# Exercise : 3

(b) 2 : 1

(c) 3 : 1

(d) Data inadequate

(e) None of these

**Ans.b**

Area of ΔEBC = ^{1}⁄_{2} × BC × EF

= ^{1}⁄_{2} × BC × AB[Since, EF = AB]

Area of ΔEBC = ^{1}⁄_{2} × area of ΔABCD

\ Required ratio = 2 : 1.

(b) 30 metres

(c) 24 metres

(d) 25 metres

(e) None of these

**Ans.d**

XYZ is a right-angled triangle

XZ = \(\sqrt{15^{2} + 20^{2}} = \sqrt{625} = 25 \; m\)

(b) 77 sq metres

(c) 308 sq metres

(d) 22 sq metres

(e) None of these

**Ans.b**

Area of semicircle = ½πr^{2}

= ½ × ^{22}⁄_{7} × 7 × 7 = 77 m^{2}

(b) 30 cm

(c) 15 cm

(d) Data inadequate

(e) None of these

**Ans.d**

Let the base and height of triangle, and length and breadth of rectangle be L and h, and L_{1} and b_{1} respectively.

Then ^{1}⁄_{2} × L × h = ^{2}⁄_{3} × L_{1} × b_{1} .....(i)

L = ^{4}⁄_{5}b_{1} .....(ii)

and L_{1} + b_{1} = 100 .....(iii)

In the above we have three equations and four unknowns. Hence the value of ‘h’ can’t be determined.

(b) 5 metres

(c) 7.5 metres

(d) Data inadequate

(e) None of these

**Ans.b**

L × B = 15 × B

∴ L = 15 m

and L – B = 10

∴ B = 15 – 10 = 5 m

(b) 4 : 3

(c) 5 : 4

(d) Data inadequate

(e) None of these

**Ans.d**

Area of shaded portion = 600 m.

∴ (l + 10)(b + 10) – lb = 600

or, lb + 10b + 10l + 100 – lb = 600

or, 10 (b + (l) = 500

∴ b + l = 50

From this equation we can’t get the required ratio.

^{2}

(b) 1600 cm

^{2}

(c) 1800 cm

^{2}

(d) 2500 cm

^{2}

(e) None of these

**Ans.e**

The four sheets are BMRN, AMQL, NSKC and DLPK

∴ Side of the new square sheet = 50 + 5 = 55 cm and the side of the inner part of the square (55 – 10) = 45 cm

Hence, area = (45)^{2} = 2025 sq.cm.

(b) 2600 cm

^{2}

(c) 2500 cm

^{2}

(d) 3025 cm

^{2}

(e) None of these

**Ans.e**

Let the side of the square be x m.

∴ Perimeter of the square = 48 × 5 = 4x ∴ x = 60 m

∴ Area = (60)^{2} = 3600 m^{2}

^{3}

(b) 86400 cm

^{3}

(c) 720 cm

^{3}

(d) Cannot be determined

(e) None of these

**Ans.c**

Volume of the box = \(\sqrt{120 \times 72 \times 60}\) = 720 cm^{3}

^{2}

(b) 154 cm

^{2}

(c) 1250 cm

^{2}

(d) 616 cm

^{2}

(e) None of these

**Ans.d**

Perimeter of the circle = 2πr = 2(18 + 26)

⇒ 2 × ^{22}⁄_{7} × r = 88 ⇒ r = 14

∴ Area of the circle

= πr^{2} = ^{22}⁄_{7} × 14 × 14 = 616 cm^{2}