the length of the side of a triangle are 4cm ,6cm and 8cm the length of the perp...

mathdude500

Question :-

The length of the side of a triangle are 4cm ,6cm and 8cm. The length of the perpendicular from opposite vertex whose length are 8 cm is equal to ............

Answer

Given :-

A triangle having sides 4 cm, 6 cm, and 8 cm.

To Find :-

The length of the perpendicular from opposite vertex whose length are 8 cm.

Formula used :-

$\underline{\boxed{\bf Perimeter \ of \ a \ triangle=a+b+c}}$

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

where,

s = Semi perimeter of triangle

$\boxed{\bf \:Area \: of \: triangle = \dfrac{1}{2} \times base \times height}$

Solution :-

Let the sides be represented as

a = 4 cm

b = 6 cm

c = 8 cm

Semi perimeter,

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

$\bf\implies \:s = \dfrac{4 + 6 + 8}{2}$

$\bf\implies \:s = \dfrac{18}{2} = 9 \: cm$

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

On substituting the values of s, a, b, c, we get

$\bf Area \ of \ triangle= \sqrt{9(9 - 4)(9 - 6)(9 - 8)}$

$\bf Area \ of \ triangle= \sqrt{9 \times 5 \times 3 \times 1}$

$\bf Area \ of \ triangle=3 \sqrt{15} \: {cm}^{2}$

Let the length of the perpendicular from opposite vertex whose length are 8 cm be 'h' cm.

$\boxed{\bf \:Area \: of \: triangle = \dfrac{1}{2} \times base \times height}$

$\bf\implies \:3 \sqrt{15} = \dfrac{1}{2} \times 8 \times h$

$\bf\implies \:3 \sqrt{15} = 4h$

$\bf\implies \:h = \dfrac{3}{4} \sqrt{15} \: cm$

_________________________________________

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ak79392336245 sir please see my first question

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paragtiwari You need help

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