Question

5. Find the area of a triangular field whose sides are 91 m, 98 m and 105
in length. Find the height corresponding to the longest side.​

Answer 1

Step-by-step explanation:

S=

2

91+98+105

=147

Δ=

s(s−a)(s−b)(s−c)

Δ=

147×56×49×42

Δ=4116m

2

Base=105m

Height =h

Δ=

2

1

× base× height

4116=

2

1

×105×h

h=78.4m

Answer 2

Answer:-

The height corresponding to the longest side is 39.2cm

Step-by-step explanation:-

To Find:-

• The height corresponding to the longest side

Solution:-

Here, a = 91 m ; b = 98m and 105m

$\setlength{\unitlength}{10mm}\begin{picture}(20,15)\thicklines\qbezier(1,1)(1,1)(5,1)\qbezier(1,1)(1,1)(3,4)\qbezier(5,1)( 5,1)(3,4)\put(2,0.3){\bf{105 m}}\put(4,3){ \bf{98 m}}\put( 1,3){ \bf{91 m}}\end{picture}$

So, Let's first find the Semi-perimeter:-

Semi-perimeter = ( 91 + 98 + 105 ) / 2

= 294 / 2

= 147cm

Now, Area, we know that,

$\sqrt{s(s - a)(s - b)(s - c)}$

By putting the values,

$\sqrt{147(147 - 91)(147 - 98)(147 - 102)}$

$= \sqrt{147(56)(49)(42)}$

$= 4116 {m}^{2}$

Now, the height of corresponding to largest side:-

= 4116 / 105

= 39.2m

Required Answer:-

The height of corresponding to largest side is 39.2m