Mensuration Part 15

Problem 1: The ratio of the radii of two circles is 5:6. If the area of the larger circle is 1296cm2, then what is the area of the smaller circle?

Solution:

\Pi \times 36\times x^{2}=1296

x^{2}=\frac{36}{\Pi }

\Pi \times 25x^{2}=\Pi \times 25\times \frac{36}{\Pi }=900

Problem 2: How many 2cm × 2cm size stickers can be cut from 9cm × 16cm size paper?

Solution:

n=\frac{9\times 16}{2\times 2}=36

Problem 3: What is the area of the largest triangle that can be constructed inside a semicircle with radius Rcm?

Solution:

\frac{1}{2}\times 2r\times r=r^{2}

Problem 4: How much distance a wheel of a diameter 1.26m will cover in 500 revolutions?

Solution:

Diameter = 2\Pi rn = \frac{22}{7}\times \frac{126}{100}\times 500=1980m

Mensuration Part 14

Problem 1: Find the total surface area of a tetrahedron, length of whose sides are 6 cm?

Solution:

Total surface area = \sqrt{3}a^{2}=36\sqrt{3}

Problem 2: If the radius and height of the cylinder are 7cm and 6cm respectively then find the curved surface area of the cylinder?

Solution:

Curved surface area = 2\Pi rh=2\times \frac{22}{7}\times 7\times 6=264

Problem 3: Ratio between height and radius of a right circular cone is 3: 1 and its volume is 1078cm3. Find the height of the cone.

Solution:

h : r = 3 : 1

Volume = \frac{1}{3}\Pi r^{2}h

1078=\frac{1}{3}\times \frac{22}{7}\times x^{2}\times 3x

49\times 7=x^{3}

x=7

Height = 3*x = 3*7 = 21

Problem 4: If the volume of a sphere is numerically equal to its surface area then its diameter is?

Solution:

\frac{4}{3}\Pi r^{3}=4\Pi r^{2}

r=3

Diameter = 2r =2*3 = 6

Problem 5: The diameters of two solid iron spheres are 4cm and 8cm respectively. These two spheres are melted to make a new solid cone whose radius is 6cm. Find its height?

Solution:

Radius of first sphere = 2

Radius of second sphere = 4

v1+v2 = v3

\frac{4}{3}\Pi (8+64)=\frac{1}{3}\Pi \times 36\times h

h=8

Mensuration Part 13

Problem 1: The base of a right prism is a triangle. If the lengths of the sides are 5cm., 12cm. and 13cm. and volume of the prism is 450cm3, then the total surface area of the prism is?

Solution:

Area  = \frac{1}{2}\times Perpendicular \times base

=\frac{1}{2}\times 5\times 12=30

Volume =Area * Height

450 = 30*h

h=15cm

Lateral surface area = Perimeter of base * Height

=(5+13+12)*15

=30*15 = 450

Total surface area of prism =Lateral surface area+2(area of base)

450 + 2*30 =510cm^{2}

 

Problem 2: Find the total surface area of the cube of side 3.5 cm?

Solution:

Total surface area = 6a^{2} =6\times 3.5\times 3.5 = 73.5

Problem 3: The diagonal of a cube is 5\sqrt{3}m. Find its volume?

Solution:

a\sqrt{3}=5\sqrt{3}

a=5

a^{3}=5^{3}=125

 

Problem 3: A cuboidal brick has a length, breadth, and height of 25cm, 15cm and 5cm respectively. Find its total surface area.

Solution:

Total surface area of cuboid = 2(lb+bh+lh)

= 2(375+75+125)

= 2*575 = 1150

 

Problem 4: The base of a right pyramid is a square of side 40cm. If the volume of the pyramid is 8000cm3, then its height is?

Solution:

Volume = \frac{1}{3}\times ArearOfBase\times Height

Area of base = Area of square = 40*40 = 1600

8000=\frac{1}{3}\times 1600\times h

h = 15

Mensuration Part 11

Problem 1: Radius of a circle is 14cm. and the angle at the center formed by a sector is 45 degrees. Find the area of the sector?

Solution:

Area of a sector = \Pi r^{2}\frac{\Theta }{360}

=\frac{22}{7}\times 14\times 14\times \frac{45}{360}

=77

Problem 2: A rectangular sheet of 30 m × 20 m dimension is cut from its every corner as a quadrant of radius 3.5m. Find the area of remaining portion?

Solution:

4 sectors = 1 circle

Area of rectangle = 30*20=600

\Pi r^{2}=\frac{22}{7}\times \frac{7}{2}\times \frac{7}{2}=38.5

600-38.5= 561.5

Problem 3: Smaller diagonal of a rhombus is equal to the length of its sides. If the length of each side is 4 cm, then what is the area (in cm2) of an equilateral triangle with side equal to the bigger diagonal of the rhombus?

Solution:
Area of equilateral triangle = \frac{\sqrt{3}}{4}a^{2}=12\sqrt{3}