Question

The figure of a plot and its measures are given. l(LM) = 60 m. l(MN) = 60 m. l(LN) = 96 m. l(OP) = 70 m. find the area of the plot.

Answer 1

Dear student,

Solution:

Area of triangle: 1/2 Base * height

Area of triangle with half perimeter formula,i.e. Herion's formula √s(s-a)(s-b)(s-c)

here a,b,c are three sides of triangle and s = (a+b+c)/2

Δ LMN : There is no right angle.

LM = 60 m

MN = 60 m

NL = 96 m

s = (60+60+96)/2

= 216/2

= 108

Area of Δ LMN =$\sqrt{s(s-a)(s-b)(s-c)}\\ \\ = \sqrt{108(108-60)(108-60)(108-96)}\\ \\ =\sqrt{108 *48*48*12} \\ \\ = 48\sqrt{1296} \\ \\ = 48*36\\ \\ = 1728 m^{2}$

Now for second triangle LNO

since it is right angle triangle.

Base LN = 96 m

Height OP = 70 m

Area of Triangle LNO = $\frac{1}{2}*96*70\\ \\ = 48*70\\ \\ =3360 m^{2}$

Now total area of plot LMNO addition of area of both the triangle ΔLMN and Δ LNO

Area of plot = 1728 + 3360 $m^{2}$

Area of plot LMNO = 5088 $m^{2}$

hope it helps you.

Answer 2

Hi ,

1 ) From the figure ,

LM = n = 60 m

MN = l = 60 m

LN = m = 96 m

*********************************************
Heron's formula:

Area of the triangle LMN = A1

A1 = sqrt[ s( s - n)( s - l )( s - m )

where s = ( n + l + m )/2
*********************************************
s = ( 60 + 60 + 96)/2

s = 108

s - n = 108 - 60 = 48

s - l = 108 - 60 = 48

s - m = 108 - 96 = 12

A1 = sqrt [ 108 × 48 × 48 × 12 ]

A1 = sqrt [ 3 × 3 × 12 ×12 × 48 × 48 ]

= 3 × 12 × 48

A1 = 1728 sq m -----( 1 )

2 ) From triangle LNO ,

base = LN = 96 m

altitude = OP = 70 m

area of LNO = A2

A2 = ( base × altitude )/2

A2 = ( 96 × 70 )/2

A2 = 3360 sq m ------( 2 )

3 ) Total area of the plot = A1 + A2

= 1728 + 3360 [ from ( 1 ) and ( 2 ) ]

= 5088 sq m

I hope this helps you