Ans .
120 \(m^2\)
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.
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 \(cm^2\)
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
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%.
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
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
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
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 \(m^2\)
Area of the square = (1/2)* \((diagonal)^2\) = [(1/2)*3.8*3.8 ]\(m^2\) = 7.22 \(m^2\) .
Ans .
4 : 25.
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%
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
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=6Let 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 \(cm^2\)..
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\(cm^2\)
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
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 \(cm^2\)
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.
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.
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
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 \(cm^2\)
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
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
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
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.
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.
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
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
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.
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.
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%
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%