Question

find the area of a triangle field whose sides are 30m, 40m and 50m. also, find the length of altitude drawn on the longest side.​

Answer 1

Given :-

Sides:-

• 30 m
• 40 m
• 50 m

To find :-

• Area of triangle.
• Length of altitude drawn to the largest side.

Solution :-

By using heron's formula we will find the area of triangle.

$\sf\boxed{Heron's~formula~=~\sqrt{s(s-a)(s-b)(s-c)}}$

To find s :-

$\sf{s = \dfrac{a + b + c}{2} }$

$\sf{ s = \dfrac{30 + 40 + 50}{2}}$

$\sf{s = \dfrac{120}{2}}$

$\sf{s = 60}$

$\sf{ \sqrt{60(60 - 30)(60 - 40)(60 - 50)}} \\ = \sf{\sqrt {60(30)(20)(10)}} \\ = \sf{\sqrt{360000}} \\ \sf{ = 600 \: {m}^{2}}$

• Area of triangle = 600 metre square.

2. Length of altitude drawn to the largest side.

Largest side = AC = 50 m

As we know,

$\sf\boxed{Area~of~triangle~=~\dfrac{1}{2} \times base \times height}$

Put the given values.

$\sf{600 = \dfrac{1}{2} \times 50 \times h}$

$\sf{= \dfrac{600 \times 2}{50}}$

Height = 24 m

• Length of altitude drawn to the largest side = 24 m

Figure :-

$\setlength{\unitlength}{1 cm}\begin{picture}(0,0)\linethickness{0.7mm}\qbezier(1, 0)(1,0)( 3,3)\qbezier(5,0)(5, 0)(3,3)\qbezier(5,0)(1,0)(1,0)\qbezier(1,0)(1,0)(4,1.5)\put(2.8,3.2){ \sf A }\put(2.5, - 0.3){40 m}}\put(0.5,-0.2){ \sf B }\put(4.5,1){50 m}\put(5.1, - 0.3){ \sf C }\put(1.1,1.8){30 m}\put(4.2, 1.5){ \sf D }\end{picture}$

Answer 2

Given :-

• Sides:-

• 30 m

• 40 m

• 50 m

To find :-

• Area of triangle.

• Length of altitude drawn to the largest side.

Solution :-

Figure :-

$\setlength{\unitlength}{1 cm}\begin{picture}(0,0)\linethickness{0.7mm}\qbezier(1, 0)(1,0)( 3,3)\qbezier(5,0)(5, 0)(3,3)\qbezier(5,0)(1,0)(1,0)\qbezier(1,0)(1,0)(4,1.5)\put(2.8,3.2){ \sf A }\put(2.5, - 0.3){40 m}}\put(0.5,-0.2){ \sf B }\put(4.5,1){50 m}\put(5.1, - 0.3){ \sf C }\put(1.1,1.8){30 m}\put(4.2, 1.5){ \sf D }\end{picture}$

• By using heron's formula we will find the area of triangle.

• $\sf\boxed{Heron's~formula~=~\sqrt{s(s-a)(s-b)(s-c)}}$

• To find s :-

$\sf{s = \dfrac{a + b + c}{2} }$

$\sf{ s = \dfrac{30 + 40 + 50}{2}}$

$\sf{s = \dfrac{120}{2}}$

$\sf{s = 60}$

$\begin{gathered}\sf{ \sqrt{60(60 - 30)(60 - 40)(60 - 50)}} \\ = \sf{\sqrt {60(30)(20)(10)}} \\ = \sf{\sqrt{360000}} \\ \sf{ = 600 \: {m}^{2}}\end{gathered}$

• Area of triangle = 600 metre square.

2. Length of altitude drawn to the largest side.

Largest side = AC = 50 m

• As we know,

$\sf\boxed{Area~of~triangle~=~\dfrac{1}{2} \times base \times height}$

• Put the given values.

$\sf{600 = \dfrac{1}{2} \times 50 \times h}$

$\sf{= \dfrac{600 \times 2}{50}}$

• Height = 24 m

• Length of altitude drawn to the largest side = 24 m