Question

the perimeter of a triangular field is 240 decimeter if two sides of its sides are 78 and 50 decimeter find the length of the perpendicular on the side of length 50 cm from the opposite vertex​

Answer 1

Given :-

Perimeter of a rectangle = 240 dm

The two sides of the rectangular field = 78 dm and 50 dm

To Find :-

The length of the perpendicular on the side of length 50 cm from the opposite vertex​

Analysis :-

Subtract the sum of the two sides from the perimeter.

Then you'll get the 3 sides, once you get them divide the sum of those three sides by 2.

Then using the Heron's formula, find the area.

Use the formula of area of triangle, substitute the value we have got and find the height accordingly.

Solution :-

We know that,

• l = Length
• h = Height
• dm = Decimeter

Given that,

Perimeter = 240 dm

Two sides = 78 dm and 50 dm

According to the question,

Third side of the triangle = $\sf 240+(78+50)$

$\sf =240-128$

$\sf =112 \ dm$

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

Substituting the sides we got,

$\sf Side=\dfrac{78+50+112}{2}$

$\sf Side=\dfrac{240}{2}$

$\sf Side=120 \ dm$

$\underline{\boxed{\sf Area \ of \ the \ triangle = \sqrt{s(s-a)(s-b)(s-c)} }}$

Substituting their values, we get

$\implies \sf \sqrt{120(120-50)(120-78)(120-112)}$

$\implies \sf \sqrt{120 \times 70 \times 42 \times 8}$

$\implies \sf \sqrt{2822400}$

$\implies \sf 1680 \ dm^{2}$

Area of the triangle = 1680 dm²

Now,

$\underline{\boxed{\sf Area \ of \ triangle=\dfrac{1}{2}\times Breadth \times Height }}$

$\therefore \: \underline{\boxed{\sf Height= \dfrac{2 \times Area}{Base} }}$

Substituting these values in the formula,

$\sf Height=\dfrac{2 \times 1680}{50}$

$\sf Height =\dfrac{336}{5}$

$\sf Height \implies 67.2 \ dm$

Therefore, the height is 67.2 dm

Answer 2

Answer:-

Given:

• Perimeter of the triangle = 240 dm

• Two of its sides are 50 dm and 78 dm

Solution:

Third side of the triangle = 240-(78+50)

= 240 dm - 128 dm

= 112 dm

now,

$\bf \large \: s \: = \frac{(a + b + c)}{2} \\ \\ \bf \large \: s \: = \frac{(78 + 50 + 112)}{2} \\ \\ \bf \large \: s \: = \frac{240}{2} \\ \\ \bf \large \: s \: = 120 \: dm$

$\bf \large \: Area \: \: of \: \: the \: \: triangle \: \: \: = \sqrt{s(s - a)(s - b)(s - c)} \\ \\\bf \large \implies \: \sqrt{120(120 - 50)(120 - 78)(120 - 112)} \\ \\\bf \large \implies \: \sqrt{120 \times 70 \times 42 \times 8} \\ \\\bf \large \implies \: \sqrt{2822400} \\ \\\bf \large \implies \:1680 \: \: sq \: \: dm. \\ \\ \bf \large \: Area \: \: of \: \: the \: \: triangle \: \: \: =1680 \: \: sq \: \: dm.$

Then,

$\bf \large \: Area \: \: of \: \: the \: \: triangle \: \: \: = \frac{1}{2} bh$

$\bf\large\implies \: H \: \: = \: \: \frac{2 \times area}{b} \\ \\\bf\large\implies \: H \: \: = \: \: \frac{2 \times 1680}{50} \\ \\ \bf\large\implies \: H \: \: = \: \: \frac{336}{5} \\ \\ \bf\large\implies \: H \: \: = \: \:67.2 \: \: dm$