Area - Study Mode
[#111] Three plots having areas 110, 130 and 190 square metres are to be subdivided into flower beds of equal size. If the breadth of a bed is 2 metres, the maximum length of a bed van be :
Correct Answer
(A) 5 m
Explanation
Solution: Maximum possible size of a flower bed : = (H.C.F. of 110, 130, 190) sq. m = 10 sq. m ∴ Maximum possible length = $$left( {frac{{10}}{2}}
ight)$$ m = 5 m
[#112] If the sides of a square be double find the increase of percentage in area :
Correct Answer
(C) 300%
Explanation
Solution: $$eqalign{
& {A_1} = {x^2}{ ext{ and }}{A_2} = {left( {2x}
ight)^2} = 4{x^2} cr
& { ext{Increase in area}} = left( {4{x^2} - {x^2}}
ight) cr
& ,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,, = 3{x^2} cr
& { ext{Increase % }} = left( {frac{{3{x^2}}}{{{x^2}}} imes 100}
ight)\% cr
& ,,,,,,,,,,,,,,,,,,,,,,,,,,, = 300\% cr} $$
[#113] A hall, whose length is 16 m and the breadth is twice its height, takes 168 m of paper with 2 m as its width to cover its four walls. The area of the floor is :
Correct Answer
(C) 192 m 2
Explanation
Solution: Let the height of the room be x metres Then, breadth of the room = (2x) metres Area of 4 walls : $$eqalign{
& = left[ {2left( {16 + 2x}
ight) imes x}
ight]{m^2} cr
& = left( {32x + 4{x^2}}
ight){m^2} cr} $$ $$eqalign{
& herefore left( {32x + 4{x^2}}
ight) = 168 imes 2 cr
& Rightarrow {x^2} + 8x - 84 = 0 cr
& Rightarrow {x^2} + 14x - 6x - 84 = 0 cr
& Rightarrow xleft( {x + 14}
ight) - 6left( {x + 14}
ight) = 0 cr
& Rightarrow left( {x + 14}
ight)left( {x - 6}
ight) = 0 cr
& Rightarrow x = 6 cr} $$ Area of the floor : = (16 × 12) m 2 = 192 m 2
[#114] In a triangle ABC, a line XY is drawn parallel to BC meeting AB in X and AC in Y. The area of the triangle AXY is half of the area of the triangle ABC. XY divides AB in the ratio of :
Correct Answer
(C) $$1:left( {sqrt 2 - 1}
ight)$$
Explanation
Solution: Note : The ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides. Since, XY || BC, we have : $$eqalign{
& angle AXY = angle ABC{ ext{ and }} cr
& angle AYX = angle ACB cr
& { ext{Aslo, }}angle A = angle A,,{ ext{(common)}} cr
& { ext{So, }}vartriangle AXY sim ,vartriangle ABC cr
& { ext{Let area (}}vartriangle ABC{ ext{)}} = x, ext{sq. units} cr
& { ext{Then,}} cr
& { ext{Area (}}vartriangle AXY{ ext{) = }}frac{x}{2} ext{sq. units} cr
& frac{{{{left( {AB}
ight)}^2}}}{{{{left( {AX}
ight)}^2}}} = frac{x}{{left( {frac{x}{2}}
ight)}} cr
& Rightarrow frac{{AB}}{{AX}} = sqrt 2 cr
& Rightarrow frac{{AX + BX}}{{AX}} = sqrt 2 cr
& Rightarrow 1 + frac{{BX}}{{AX}} = sqrt 2 cr
& Rightarrow frac{{BX}}{{AX}} = left( {sqrt 2 - 1}
ight) cr
& Rightarrow frac{{AX}}{{BX}} = frac{1}{{left( {sqrt 2 - 1}
ight)}}{ ext{ Or 1}}:left( {sqrt 2 - 1}
ight) cr} $$
[#115] If an equilateral triangle of area X and a square of area Y have the same perimeter, then X is :
Correct Answer
(C) less than Y
Explanation
Solution: Let the side of the triangle be a cm and each side of the square be b cm Then, $$eqalign{
& X = frac{{sqrt 3 }}{4}{a^2}{ ext{ and }}Y = {b^2} cr
& { ext{Where }}3a = 4b,i,e.,b = frac{{3a}}{4} cr} $$ $$ herefore X = frac{{sqrt 3 {a^2}}}{4}{ ext{ and }}Y = frac{{9{a^2}}}{{16}}$$ xa0 xa0 $$left[ {x08ecause b = frac{{3a}}{4}}
ight]$$ $$eqalign{
& { ext{Now,}} cr
& frac{{sqrt 3 {a^2}}}{4} = frac{{1.732{a^2}}}{4} = 0.433{a^2} cr
& { ext{And }}frac{{9{a^2}}}{{16}} = 0.5625{a^2} cr
& herefore X < Y cr} $$