Exercise : 6


1. The altitude drawn to the base of an isosceles triangle is 8 cm and the perimeter is 32 cm. The area of the triangle is
(a) 72 cm2
(b) 60 cm2
(c) 66 cm2
(d) 65 cm2
(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).

quant

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

In ΔADC, AC2 = AD2 + DC2

⇒ x2 = (8)2 + (16 – x)2

⇒ 32x = 320 ⇒ x = 10.

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

Hence, required area = (12 × BC × AD)

= (12 × 12 × 10) cm2 = 60 cm2

2. The cross section of a canal is a trapezium in shape. If the canal is 7 metres wide at the top and 9 metres at the bottom and the area of cross-section is 1280 square metres, find the length of the canal.
(a) 160 metres
(b) 172 metres
(c) 154 metres
(d) 165 metres
(e) None of these
Ans.a

quant

Let the length of canal = h m.

Then,area of canal = 12 × h(9 + 7)

or 1280 = 12h(16)

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

3. It is required to fix a pipe such that water flowing through it at a speed of 7 metres per minute fills a tank of capacity 440 cubic metres in 10 minutes. The inner radius of the pipe should be :
(a) √2 m
(b) 2 m
(c) 12 m
(d) 1√2 m
(e) None of these
Ans.a

Let inner radius of the pipe be r.

Then, 440 = 227 × r2 × 7 × 10

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

or r = √2 m

4. The area of a square field is 576 km2. How long will it take for a horse to run around at the speed of 12 km/h ?
(a) 12 h
(b) 10 h
(c) 8 h
(d) 6 h
(e) None of these
Ans.c

Area of field = 576 km2. 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 = \(\frac{distance}{speed} = \frac{96}{12} = 8 \; h\)

5. The area of a rectangular field is 144 m2. If the length had been 6 metres more, the area would have been 54 m2 more. The original length of the field is
(a) 22 metres
(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 m2

∴ x = 144y .....(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

(144y + 6)y = 198

⇒ 144 + 6y = 198

⇒ 6y = 54 ⇒ y = 9m

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

6. A rectangular parking space is marked out by painting three of its sides. If the length of the unpainted side is 9 feet, and the sum of the lengths of the painted sides is 37 feet, then what is the area of the parking space in square feet?
(a) 46
(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.

7. A farmer wishes to start a 100 square metres rectangular vegetable garden. Since he has only 30 m barbed wire, he fences three sides of the garden letting his house compound wall act as the fourth side fencing. The dimension of the garden is:
(a) 15 m × 6.67 m
(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 m2 ⇒ l × b = 100 ⇒ b(30 – 2b) = 100

⇒ b2 – 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.

8. A rectangular tank measuring 5 m × 4.5 m × 2.1 m is dug in the centre of the field measuring 13.5 m × 2.5. The earth dug out is spread evenly over the remaining portion of a field. How much is the level of the field raised ?
(a) 4.0 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 m2

Area covered by the rectangular tank

= 5 × 4.5 = 22.50 m2

Area of the field on which the earth dug out is to be spread = 33.75 – 22.50 = 11.25 m2

Let the required height be h.

Then, 11.25 × h = 5 × 4.5 × 2.1

or h = 4.2 m

9. A rectangular paper, when folded into two congruent parts had a perimeter of 34 cm for each part folded along one set of sides and the same is 38 cm when folded along the other set of sides. What is the area of the paper?
(a) 140 cm2
(b) 240 cm2
(c) 560 cm2
(d) 160 cm2
(e) None of these
Ans.a

When folded along breadth, we have :

2(l2 + b) = 34 or l + 2b = 34 ....(i)

When folded along length, we have :

2(l + b2) = 38 or 2l + b = 38 ....(ii)

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

l = 14 and b = 10.

∴ Area of the paper = (14 × 10) cm2 = 140 cm2

10. The length and breadth of the floor of the room are 20 feet and 10 feet respectively. Square tiles of 2 feet length of different colours are to be laid on the floor. Black tiles are laid in the first row on all sides. If white tiles are laid in the one-third of the remaining and blue tiles in the rest, how many blue tiles will be there?
(a) 16
(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 = (13 × 96) sq. ft = 32 sq. ft.

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

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

11. Four equal circles are described about the four corners of a square so that each touches two of the others. If a side of the square is 14 cm, then the area enclosed between the circumferences of the circles is :
(a) 24 cm2
(b) 42 cm2
(c) 154 cm2
(d) 196 cm2
(e) None of these
Ans.b

quant

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 × 14π(7)2

= 196 - 227 × 49 = 196 - 154 = 42 cm2