Area

An error 2% in excess is made while measuring the side of a square. The percentage of error in the calculated area of the square is:

Answer: (d)

100 cm is read as 102 cm.

A1 = (100 × 100) cm2  and

A2 = (102 × 102) cm2

⇒ (A2 - A1) = [1022 - 1002]

⇒ (102 + 100) × (102 - 100) = 404 cm2

∴ Percentage error = \(\left ( \frac{404}{100\times 100}\times 100 \right )\) % = 4.04%

If the diagonal of a rectangle is 17cm long and its perimeter is 46 cm. Find the area of the rectangle.

Answer: (b)

let length = x and breadth = y then

2 (x + y) = 46

⇒ x + y = 23

⇒ x²+y² = 17² = 289

now (x + y)² = 23²

⇒ x² + y² + 2xy = 529

⇒ 289 + 2xy = 529

⇒ xy = 120

∴ area = xy = 120 sq. cm

A rectangular park 60 m long and 40 m wide has two concrete crossroads running in the middle of the park and rest of the park has been used as a lawn. If the area of the lawn is 2109 sq. m, then what is the width of the road?

Answer: (c)

Area of the park = (60 x 40) = 2400 m2

Area of the lawn = 2109 m2

Area of the crossroads = (2400 - 2109) = 291 m2

Let the width of the road be x metres. Then,

60x + 40x − x2 = 291

⇒ x2 − 100x + 291 = 0

⇒ (x − 97) (x − 3) = 0

⇒ x = 3.

A circular swimming pool is surrounded by a concrete wall 4 ft wide. If the area of the concrete wall surrounding the pool is 11/25 that of the pool, then the radius of the pool is?

Answer: (b)

let the radius of the pool be R ft.

radius of the pool including the wall = (R+4)ft

area of the concrete wall =

π[(R + 4)2 - R2]

= π[(R + 4 + R)(R + 4 - R)]

= 8π(R + 2) sq feet

⇒ 8π(R + 2) = 11/25πR2  ⇒ 11 R2 = 200(R + 2)

Radius of the pool R = 20 ft

Find the ratio of the areas of the incircle and circumcircle of a square.

Answer: (b)

Let the side of the square be x.

Then, its diagonal =√(2x² ) = √2(x)

Radius of in circle = (x/2)

Radius of circum circle=  √2(x/2) = (x/√2)

Required ratio = (πx²/4) : (πx²/2)

= ( 1/4 ) : (1/2)

The length of the room is 5.5 m and width is 3.75 m. Find the cost of paving the floor by slabs at the rate of Rs.800 per sq. meter

Answer: (a)

l = 5.5 m w = 3.75 m

area of the floor = 5.5 × 3.75 = 20.625 sq. m

cost of paving = 800 × 20.625 = Rs. 16500

The length of a rectangle is halved, while its breadth is tripled. What is the percentage change in area?

Answer: (d)

Let original length = x and original breadth = y.
Original area = xy.

New length = x/2 and New breadth=3y

New area =\(\frac{3}{2}xy\)

Increase in area = New area − original area =\(\frac{3}{2}xy - xy = \frac{1}{2}xy\)

Increase % = \(\frac{Increase \; in \; area}{Original \; area}\times 100 = \frac{\frac{1}{2}xy}{xy}\times 100\) = 50

The ratio between the perimeter and the breadth of a rectangle is 5 : 1. If the area of the rectangle is 216 sq. cm, what is the length of the rectangle?

Answer: (b)

\(\frac{2\left ( l+b \right )}{b} = \frac{5}{1}\)

⇒ 2l + 2b = 5b

⇒ 3b = 2l

⇒ b = (2/3)l

Then, Area = 216 cm2

⇒ l × b = 216

⇒ l × (2/3)l =216

⇒ l2 = 324

⇒ l = 18 cm.

A parallelogram has sides 30m and 14m and one of its diagonals is 40m long. Then its area is

Answer: (c)

let ABCD be the given parallelogram

area of parallelogram ABCD = 2 × (area of triangle ABC)

now a = 30m, b = 14m and c = 40m

\(S = \frac{1}{2} \times \left ( 30+14+40 \right ) = 42\)

Area of triangle ABC = \(\sqrt{s\left ( s-a \right )\left ( s-b \right )\left ( s-c \right )}\)

= \(\sqrt{s \left ( 12 \right )\left ( 28 \right )\left ( 2\right )}\) = 168 sq. m

area of parallelogram ABCD = 2 × 168 = 336 sq. m

One side of a rectangular field is 15m and one of its diagonal is 17m. Find the area of field?

Answer: (b)

Other side = [(17 × 17) - (15 × 15)]

= (289 − 225) = 8m

∴ Area = 15 × 8 = 120 sq. m