2

Explanation

Solution: Let the radii of the two circle be 3r and 2r respectively Then, required ratio : $$eqalign{
& = frac{{2pi left( {3r}
ight)}}{{2pi left( {2r}
ight)}} cr
& = frac{3}{2} cr
& = 3:2 cr} $$

[#124] A circular grassy plot of land, 42 cm is diameter, has a path 3.5 m wide running around it outside. The cost of gravelling the path at Rs. 4 per square metre is :
Correct Answer

(C) Rs. 2002

Explanation

Solution: Radius of the plot = 21 m Area of the path : $$eqalign{
& = pi left[ {{{left( {24.5}
ight)}^2} - {{left( {21}
ight)}^2}}
ight]{m^2} cr
& = left[ {pi left( {24.5 + 21}
ight)left( {24.5 - 21}
ight)}
ight]{m^2} cr
& = left( {frac{{22}}{7} imes 45.5 imes 3.5}
ight){m^2} cr
& = 500.5,{m^2} cr} $$ ∴ Cost of gravelling : = Rs. (500.5 × 4) = Rs. 2002

[#125] A horse is tied at the corner of a rectangular field whose length is 20 m and width is 16 m, with a rope whose length is 14 m. Find the area which the horse can graze :
Correct Answer

(B) 154 sq. m

Explanation

Solution: Required area = Area of the quadrant with radius 14 m : $$eqalign{
& = left( {frac{{22}}{7} imes 14 imes 14 imes frac{{90}}{{360}}}
ight){m^2} cr
& = 154,{m^2} cr} $$