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1.

An isosceles trianglehas perimeter `30 c m`and each of the equalsides os `12 c mdot`Find the area of thetriangle.

Answer» Here, Perimeter of the isosceles triangle `P = 30cm`
So, Semi-perimeter of the triangle, `s = 30/2 = 15cm`
Two sides of the triangle are `12cm`. So, `a=b = 12cm`
Third side, `c = P-a-b = 30-12-12 = 6cm`
Now, Area of the triangle, `A = sqrt(s(s-a)(s-b)(s-c))`(Herons formula)
`A = sqrt(15(3)(3)(9)) = 9sqrt15cm^2`
Discussion
2.

A field is in the shape of a trapezium whose parallel sides are 25m and 10m. If its non-parallel sides are 14m and 13m, find its area.

Answer» ADBE is a parallelogram
AB= DE
AD= BE
EC= 25 - BE = 15
value of s for `/_ DEC = (13+14+15)/2 = 21 `
area = `sqrt(s(s-a)(s-b)(s-c))`
`= sqrt(21*8*7*6)`
`= 7*3*2^2`
`= 84 unit^2`
`EX= 13/(13+14) *15 `
`=65/9 `
Now, `y^2 = 13^2 - (65/9)^2`
`y= 10.8`
total area= `108+ 84`
` =192 unit^2 `
Discussion
3.

A park, in the shape of a quadrilateral ABCD, has `/_C=90^o , A B=9 m , B C=12 m , C D=5 m a n d A D=8 m`. How much area does it occupy?

Answer» We can create a diagram with the given details.
Please refer to video for the diagram.
Now, `ar(BCD) = 1/2**12**5 = 30m^2`
Also, `BD = sqrt(12^2+5^2) = sqrt169 = 13m`So, in `Delta ABD`
`a = 8m,b=9m,c=13m`
So, `s = (a+b+c)/2 = (8+9+13)/2 = 15m`
So, `ar(ABD) = sqrt(s(s-a)(s-b)(s-c))`
`ar(ABD)=sqrt(15(15-8)(15-9)(15-13)) = sqrt(15**7**6**2)=6sqrt35m^2`
So, `ar(ABCD) = ar(BCD)+ar(ABD) = 30+6sqrt35m^2`
Discussion
4.

A triangular park ABC has sides 120m, 80m and 50m. A gardener Dhania has to put a fence all around it and also plant grass inside. How much area does she need to plant? Find the cost of fencing it with barbed wire at the rate of Rs 20 per metre leaving a space 3m wide for a gate on one side.

Answer» Length of fence=perimeter of Triangle-3m
=50+80+120-3
=247m
Cost of wire=length *cost
=247*20
=4940Rs
Herons formula
area=`sqrt(s(s-a)(s-b)(s-c))`
=`sqrt(125(125-120)(125-80)(125-50)`
=`375sqrt15 m^2`
Discussion
5.

Find the area of a quadrilateral ABCD in which `A B=3 c m , B C=4 c m , C D=4 c m , D A=5 c m a n d A C=5 c m`.

Answer» We can create a diagram with the given details.
Please refer to video for the diagram.
Now, in `Delta ABC`,`5^2 = 3^2+4^2` that means it is a right angle triangle as it is satisfying Pythagoras theorem.
So, `ar(ABC) = 1/2**3**4 = 6cm^2`
Now, in `Delta ACD`,
`a=4cm,b=5cm and c = 5cm`
So, Semi-perimeter,`s = (a+b+c)/2 = 14/2 = 7cm`
So, `ar(ACD) = sqrt(s(s-a)(s-b)(s-c)) = sqrt(7**3**2**2)=2sqrt21cm^2`
Thus, `ar(ABCD) = ar(ABC)+ar(ACD) = 6+2sqrt21cm^2`
Discussion
6.

An umbrella is made by stitching 10 triangular pieces of cloth of two different colour, each piece measuring 20 cm, 50 cm and 50 cm. How much cloth of each colour is required for the umbrella?

Answer» s=(a+b+c)/2=60cm
Heroes formula
area=`sqrt(s(s-a)(s-b)(s-c))`
=`sqrt(60(60-50)(60-50)(60-20))`
=`200sqrt6`
There are total 10 piecesFirst colour=`5*200sqrt6=1000sqrt6`
Second color=`5*200sqrt6=1000sqrt6`
Discussion
7.

The sides of a triangular plot are in the ratio of `3:5:7`and its perimeteris 300 m. Find its area.

Answer» a:b:c=3:5:7
a=3x=60
b=5x=100
c=7x=140
a+b+c=300
3x+5x+7x=300
15x=300
x=20
Semiperemeter=300/2=150
Herons formula
area=`sqrt(s(s-a)(s-b)(s-c))`
=`sqrt(150(150-60)(150-100)(150-140))`
=`1500sqrt3m^2`
Discussion
8.

Sides of a triangle are in the ratio of `12 : 17 : 25`and its perimeter is 540cm. Find its area.

Answer» Let the sides of the triangle be `12x,17x `and `25x`
perimeter of triangle=`540=12x+17x+25x=54x`
=> `x=10`
The sides of the triangle are 120 cm,170cm and250cm.
Area of triangle=`sqrt(s(s-a)(s-b)(s-c))` where `s` is the semi perimeter and a,b,c are the sides of the triangle.
substituting values=>`sqrt(270(150)(100)(20))=9000(cm)^2`
Discussion
9.

Students of a school staged a rally for cleanliness campaign. They walked through the lanes in two groups. One group walked through the lanes AB, BC and CA; while the other through AC, CD and DA (see Fig. 12.12). Then they cleaned the area enclos

Answer» areaABC=1/2*BC*AB
1/2*40*9=180`m^2`
By Pythagoras Theorem
`AC^2=AB^2+BC^2`
`AC^2=9^2+40^2`
`AC^2=1681`
`AC=41m`
s=(a+b+c)/2=84/2=42
Herons formula
areaACD=`sqrt(s(s-a)(s-b)(s-c))`
`=sqrt(42(42-15)(42-41)(42-28))`
=126`m^2`
Group 1 has cleared more area of 54`m^2`.
Discussion