Area - Study Mode

[#51] The sides of a triangle are consecutive integers. The perimeter of the triangle is 120 cm. Find the length of the greatest side :
Correct Answer

(C) 41 cm

Explanation

Solution: Let the sides of the triangles be x cm, (x + 1) cm and (x + 2) cm respectively. Then, $$eqalign{
& x + left( {x + 1}
ight) + left( {x + 2}
ight) = 120 cr
& Rightarrow 3x + 3 = 120 cr
& Rightarrow 3x = 117 cr
& Rightarrow x = 39 cr} $$ ∴ Length of greatest side : = (39 + 2) cm = 41 cm

[#52] ABCD is a square. E is the mid-point of BC and F is the mid-point of CD. The ratio of the area of triangle AEF to the area of the square ABCD is :
Correct Answer

(D) 3 : 8

Explanation

Solution: Let the length of side of the square be a units Then, $$BE = EC = DF = FC = frac{a}{2}$$ $$eqalign{
& AE = sqrt {{{left( {AB}
ight)}^2} + {{left( {BE}
ight)}^2}} cr
& ,,,,,,,,,, = sqrt {{a^2} + {{left( {frac{a}{2}}
ight)}^2}} cr
& ,,,,,,,,,, = sqrt {{a^2} + frac{{{a^2}}}{4}} cr
& ,,,,,,,,,, = sqrt {frac{{5{a^2}}}{4}} cr
& ,,,,,,,,,, = frac{{sqrt 5 a}}{2} cr} $$ Similarly, $$AF = frac{{sqrt 5 a}}{2}$$ $$eqalign{
& EF = sqrt {{{left( {CE}
ight)}^2} + {{left( {CF}
ight)}^2}} cr
& ,,,,,,,,,,, = sqrt {{{left( {frac{a}{2}}
ight)}^2} + {{left( {frac{a}{2}}
ight)}^2}} cr
& ,,,,,,,,,,, = sqrt {frac{{2{a^2}}}{4}} cr
& ,,,,,,,,,,, = frac{a}{{sqrt 2 }} cr
& EX = frac{1}{2}EF = frac{a}{{2sqrt 2 }} cr
& AX = sqrt {{{left( {AE}
ight)}^2} - {{left( {EX}
ight)}^2}} cr
& ,,,,,,,,,,, = sqrt {{{left( {frac{{sqrt 5 a}}{2}}
ight)}^2} - {{left( {frac{a}{{2sqrt 2 }}}
ight)}^2}} cr
& ,,,,,,,,,,, = sqrt {frac{{5{a^2}}}{4} - frac{{{a^2}}}{8}} cr
& ,,,,,,,,,,, = sqrt {frac{{9{a^2}}}{8}} cr
& ,,,,,,,,,,, = frac{{3a}}{{2sqrt 2 }} cr} $$ $$eqalign{
& herefore ,{ ext{Area of }}left( {vartriangle AEF}
ight): cr
& = frac{1}{2} imes EF imes AX cr
& = frac{1}{2} imes frac{a}{{sqrt 2 }} imes frac{{3a}}{{2sqrt 2 }} cr
& = frac{{3{a^2}}}{8} cr} $$ $$eqalign{
& { ext{Required ratio :}} cr
& = frac{{3{a^2}}}{8}:{a^2} cr
& = 3:8 cr} $$

[#53] The area of a rectangle is 252 cm 2 and its length and breadth are in the ratio of 9 : 7 respectively. What is its perimeter ?
Correct Answer

(A) 64 cm

Explanation

Solution: Let the length and breadth of the rectangle be (9x) cm and (7x) cm respectively. Then, $$eqalign{
& 9x imes 7x = 252 cr
& Rightarrow 63{x^2} = 252 cr
& Rightarrow {x^2} = 4 cr
& Rightarrow x = 2 cr} $$ So, length = 18 cm, breadth = 14 cm ∴ Perimeter : = 2(18 + 14) cm = 64 cm

[#54] A rectangular carpet has an area of 60 sq.m. If its diagonal and longer side together equal 5 times the shorter side, the length of the carpet is :
Correct Answer

(B) 12 m

Explanation

Solution: We have : $$lb$$ = 60 and $$sqrt {{l^2} + {b^2}} + l = 5b$$ Now, $$eqalign{
& {l^2} + {b^2} = {left( {5b - l}
ight)^2} cr
& Rightarrow 24{b^2} - 10lb = 0 cr
& Rightarrow 24{b^2} - 600 = 0 cr
& Rightarrow {b^2} = 25 cr
& Rightarrow b = 5 cr
& herefore l = left( {frac{{60}}{b}}
ight) cr
& ,,,,,,,, = left( {frac{{60}}{5}}
ight)m cr
& ,,,,,,,, = 12,m cr} $$ So, length of the carpet = 12 m

[#55] A rectangular garden (60 m × 40 m) is surrounded by a road of width 2 m, the road is covered by tiles and the garden is fenced. If the total expenditure is Rs. 51600 and rate of fencing is Rs. 50 per metre, then the cost of covering 1 sq. m of road by tiles is :
Correct Answer

(C) Rs. 100

Explanation

Solution: Length of the fence : = 2(60 + 40) m = 200 m Cost of fencing : = Rs. (200 × 50) = Rs. 10000 Area of the road : = [(64 × 44) - (60 × 40)] m 2 = (2816 - 2400) m 2 = 416 m 2 Let the cost of tiling the road be Rs. x per sq.m ∴ 416x + 10000 = 51600 ⇒ 416x = 41600 ⇒ x = Rs. 100