# Exercise : 4

(b) 7.5 m

(c) 8 m

(d) 8.5 m

(e) None of these

**Ans.b**

Length of the carpet = \(\frac{Total \; Cost}{Rate/m}\)

= (^{8100}⁄_{45})m = 180 m.

Area of the room = Area of the carpet

= (180 × ^{75}⁄_{100}) m^{2} = 135 m^{2}

∴ Breadth of the room = \(\frac{Area}{Length} = \frac{135}{18} \; m\)

= 7.5 m

^{2}

(b) 16 cm

^{2}

(c) 20 cm

^{2}

(d) 24 cm

^{2}

(e) None of these

**Ans.a**

In a rectangle,

\(\frac{\left(Perimeter \right)^{2}}{4} = \left(diagonal \right)^{2} + 2 \times area\)⇒ \(\frac{\left(14 \right)^{2}}{4} = 5^{2} + 2 \times area\)

49 = 25 + 2 × area

∴ Area = \(\frac{49 - 25}{2} = \frac{24}{2} = 12 \; cm^{2}\)

^{2}

(b) 8,750 m

^{2}

(c) 7,980 m

^{2}

(d) 6,890 m

^{2}

(e) None of these

**Ans.a**

In an isoscele right angled triangle,

Area = 23.3 × perimeter^{2}

= 23.3 × 20^{2} = 9320 m^{2}

(b) circumference

(c) 4

(d) 2π

(e) None of these

**Ans.c**

According to question, circumference of circle = Area of circle

or πd = π(^{d}⁄_{2})^{2} [where d = diameter]

∴ d = 4

^{2}

(b) 540 m

^{2}

(c) 680 m

^{2}

(d) 574 m

^{2}

(e) None of these

**Ans.a**

In a parallelogram.

Area = Diagonal × length of perpendicular on it.

= 30 × 20 = 600 m^{2}

^{22}⁄

_{7})

^{2}

(b) 44 m

^{2}

(c) 48 m

^{2}

(d) 36 m

^{2}

(e) None of these

**Ans.b**

Required area covered in 5 revolutions

= 5 × 2πrh = 5 × 2 × ^{22}⁄_{7} × 0.7 × 2 = 44 m^{2}

^{2}. If one of its sides is 123 metre, find the length of the perpendicular dropped on that side from opposite vertex.

(b) 12 metres

(c) 10 metres

(d) 9 metres

(e) None of these

**Ans.c**

In a triangle,

Area = ^{1}⁄_{2} × length of perpendicular × base

or 615 = ^{1}⁄_{2} × length of perpendicular × 123

∴ Length of perpendicular = \(\frac{615 \times 2}{123} = 10 \; m\)

^{2}

(b) 308 m

^{2}

(c) 150 m

^{2}

(d) 407 m

^{2}

(e) None of these

**Ans.a**

Area of the shaded portion

= ^{1}⁄_{4} × π × (14)^{2} = 154 m^{2}

^{2}for each plant ?

(b) 750

(c) 24

(d) 120

(e) None of these

**Ans.a**

Circumference of circular bed = 30 cm

Area of circular bed = ^{(30)2}⁄_{4π}

Space for each plant = 4 cm^{2}

∴ Required number of plants

= ^{(30)2}⁄_{4π} ÷ 4 = 17.89 = 18(approx)

^{4}⁄

_{5}

(b)

^{3}⁄

_{5}

(c)

^{5}⁄

_{6}

(d)

^{6}⁄

_{7}

(e) None of these

**Ans.a**

Area of the square = (10)^{2} = 100 cm^{2}

The largest possible circle would be as shown in the figure below :

Area of the circle = ^{22}⁄_{7} × (5)^{2} = \(\frac{22 \times 25}{7}\)

Required ratio = \(\frac{22 \times 25}{7 \times 100} = \frac{22}{28} = \frac{11}{14}\)

= 0.785 ≈ 0.8 = ^{4}⁄_{5}