# AREA

Ans .

120 $$m^2$$

1. Explanation :

Other side = $${17}^ 2$$- $$15^2{^{(1/2)}}$$ = $$(289- 225)^{1/2}$$ = $${64}^{1/2}$$ = 8 m.

Area = (15 x 8) $$m^2$$ = 120 $$m^2$$

Ans .

33/3,50

So, 2x * 3x = 5000/3 <=> $$x^2$$ = 2500/9 <=> x = 50/3

therefore Length = 2x = (100/3) m = 33(1/3) m and Breadth = 3x = 3(50/3) m = 50m.

Ans .

Rs. 1934.40.

1. Explanation :

Area of the carpet = Area of the room = (13 * 9)$$m^2$$ = 117 $$m^2$$. Length of the carpet = (area/width) = 117 *(4/3) m = 156 m. Therefore Cost of carpeting = Rs. (156 * 12.40) = Rs. 1934.40.

Ans .

120 $$cm^2$$

1. Explanation :

Let length = x and breadth = y. Then,

2 (x + y) = 46 or x + y = 23 and $$x^2$$ + $$y^2$$ = $$17^2$$ = 289.

Now, $$(x+y)^2$$ = $$23^2$$ <=> ( $$x^2$$ + $$y^2$$ ) + 2xy = 529 <=> 289 + 2xy = 529 xy=120

Area = xy = 120 $$cm^2$$ .

Ans .

20 cm

1. Explanation :

Let breadth = x. Then, length = 2x. Then,

(2x - 5) (x + 5) - 2x * x = 75 <=> 5x - 25 = 75 <=> x = 20.

Length of the rectangle = 20 cm.

Ans .

0.8%.

1. Explanation :

Let x and y be the sides of the rectangle. Then, Correct area = xy.

Calculated area = (105/100)*x * (96/100)*y = (504/500 )(xy)

Error In measurement = (504/500)xy- xy = (4/500)xy

Error % = [(4/500)xy *(1/xy) *100] % = (4/5) % = 0.8%.

Ans .

Rs. 680

1. Explanation :

Area of the plot = (110 x 65) $$m^2$$ = 7150 $$m^2$$

Area of the plot excluding the path = [(110 - 5) * (65 - 5)] $$m^2$$ = 6300 $$m^2$$ .

Area of the path = (7150 - 6300) $$m^2$$ = 850 $$m^2$$ .

Cost of gravelling the path = Rs.850 * (80/100)= Rs. 680

Ans .

24 cm

1. Explanation :

Side of first square = (40/4) = 10 cm;

Side of second square = (32/4)cm = 8 cm.

Area of third square = [(10) 2 - (8) 2] $$cm^2$$ = (100 - 64) $$cm^2$$ = 36 $$cm^2$$

Side of third square = (36)(1/2) cm = 6 cm.

Required perimeter = (6 x 4) cm = 24 cm.

Ans .

176

1. Explanation :

Area of the room = (544 x 374) $$cm^2$$.

Size of largest square tile = H.C.F. of 544 cm and 374 cm = 34 cm.

Area of 1 tile = (34 x 34) $$cm^2$$.

Number of tiles required =(544*374)/(34*34)=176

Ans .

7.22 $$m^2$$

1. Explanation :

Area of the square = (1/2)* $$(diagonal)^2$$ = [(1/2)*3.8*3.8 ]$$m^2$$ = 7.22 $$m^2$$ .

Ans .

4 : 25.

1. Explanation :

Let the diagonals of the squares be 2x and 5x respectively.

Ratio of their areas = (1/2)*$$(2x) ^2$$ : (1/2)*$$(5x)^ 2$$ = 4$$x^2$$ : 25$$x^2$$ = 4 : 25.

Ans .

56.25%

1. Explanation :

Let each side of the square be a. Then, area = $$a^2$$.

New side =(125a/100) =(5a/4). New area = $$(5a/4) ^2$$ =(25$$a^2$$)/16.

Increase in area = ((25 $$a^2$$)/16)-$$a^2$$ =(9$$a^2$$)/16.

Increase% = [((9$$a^2$$)/16)*(1/$$a^2$$)*100] % = 56.25%.

Ans .

50 cm

1. Explanation :

Let x and y be the length and breadth of the rectangle respectively.

Then, x - 4 = y + 3 or x - y = 7 ----(i)

Area of the rectangle =xy; Area of the square = (x - 4) (y + 3)

(x - 4) (y + 3) =xy <=> 3x - 4y = 12 ----(ii)

Solving (i) and (ii), we get x = 16 and y = 9.

Perimeter of the rectangle = 2 (x + y) = [2 (16 + 9)] cm = 50 cm.

Ans .

l=9,b=6,h=6

1. Explanation :

Let breadth = x metres, length = 3x metres, height = H metres.

Area of the floor=(Total cost of carpeting)/(Rate/m2)=(270/5)m2=54m2.

x* (3x/2) = 54 <=> $$x^2$$= (54*2/3) = 36 <=> x = 6.

So, breadth = 6 m and length =(3/2)*6 = 9 m.

Now, papered area = (1720/10)$$m^2$$ = 172 $$m^2$$.

Area of 1 door and 2 windows = 8 $$m^2$$.

Total area of 4 walls = (172 + 8)$$m^2$$ = 180 $$m^2$$

2*(9+ 6)* H = 180 <=> H = 180/30 = 6 m.

Ans .

84 $$cm^2$$..

1. Explanation :

Let a = 13, b = 14 and c = 15. Then, S = (1/2)(a + b + c) = 21.

(s- a) = 8, (s - b) = 7 and (s - c) = 6.

Area = $$(s(s- a) (s - b)(s - c))^{1/2}$$ = $$(21 *8 * 7*6)^{1/2}$$ = 84 $$cm^2$$.

Ans .

30$$cm^2$$

1. Explanation :

Height of the triangle = [$$13^2$$ - $$12^{2 ^{1/2}}$$] cm = $$25^{1/2}$$ cm = 5 cm.

Its area = (1/2)* Base * Height = ((1/2)*12 * 5) $$cm^2$$ = 30$$cm^2$$ .

Ans .

Base = 900 m and Altitude = 300 m

1. Explanation :

Area of the field = Total cost/rate = (333.18/25.6)hectares = 13.5 hectares

(13.5 x 10000) $$m^2$$ = 135000 $$m^2$$

Let altitude = x metres and base = 3x metres.

Then, (1/2)* 3x * x = 135000 <=>$$x^2$$ = 90000 <=>x = 300.

Base = 900 m and Altitude = 300 m.

Ans .

60 $$cm^2$$

1. Explanation :

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 triangle ADC, $$AC^2$$= AD +$$DC^2$$ =>$$x^2$$ =(82)+$$(16-x) ^2$$

=>32x = 320 =>x= 10.

BC = (32- 2x) = (32 - 20) cm = 12 cm.

Hence, required area = ((1/2)x*BC * AD) = ((1/2)*12 *10)$$cm^2$$ = 60 $$cm^2$$

Ans .

4.5 cm.

1. Explanation :

Area of the triangle = ($$\sqrt{3}$$/4) x (3$$\sqrt{3}$$)2 = 27$$\sqrt{3}$$. Let the height be h.

Then, (1/2) x 3$$\sqrt{3}$$ x h = (27$$\sqrt{3}$$/4) X(2/$$\sqrt{3}$$) = 4.5 cm.

Ans .

16 : 9.

1. Explanation :

Let the bases of the two triangles be x and y and their heights be 3h and 4h respectively.

Then,

((1/2) X x X 3h)/(1/2) X y X 4h) =4/3 => x/y =(4/3 X 4/3)=16/9

Required ratio = 16 : 9.

Ans .

6 cm

1. Explanation :

Let the height of the parallelogram be x. cm. Then, base = (2x) cm.

2x X x =72 2$$x^2$$ = 72 $$x^2$$ =36 x=6

Hence, height of the parallelogram = 6 cm.

Ans .

384 $$cm^2$$

1. Explanation :

Let other diagonal = 2x cm.

Since diagonals of a rhombus bisect each other at right angles, we have:

$$20^2$$ = $$12^2$$ + $$x^2$$ x= $$\sqrt{20^2-{12^2}}$$= 256= 16 cm.

So, other diagonal = 32 cm.

Area of rhombus = (1/2) x (Product of diagonals) = ((1/2)x 24 x 32) $$cm^2$$ = 384 $$cm^2$$

Ans .

27 cm and 23 cm

1. Explanation :

Let the two parallel sides of the trapezium be a em and b em.

Then, a - b = 4

And, (1/2) x (a + b) x 19 = 475 --> (a + b) =((475 x 2)/19) --> a + b = 50

Solving (i) and (ii), we get: a = 27, b = 23.

So, the two parallel sides are 27 cm and 23 cm

Ans .

paste right option

1. Explanation :

Clearly, the cow will graze a circular field of area 9856 sq. metres and radius equal to the length of the rope. Let the length of the rope be R metres. Then, $$\prod R^2$$= (9856 X (7/22)) = 3136 --> R = 56. Length of the rope = 56 m.

Ans .

Rs. 5808

1. Explanation :

Area = (13.86 x 10000) $$m^2$$ = 138600 $$m^2$$ .

($$R^2$$ = 138600 ($$R^2$$ = (138600 x (7/22)) R = 210 m.

Circumference = 2$$\prod$$R = (2 x (22/7) x 210) m = 1320 m.

Cost of fencing = Rs. (1320 x 4.40) = Rs. 5808.

Ans .

250.

1. Explanation :

Distance to be covered in 1 min. = $$\frac{66 * 1000}{60}$$ m = 1100 m.

Circumference of the wheel = (2 x (22/7) x 0.70) m = 4.4 m.

Number of revolutions per min. =(1100/4.4) = 250.

Ans .

14 m.

1. Explanation :

Distance covered in one revolution =$$\frac{88 X 1000}{1000}$$= 88m.

2$$\prod$$R = 88 --> 2 x (22/7) x R = 88 --> R = 88 x (7/44) = 14 m.

Ans .

84 m

1. Explanation :

Let inner radius be r metres. Then, 2$$\prod$$r = 440 --> r = (440 x (7/44))= 70 m.

Radius of outer circle = (70 + 14) m = 84 m.

Ans .

4 m

1. Explanation :

Let the inner and outer radii be r and R metres.

Then 2$$\prod$$r = (352/7) -->r =((352/7) X (7/22) X (1/2))=8m.

2$$\prod$$R=(528/7) --> R=((528/7) X (7/22) X (1/2))= 12m.

Width of the ring = (R - r) = (12 - 8) m = 4 m.

Ans .

3 cm.

1. Explanation :

Let the radius of the circle be r cm. Then,

$$\frac{\prod r^{2}\theta }{360}$$=(66/7) --> (22/7) X $$r^2$$ X(120/360)= (66/7)

$$r^2$$=((66/7) X (7/22) X 3) --> r=3. Hence, radius = 3 cm.

Ans .

1 : 2.

1. Explanation :

Radius of incircle = (x/2)

Radius of circum circle= ($$\sqrt{2}$$x/2) =(x/$$\sqrt{2}$$)

Required ratio = (($$\frac{\prod r^2}{4}$$ ) : (($$\frac{\prod r^2}{2}$$ ) = (1/4) : (1/2) = 1 : 2.

Ans .

75%

1. Explanation :

Let original radius = R. New radius =(50/100) R = (R/2)

Original area=$$\prod {r/2}^2$$= and new area= $$\prod {r/2}^2$$= (($$\frac{\prod r^2}{4}$$ )

Decrease in area =((3$$\frac{\prod r^2}{4}$$ ) X (1/$$\prod {r/2}^2$$ X 100) % = 75%