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 b\times h=\frac{1}{2}\times 2r\times 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 12\times 16=\frac{1}{2}\times 20\times h

=9.6m^{2}

Problem 3:

 

Mensuration part 11

Problem 1: Radius of a circle is 14cm. and the angle at centre 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 each corner as 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 length of its sides. If 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}

Mensuration part 10

Problem 1: A rectangular park whose length and breadth are 30m and 20m, is surrounded by 2m broad road. Find the area of the road ?

Solution:

Area =(l+b+2w)2w

(30+20+4)*4)=216

Problem 2: A rectangular field whose sides are 50m × 40m has two straight road of width 2m and connects their opposite sides. Find the area of the road?

Solution:

Area of shaded region = (50+40-2)*2 = 88*2 =176

Problem 3: The area of a rectangle whose one side is ‘a’ is ‘2a2’. What is the area of a square having one of the diagonals of the rectangle as side?

Solution:

2a^{2}=a\times b

b=2a

Diagonal =\sqrt{a^{2}+4a^{2}}=\sqrt{5a^{2}}

Area of a square = (\sqrt{5a^{2}})^{2}=5a^{2}

Mensuration part 8

Important formula

CUBE:

Volume = a^{3}

Curved surface area = 4a^{2}

Total surface area = 6a^{2}

CUBOID:

Volume = l*b*h

Curved surface area = 2(l+b)*h

Total surface area = 2(lb+bh+hl)

diagno1 =\sqrt{l^{2}+b^{2}+h^{2}}

CYLINDER:

Volume =\Pi r^{2}h

Curved surface area =2\Pi rh

Total surface area =  2\Pi r(h+r)

CONE:

l=slant height = \sqrt{h^{2}+r^{2}}

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

Curved surface area = \Pi rl

Total surface area = \Pi r(l+r)

SPHERE:

Example is ball

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

Curved surface area = 4\Pi r^{2}

Total surface area = 4\Pi r^{2}

HEMISPHERE:

Volume = 2\Pi r^{3}

Curved surface area = 2\Pi r^{2}

Total surface area = 3\Pi r^{2}

FRUSTUM:

Volume = \frac{1}{3}\Pi (R^{2}+r^{2}+rR)h

Curved surface area = \Pi (R+r)l

Total surface area = \Pi (R^{2}+r^{2}+Rl+rl)

l=\sqrt{h^{2}+(R-r)^{2}}

PRISM:

         

Volume=Area of base * Height

Lateral surface area = Perimeter of base * height

Total surface area = L.S.A+2*Area of base

 

Mensuration part 7

Problem 1: The perimeter of an isosceles triangle is 32 cm and each of the equal sides is 5/6 times of the base. What is the area (in cm2) of the triangle?

Solution:

2a+b=32

2\times \frac{5}{6}b+b=32

\frac{10b+6b}{6}=32

b=12

2a=32-12=20

a=10

Area =\frac{b}{4}\sqrt{4\times 10^{2}-12^{2}}

\frac{12}{4}\times 16=48

Problem 2: Sides of a parallelogram are in the ratio 5 : 4. Its area is 1000 sq. units. Altitude on the greater side is 20 units. Altitude on the smaller side is?

Solution:

h_{1}\times b=h_{2}\times l

5x*20=4x*h

Height =25

Mensuration part 6

Problem 1: If the length of a rectangular field is increased by 20 % and the breadth is reduced by 20 %. The area of the rectangle will be 192 m2. What is the area of the original rectangle ?

Solution:

Length       Breadth————->Area

10                       10      ————->100

12                        8       ————->96

96——->192

100——->?\frac{192\times 100}{96}=200

The original area of rectangle is 200

Problem 2: An order was placed for supply of carpet of breadth 3 meters. The length of carpet was 1.44 times of breadth. Subsequently, the breadth and length were increased by 25 and 40 percent respectively. At the rate of Rs. 45 per square meter, what would be the increase in the cost of the carpet?

Solution:

Breadth = 3m

Length =3*1.44 = 4.32

Breadth=3\times \frac{125}{100}=3.75

Length = 4.32\times \frac{140}{100}=6.048

=3.75*6.048*45 = 1020.6

Problem 3: The surface area of water in a swimming pool forms a rectangle with length 40 m and breadth 15m. The depth of water increases uniformly from 1.2m to 2.4m at the other end. The volume ( in m3) of water in the pool is ?

Solution:

Volume = Length * Breadth * (\frac{h_{1}+h_{2}}{2})

= 40 * 15 *( \frac{1.2+2.4}{2})

= 600*1.8

=  1080

Mensuration part 5

problem 1: The area of a field in the shape of a trapezium measures 1440m2. The perpendicular distance between its parallel sides is 24 m. If the ratio of the parallel sides is 5 : 3 , the length of the longer parallel sides is?

Solution:

Area of trapezium =\frac{1}{2}(a+b)\times h

1440=\frac{1}{2}\times (5x+3x)\times 24

1440=\frac{1}{2}\times (8x)\times 24

x=15

5x=5*x=75

Problem2:What would be the cost of building a 7 meters wide garden around a circular field with diameter equal to 280 meters if the cost per square meter for building the garden is Rs. 21 ?

Solution:

\Pi r^{2}-\Pi r^{2}

\Pi (147^{2}-140^{2})

\frac{22}{7}\times 287\times 7=6314

1m———–>21

6341m—–>? 6341*21 = 132594

Problem 3: There is a rectangular plot whose length is 36 meter and breadth is 28 meter. There are two paths middle of the plot and parallel to length and breadth of plot. The remaining part is lawn whose area is 825 square meter. What is the area of the paths ?

Solution:

Area of rectangle=36*28=1008

Given area of lawn = 1008-825 = 183

Mensuration part 4

Problem 1: If the perimeter of a rectangle and a square are equal and the ratio of two adjacent sides of the rectangle is 1 : 2 then the ratio of area of the rectangle and that of the square is?

Solution:

Length =x , Breadth = y

If the perimeter of a rectangle and a square are equal
2(l+b) = 4a
2(x+2x) = 4a
3x=2a
a=\frac{3}{2}x
2x^{2}:\frac{9}{4}x^{2}
8  :  9
Problem 2: The cost of fencing a circular plot at the rate of Rs. 15 per meter is Rs. 3300. What will be the cost of flooring the plot at the rate of Rs. 100 per square meter ?
Solution:
RS. 15 —–> 1meter
RS. 3300—–>?\frac{3300}{15}=220m
Perimeter = 220
2\Pi r=220
r=220\times \frac{7}{22}\times \frac{1}{2}
r=35
Area = \Pi r^{2}
\frac{22}{7}\times 35\times 35=3850 m^{2}
1m——->RS.100
3850m—->? 3850*100 = 385000

Mensuration part 3

Problem 1: The ratio of three angles of a quadrilateral are in the ratio 1: 6 : 2. The measure of fourth angle is 450. What is the difference between the largest and smallest angle?

Solution:

Sum of four angles in a quadrilateral is 360

x+6x+2x+45 = 360

9x = 315

x = 35

Difference between smallest and largest angle is 6x-x = 5x

5x = 5*35 = 175

Problem 2: On increasing the diameter of a circle by 75 %, what will be the percentage increase in the perimeter?

Solution:

If radius is increased perimeter is also increase.

75%=\frac{3}{4}

actual diameter        increased diameter

4                                           7

16                                        49

\frac{33}{16}\times 100=87.5%

Problem 3: If the longer side of a rectangle is doubled and the other reduced to half, then find what will be the percentage increment in the area of the new rectangle?

Solution:

Let       length =10     breadth = 10  ——->10*10=100

double length =20   half breadth=5——->20*5=100

no increment in the area  0%