Let hypotenuse = x cm Then, by Pythagoras theorem: x^{2} = (48020)^{2} + (36015)^{2} x so 60025 cm
Let one side of the △ be = a Perimeter of equilateral triangle = 3a 3a = 72\( \sqrt{3} \) = a = 24\( \sqrt{3} \) cm Height = AC; by Pythagoras theorem AC^{2} = a^{2} – (a/2)^{2} AC = 36 cm
Let inner radius = A; then 2pr = 440 so p = 70 Radius of outer circle = 70 + 14 = 84 cm Outer diameter = 2 × Radius = 2 × 84 = 168
Let inner radius = r and outer radius = R Width = R – r = 396/2π - 352/2π so (R – r) = 44/2π = 7 meters
Let outer radius = R; then inner radius = r = R – 7 2pR = 220 so 35m; r = 35 – 7 = 28 m Area of torch = pR^{2} – pr^{2} p(R^{2} – r^{2}) = 1386 m2 Cost of traveling it = 1386 × = ` 693
Circumference of circle = 2pr = 44 = r = 7 cm Area of a quadrant = π * r^{2} / 4 = 38.5 cm2
Volume of soil removed = l × b × h = 7.5 × 6 × 1.5 = 67.5 m3
The longest pole can be placed diagonally (3-dimensional) BC = \( \sqrt{18^2 + 24^2} \)= 30 AC = \( \sqrt{30^2 + 16^2} \) = 34 m
(d) Let the common ratio be = x Then; length = 3x, breadth = 2x and height = x Then; as per question 3x * 2x * x = 1296 so 6x^{3} = 1296 fi x = 6 m Breadth = 2x = 12 m
Data is inadequate as it’s not mentioned that what part of the cube is melted to form cylinder
(b) Let the common ratio be = x Then, length = 4x, breadth = 3x and height = 2x As per question; 2(4x * 3x + 3x * 2x + 2x * 4x) = 8788 2(12x^{2} + 6x^{2} + 8x^{2}) = 8788 fi 52x^{2} = 8788 fi x = 13 Length = 4x = 52 cm
The total volume will remain the same, let the side of the resulting cube be = a. Then, 6^{3} + 8^{3} + 10^{3} = a^{3} fi a = cube root of 1728 = 12 cm
Slant length = l = \( \sqrt{6^2 + 8^2} \)= 10 cm Then curved surface area = prl = p × 6 × 10 so 60p And total surface area = prl + pr^{2} so p((6 × 10) + 6^{2}) = 96p
Volume of a cone = π * r^{2} * h / 3 Then;100p = π * r^{2} * 12 / 3 = fi r = 5 cm Curved surface area = prl l = √ h^{2} + r^{2} = √ 5^{2} + 12^{2} = 13 then,prl = p × 13 × 5 = 65p cm2
(d) Let the radius of the two cones be = x cm Let slant height of 1st cone = 5 cm and Slant height of 2nd cone = 7 cm Then ratio of covered surface area = 5π / 7π = 5 : 7
Radius = π * r * l / π * l = 2376 / 3.14 * 18 = 42 cm Diameter = 2 × Radius = 2 × 42 = 84 cm
Let the radius of cylinder = 1(r) Then the radius of cone be = 2(R) Then as per question = π * r^{2}*h / π * R^{2}*h / 3 = 3 * π * r^{2}*h / π * R^{2}*h so 3 : 4
The perimeter would remain the same in any case. Let one side of a square be = a cm Then a^{2} = 484 fi a = 22 cm \ perimeter = 4a = 88 cm Let the radius of the circle be = r cm Then 2pr = 88 fi r = 14 cm Then area = pr^{2} = 616 cm2
Let the radius of the circle be = p Then 2pr – 2r = 16.8 fi r = 3.92 cm Then 2pr = 24.6 cm
Let the radius of the wheel be = p Then 5000 × 2pr = 1100000 cm fi r = 35 cm
Let the slant height be = l Let radius = r Then v = π * r^{2} * h/3 = so r = √ 3v/πh = fi = 5 cm l = √(h^{2} + l^{2}) = √(12^{2} + 5^{2}) = 13 cm
In 4 days, the short hand covers its circumference 4 × 2 = 8 times long hand covers its circumference 4 × 24 = 96 times Then they will cover a total distance of:- (2 × p × 4)8 + (2 × p × 6)96 fi 3818.24 cm
Let the radius of the smaller sphere = r Then, the radius of the bigger sphere = R Let the surface area of the smaller sphere = 1 Then, the surface area of the bigger sphere = 4 Then, as per question \( \frac{4\pir^2}{4\piR^2} = 1/4 , r/R = 1/2 \) and hence volumes = \( \frac{4\pir^3}{3} * \frac{3}{4\pi(2r)^3} = 1/8 \)
Inner radius(p) = 9/2 = 4.5 cm Outer radius (R) = 10/2 = 5 cm Volume of metal contained in the shell = 4πR^{3} – 4πr^{3} = 141.9 fi 141.9 cm3
Let smaller radius (r) = 1 Then bigger radius (R) = 2 Then, as per question 4πR^{2} / 4πr^{2} = (1/2)^{2} = 1 : 4
As per question 4πr^{3}/3 = πr^{2}h / 3 = 4r
Volume of wall = 1200 × 500 × 25 = 15000000 cm3 Volume of cement = 5% of 15000000 = 750000 cm3 Remaining volume = 15000000 – 750000 = 14250000 cm3 Volume of a brick = 25 × 12.5 × 7.5 = 2343.75 cm3 Number of bricks used = 14250000 / 2343.75 = 6080
Let the inner radius = r Then 2pr = 352 m. Then r = 56 Then outer radius = r + 7 = 63 = R Now,pR^{2} – pr^{2} = Area of road = p(R^{2} – r^{2}) = 2618 m2
1 hectare = 10000 m2 Height = 10 cm = 1/10m Volume = 10000 × 1/10 = 1000 m3
Total surface area of 7 cubes so 7 × 6a^{2} = 1050 But on joining end to end, 12 sides will be covered. So there area = 12 × a^{2} so 12 × 25 = 300 So the surface area of the resulting figure = 1050 – 300 = 750
Let the rise in height be = h Then, as per the question, the volume of water should be equal in both the cases. Now, 90 × 40 × h = 150 × 8 h = 150*8 / 90*40 = 1/3m = 100/3cm = 33.33 cm
Slant height (l) = √(7^{2} + 24^{2})= 25 m Area of cloth required = covered surface area of cone = prl = 22/7 × 7 × 25 = 550 m2 Amount of cloth required = 550/5 = 110 m
If the ratio of their diameters = 2 : 1, then the ratio of their radii will also be = 2 : 1 Let the radii of the broader cone = 2 and height be = 1 Then the radii of the smaller cone = 1 and height be = 2 Ratio of volumes = (π2^{2}*1 / 3) / (π1^{2}*2 / 3) = 4π/3 * 3/2π = 2 : 1
Area of base = 6 × 10 = 60 m2 Volume of tent = 30 × 10 = 300 m3 Let the radius be = r, height = h, slant height = l pr 2 = 60 fi r = √(60/π) 300 = πr^{2}h/3 = 900 = p *60/π* h = h = 15 m
Volume of wood used = External volume – Outer Volume fi (10 × 8 × 6) – (10 – 1) × (8 – 1) × (6 – 1) fi 480 – (9 × 7 × 5) = 165 cm2
Total volume in both the cones will be equal. Let the number of smaller cubes = x x * 3^{3} = 24 × 9 × 8 fi x = 24*72/27 = 64
Let one side of the cube = a Then a^{3} = 216 fi a = 6 m Area of the resultant figure = Area of all 3 cubes – Area of covered figure fi 216 × 3 – (4 × a^{2}) fi 648 – 144 fi 504 m2
Volume of metal used = 4πR^{3}/3 - 4πr^{3}/3 = 4π/3(12^{3} – 10^{3}) = 3047.89 cm3 Weight = volume × densityfi 4.9 × 3047.89 fi 14942.28 gm
Volume of cube = 7 3 = 343 cm3 Radius of cone = 7/2 = 3.5 cm Height of cone = 7 Ratio of volumes = πr^{2}h/3/343 = 11:42
The volume in both the cases will be equal. Let the height of cone be = h 4 × 22/7 × (14)^{3} × 1/3 = 22/7 * h/3 * (35/2)^{2} so 4(14)^{3} = h(35/2)^{2} = h = 35.84 cm
Diameter of circle = diagonal of square = √(10^{2} + 10^{2})= 10√2 ◊.◊ Radius = 5√2 Area of circle = pr^{2} = 50p = 50 × 3.14 = 157.14 cm3
Area of triangle = rS; where r = inradius S = 15+8+7/2 = 20 cm △ = \( \sqrt{s(s-a)(s-b)(s-c)} \) △ = 60 cm2 Area of triangle = rS; r = 3 cm
Circumference of the circular face of the cylinder = 2pr fi 2 × 22/7 × 35/100 = 2.2 m Number of revolutions required to lift the bucket by 11 m = 11/2.2 = 5
Surface area of the cube = 6a^{2} = 6 × (20)^{2} = 2400 Area of 6 circles of radius 10 cm = 6pr^{2} = 6 × p × 100 = 1885.71 Remaining area = 2400 – 1884 = 514.28
x * y * z = lb × bh × lh = (lbh)^{2} (V) Volume of a cuboid = lbh So V^{2} = (lbh)^{2} = xyz
Diameter of the circle = diagonal of rectangle = √(8^{2} + 6^{2}) = 10 cm Radius = 10/2 = 5 cm Area of shaded portion = pr^{2} – lb = 3.14 × 5^{2} – 8 × 6 = 30.57 cm2
Larger Radius (R) = 14 + 7 = 21 cm Smaller Radius (r) = 7 cm Area of shaded portion πR^{2}θ/360 – πR^{2}θ/360 π*θ/360 (21^{2} - 7^{2}) = 102.67 cm
Area of quadrilateral = Area of right angled triangle + Area of equilateral triangle x = √(20^{2} - 12^{2}) = 16 Area of quadrilateral = 1/2*16*12 + √3/4 × 20 × 20 = 269 units2
h = √(24^{2} - 13^{2}) = √407 volume = Area of base * height/3 = 18*26*√407/3 = 156√407
The perimeter would remain the same in both cases. Circumference of circle = 2pr = 2 × 22/7 × 28 = 176 cm Perimeter of square = 176 Greatest side possible = 176/4 = 44 cm Length of diagonal = √(44^{2} + 44^{2}) = 62.216 = 88/2 * √2 = 44√2