Area - Study Mode
[#176] The ratio of the area of a square to that of the square drawn on diagonal is :
Correct Answer
(C) 1 : 2
Explanation
Solution: $$eqalign{
& { ext{Required ratio}} cr
& = frac{{{a^2}}}{{{{left( {sqrt 2 a}
ight)}^2}}} cr
& = frac{{{a^2}}}{{2{a^2}}} cr
& = frac{1}{2} cr
& = 1:2 cr} $$
[#177] The area of the four walls of a room is 120 m 2 and the length is twice the breadth. If the height of the room is 4 m, then the area of the floor is :
Correct Answer
(C) 50 m 2
Explanation
Solution: Let the breadth = x metres and length = (2x) metres Area of 4 walls = [2(2x + x)× 4] m 2 = (24x) m 2 ∴ 24x = 120 ⇒ x = 5 So, length = 10 m, and breadth = 5 m Area of the floor = (10 × 5) m 2 = 50 m 2
[#178] The area of a square with perimeter 48 cm, is :
Correct Answer
(A) 144 sq. cm
Explanation
Solution: Side of the square : $$eqalign{
& = left( {frac{{48}}{4}}
ight)cm cr
& = 12,cm cr} $$ Area of the square : $$eqalign{
& = left( {12 imes 12}
ight)c{m^2} cr
& = 144,c{m^2} cr} $$
[#179] The area of the shaded portion is :
Correct Answer
(C) 21 sq. cm
Explanation
Solution: Required area : = [(2 × 3) + (3 × 3) + (2 × 3)] cm 2 = (6 + 9 + 6) cm 2 = 21 cm 2
[#180] If the length of a certain rectangle is decreased by 4 cm and the width is increased by 3 cm, a square with the same area as the original rectangle would result. The perimeter of the original rectangle (in cm) is :
Correct Answer
(D) 50
Explanation
Solution: Let the length and width of the rectangle be $$l$$ cm and b cm respectively. Then, $$eqalign{
& left( {l - 4}
ight)left( {b + 3}
ight) = lb cr
& Rightarrow lb + 3l - 4b - 12 = lb cr
& Rightarrow 3l - 4b = 12.....(i) cr
& { ext{And, }} cr
& l - 4 = b + 3 cr
& Rightarrow l - b = 7.....(ii) cr} $$ Multiplying (ii) by 4 and subtracting (i) from it, we get : $$l$$ = 16 Putting $$l$$ = 16 in (ii), we get : b = 9 ∴ Perimeter of the original rectangle : $$eqalign{
& = 2left( {l + b}
ight) cr
& = [2left( {16 + 9}
ight)]cm cr
& = 50,cm cr} $$