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Simple Problems:
1.One side of a rectangular field is 15m and one of its diagonal is 17m.
Find the area of field?
Sol:Other side = √[(17*17) – (15*15)] = √(289-225) = 8m
Area = 15 * 8 =120 sq. m
2.A lawn is in the form of a rectangle having its sides in the ratio 2:3
The area of the lawn is 1/6 hectares.
Find the length and breadth of the lawn.
Sol: let length = 2x meters and breadth = 3x mt
Now area = (1/6 * 1000)sq m = 5000/3 sq m
2x * 3x = 5000/3 =>x * x =2500 / 9
x = 50/3
length = 2x = 100/3 m and breadth = 3x = 3*(50/3) = 50m
3.Find the cost of carpeting a room 13m long and 9m broad with a carpet 75cm wide
at the rate of Rs 12.40 per sq meter
Sol: Area of the carpet = Area of the room = 13* 9 =117 sq m
length of the carpet = (Area/width) = 117 * (4/3) = 156 m
Cost of carpeting = Rs (156 * 12.40) = Rs 1934.40
4.The length of a rectangle is twice its breadth if its length is
decreased by 5cm and breadth is increased by 5cm, the area of the rectangle is
increased by 75 sq cm. Find the length of the rectangle.
Sol: let length = 2x and breadth = x then
(2x-5) (x+5) – (2x*x)=75
5x-25 = 75 => x=20
length of the rectangle = 40 cm
5.In measuring the sides of a rectangle, one side is taken 5% in excess
and the other 4% in deficit. Find the error percent in the area,
calculate from the those measurements.
Sol: let x and y be the sides of the rectangle then
correct area = (105/100 * x) * (96 / 100 *y)
=(504/500 xy) – xy = 4/500 xy
Error% = 4/500 xy*(1/xy)*100 % = 4/5% = 0.8%
6.A room is half as long again as it is broad. The cost of
carpeting the room at Rs 5 per sq m is Rs 2.70 and the cost of papering
the four walls at Rs 10 per sq m is Rs 1720.
If a door and 2 windows occupy 8 sq cm. Find the dimensions of the room?
Sol: let breadth=x mt ,length= 3x/2 mt and height=h mt
Area of the floor = (total cost of carpeting /rate) = 270/5 sq m = 54 sq m
x * 3x/2=54 => x*x= 54*(2/3)=36 => x = 6m
so breadth = 6m and length=3/2*6 = 9m
now papered area = 1720 /10 = 172 sq m
Area of one door and 2 windows =8 sq m
total area of 4 walls = 172+8 = 180 sq m
2(9+6)*h = 180 => h=180/30 = 6m
7.The altitude drawn to the base of an isosceles triangle is 8cm and
the perimeter is 32cm. Find the area of the triangle?
Sol: let ABC be the isosceles triangle, the AD be the altitude
let AB = AC=x then BC= 32-2x
since in an isoceles triange the altitude bisects the base so BD=DC=16-x
in ∆ADC,(AC) 2 = (AD) 2 + (DC) 2
x*x=(8*8) + (16-x)*(16-x)
32x =320 => x = 10
BC = 32-2x = 32-20 = 12 cm
Hence, required area = ½ * BC * AD
= ½ * 12 * 10 = 60 sq cm
8.If each side of a square is increased by 25%, find the percentage change in its area?
Sol: let each side of the square be a , then area = a * a
New side = 125a / 100 = 5a / 4
New area =(5a * 5a)/(4*4) = (25a²/16) – a²
= 9a²/16
Increase %= 9a²/16 * 1/a² * 100%
= 56.25%
9.Find the area of a Rhombus one side of which measures 20cm and one diagonal 24cm.
Sol: Let other diagonal = 2x cm
since diagonals of a rhombus bisect each other at right angles, we have
20² = 12² + x² => x = √[20² -12²]= √256 = 16cm
so the diagonal = 32 cm
Area of rhombus = ½ * product of diagonals
= ½ * 24 * 32
= 384 sq cm
10. The area of a circular field is 13.86 hectares. Find the cost of
fencing it at the rate of Rs. 4.40 per meter.
Sol: Area = 13.86 * 10000 sq m = 138600 sq m
r²= 138600 => r² = 138600 * 7/22 => 210 m
circumference = 2r = 2 * 22/7 * 210m = 1320 m
cost of fencing = Rs 1320 * 4.40 = Rs. 5808
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