# Exercise : 6

^{2}

(b) 60 cm

^{2}

(c) 66 cm

^{2}

(d) 65 cm

^{2}

(e) None of these

**Ans.b**

Let ABC be the isosceles triangle and AD be the altitude.

Let AB = AC = x. Then, BC = (32 – 2x).

Since, in an isosceles triangle, the altitude bisects the base. So, BD = DC = (16 – x).

In ΔADC, AC^{2} = AD^{2} + DC^{2}

⇒ x^{2} = (8)^{2} + (16 – x)^{2}

⇒ 32x = 320 ⇒ x = 10.

∴ BC = (32 – 2x) = (32 – 20) cm = 12 cm.

Hence, required area = (^{1}⁄_{2} × BC × AD)

= (^{1}⁄_{2} × 12 × 10) cm^{2} = 60 cm^{2}

(b) 172 metres

(c) 154 metres

(d) 165 metres

(e) None of these

**Ans.a**

Let the length of canal = h m.

Then,area of canal = ^{1}⁄_{2} × h(9 + 7)

or 1280 = ^{1}⁄_{2}h(16)

∴ h = \(\frac{1280 \times 2}{16} = 160 \; m\)

(b) 2 m

(c)

^{1}⁄

_{2}m

(d)

^{1}⁄

_{√2}m

(e) None of these

**Ans.a**

Let inner radius of the pipe be r.

Then, 440 = ^{22}⁄_{7} × r^{2} × 7 × 10

or \(r^{2} = \frac{440}{22 \times 10} = 2\)

or r = √2 m

^{2}. How long will it take for a horse to run around at the speed of 12 km/h ?

(b) 10 h

(c) 8 h

(d) 6 h

(e) None of these

**Ans.c**

Area of field = 576 km^{2}. Then,

each side of field = \(\sqrt{576} = 24 \; km\)

Distance covered by the horse

= Perimeter of square field

= 24 × 4 = 96 km

∴ Time taken by horse = \(\frac{distance}{speed} = \frac{96}{12} = 8 \; h\)

^{2}. If the length had been 6 metres more, the area would have been 54 m

^{2}more. The original length of the field is

(b) 18 metres

(c) 16 metres

(d) 24 metres

(e) None of these

**Ans.c**

Let the length and breadth of the original rectangular field be x m and y m respectively.

Area of the original field = x × y = 144 m^{2}

∴ x = ^{144}⁄_{y} .....(i)

If the length had been 6 m more, then area will be

(x + 6) y = 144 + 54

⇒ (x + 6)y = 198… (ii)

Putting the value of x from eq (i) in eq (ii), we get

(^{144}⁄_{y} + 6)y = 198

⇒ 144 + 6y = 198

⇒ 6y = 54 ⇒ y = 9m

Putting the value of y in eq (i) we get x = 16 m

(b) 81

(c) 126

(d) 252

(e) None of these

**Ans.c**

Clearly, we have : l = 9 and l + 2b = 37 or b = 14.

∴ Area = (l × b) = (9 × 14) sq. ft. = 126 sq. ft.

(b) 20 m × 5 m

(c) 30 m × 3.33 m

(d) 40 m × 2.5 m

(e) None of these

**Ans.b**

We have : 2b + l = 30 ⇒ l = 30 – 2b.

Area = 100 m^{2} ⇒ l × b = 100 ⇒ b(30 – 2b) = 100

⇒ b^{2} – 15b + 50 = 0 ⇒ (b – 10)(b – 5) = 0

⇒ b = 10 or b = 5.

When b = 10, l = 10 and when b = 5, l = 20.

Since the garden is rectangular,

so its dimension is 20 m × 5 m.

(b) 4.1 m

(c) 4.2 m

(d) 4.3 m

(e) None of these

**Ans.c**

Area of the field = 13.5 × 2.5 = 33.75 m^{2}

Area covered by the rectangular tank

= 5 × 4.5 = 22.50 m^{2}

Area of the field on which the earth dug out is to be spread = 33.75 – 22.50 = 11.25 m^{2}

Let the required height be h.

Then, 11.25 × h = 5 × 4.5 × 2.1

or h = 4.2 m

^{2}

(b) 240 cm

^{2}

(c) 560 cm

^{2}

(d) 160 cm

^{2}

(e) None of these

**Ans.a**

When folded along breadth, we have :

2(^{l}⁄_{2} + b) = 34 or l + 2b = 34 ....(i)

When folded along length, we have :

2(l + ^{b}⁄_{2}) = 38 or 2l + b = 38 ....(ii)

Solving (i) and (ii), we get :

l = 14 and b = 10.

∴ Area of the paper = (14 × 10) cm^{2} = 140 cm^{2}

(b) 24

(c) 32

(d) 48

(e) None of these

**Ans.a**

Area left after laying black tiles

= [(20 – 4) × (10 – 4)] sq. ft. = 96 sq. ft.

Area under white tiles = (^{1}⁄_{3} × 96) sq. ft = 32 sq. ft.

Area under blue tiles = (96 – 32) sq. ft = 64 sq. ft.

Number of blue tiles = \(\frac{64}{2 \times 2} = 16.\)

^{2}

(b) 42 cm

^{2}

(c) 154 cm

^{2}

(d) 196 cm

^{2}

(e) None of these

**Ans.b**

The shaded area gives the required region.

Area of the shaded region = Area of the square – area of four quadrants of the circles

= (14)^{2} - 4 × ^{1}⁄_{4}π(7)^{2}

= 196 - ^{22}⁄_{7} × 49 = 196 - 154 = 42 cm^{2}