Area - Study Mode

[#106] Two small circular parks of diameters 16 m and 12 m are to be replaced by a Bigger circular park. What would be the radius of this new park. If the new park has to occupy the same space as the two small parks ?
Correct Answer

(A) 10 m

Explanation

Solution: Let the radius of the new park be R m Then, $$eqalign{
& pi {R^2} = pi imes {8^2} + pi imes {6^2} cr
& Rightarrow pi {R^2} = 100pi cr
& Rightarrow {R^2} = 100 cr
& Rightarrow R = 10 cr} $$

[#107] If the wheel of the engine of a train $$4frac{2}{7}$$ metres in circumference makes 7 revolutions in 4 seconds, then the speed (in km/hr) of the train is :
Correct Answer

(A) 27

Explanation

Solution: Distance covered in 4 sec : $$eqalign{
& = left( {frac{{30}}{7} imes 7}
ight)m cr
& = 30,m cr} $$ Distance covered in 1 sec = $$frac{{30}}{4}$$ Distance covered in 1 revolution : $$eqalign{
& = left( {frac{{30}}{4}}
ight)m cr
& = frac{{15}}{2}m cr} $$ ∴ Required speed : $$eqalign{
& = left( {frac{{15}}{2}}
ight)m/s cr
& = left( {frac{{15}}{2} imes frac{{18}}{5}}
ight)km/hr cr
& = 27,km/hr cr} $$

[#108] The area of a rhombus with side 13 cm and one diagonal 10 cm will be :
Correct Answer

(C) 120 cm 2

Explanation

Solution: Side of a rhombus = 13 cm Diagonal of rhombus = 10 cm In ΔAOB $$eqalign{
& AO = sqrt {{{13}^2} - {5^2}} cr
& ,,,,,,,,,, = sqrt {169 - 25} cr
& ,,,,,,,,,, = sqrt {144} cr
& ,,,,,,,,,, = 12,cm cr} $$ $$eqalign{
& Rightarrow AC = 24 cr
& { ext{Area of rhombus :}} cr
& = frac{1}{2} imes {d_1} imes {d_2} cr
& = frac{1}{2} imes 24 imes 10 cr
& = 120,sq.cm cr} $$

[#109] The length of a rectangular blackboard is 8 m more than its breadth. If its length is increased by 7 m and its breadth is decreased by 4 m, its area remains unchanged. The length and breadth of the rectangular blackboard is :
Correct Answer

(D) 28 m, 20 m

Explanation

Solution: Let the breadth = x cm Then, length = (x + 8) m $$eqalign{
& herefore left( {x + 8}
ight)x = left( {x + 15}
ight)left( {x - 4}
ight) cr
& Rightarrow {x^2} + 8x = {x^2} + 11x - 60 cr
& Rightarrow x = 20 cr} $$ So, length = 28 m and breadth = 20 m

[#110] A rectangular lawn 80 metres by 60 metres has two roads each 10 m wide running in the middle of it, one parallel to the length and the other parallel to the breadth. Find the cost of gravelling them at Rs. 30 per square metre.
Correct Answer

(D) Rs. 39000

Explanation

Solution: Area of the roads : = (80 × 10 + 60 × 10 - 10 × 10) m 2 = 1300 m 2 ∴ Cost of gravelling : = Rs. (1300 × 30) = Rs. 39000