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:

$\Pi&space;\times&space;36\times&space;x^{2}=1296$

$x^{2}=\frac{36}{\Pi&space;}$

$\Pi&space;\times&space;25x^{2}=\Pi&space;\times&space;25\times&space;\frac{36}{\Pi&space;}=900$

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

Solution:

$n=\frac{9\times&space;16}{2\times&space;2}=36$

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

Solution:

$\frac{1}{2}\times&space;2r\times&space;r=r^{2}$

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

Solution:

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

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 = $2\Pi&space;rh$=$2\times&space;\frac{22}{7}\times&space;7\times&space;6=264$

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 = $\frac{1}{3}\Pi&space;r^{2}h$

1078=$\frac{1}{3}\times&space;\frac{22}{7}\times&space;x^{2}\times&space;3x$

$49\times&space;7=x^{3}$

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:

$\frac{4}{3}\Pi&space;r^{3}=4\Pi&space;r^{2}$

$r=3$

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

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

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  = $\frac{1}{2}\times&space;Perpendicular&space;\times&space;base$

=$\frac{1}{2}\times&space;5\times&space;12=30$

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 = $6a^{2}$ =$6\times&space;3.5\times&space;3.5$ = 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 = $\frac{1}{3}\times&space;ArearOfBase\times&space;Height$

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

8000=$\frac{1}{3}\times&space;1600\times&space;h$

h = 15

Mensuration Part 12

Problem 1: The area of the largest triangle that can be inscribed in a semicircle of radius 6m is?

Solution:

$\frac{1}{2}\times&space;b\times&space;h=\frac{1}{2}\times&space;2r\times&space;r$

$r^{2}=36m^{2}$

Problem 2: The base and altitude of a right-angled triangle are 12cm and 16cm respectively the perpendicular distance of its hypotenuse from the opposite vertex is?

Solution:

$\frac{1}{2}\times&space;12\times&space;16=\frac{1}{2}\times&space;20\times&space;h$

=9.6$m^{2}$

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 = $\Pi&space;r^{2}\frac{\Theta&space;}{360}$

$=\frac{22}{7}\times&space;14\times&space;14\times&space;\frac{45}{360}$

=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

$\Pi&space;r^{2}=\frac{22}{7}\times&space;\frac{7}{2}\times&space;\frac{7}{2}=38.5$

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 = $\frac{\sqrt{3}}{4}a^{2}=12\sqrt{3}$