Question

The lengths of the side of a triangle are 5cm 12cm and 13cm. Find the length of perpendicular from the opposite vertex to the side whose length is 13cm. By herons formulae

Answer 1

$\large\bf\underline{Given:-}$

• Length of side's of triangle are 5cm , 12 cm and 13 cm.

$\large\bf\underline {To \: find:-}$

• length of perpendicular

$\huge\bf\underline{Solution:-}$

Sides of triangle are :- 5cm , 12 cm ,and 13cm

$\begin{cases} \tt \: let \: \bullet a = 5 \: cm \\\tt \: \: \bullet \: b = 12 \: cm \\\tt \: \: \bullet \: c = 13 \: cm\end{cases}$

$\dag \: \large \bf \: s = \frac{a + b + c}{2}$

$\tt \rightarrowtail \: s = \frac{5 + 12 + 13}{2} \\ \\ \tt \rightarrowtail \: s = \cancel \frac{30}{2} \\ \\ \tt \rightarrowtail \: s = 15$

✝️ Heron's Formula :-

$\bf area \: of \: triangle = \sqrt{s(s - a)(s - b)(s - c)}$

Area of triangle= √15(15-5)(15-12)(15-13)

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀»» √ 15(10)×(3)×(2)

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀»» √ 15 × 60

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀»» √ 900

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀»» 30 cm²

✝️Area of triangle = ½× base × height

• let the perpendicular be p

»» 30 = ½ × 13 × p

»» 30 × 2 = 13p

»» 60 = 13p

»» p = 60/13

• ≫ p = 4.6 cm

So,

length of the perpendicular is 4.6 cm.

Answer 2

Answer:

$\sf{The \ length \ of \ perpendicular \ from}$

$\sf{opposite \ vertex \ to \ the \ side \ whose \ length}$

$\sf{is \ 13 \ cm \ is \ 4.61 \ cm}$

Given:

• The lengths of the side of a triangle are
• 5 cm, 12 cm and 13 cm

To find:

• The length of perpendicular from the opposite vertex to the side whose length is 13 cm

Solution:

$\sf{By \ heron's \ formula}$

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

$\sf{\therefore{s=\frac{5+12+13}{2}}}$

$\sf{\therefore{s=15}}$

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

$\sf{=\sqrt{15(15-5)(15-12)(15-13)}}$

$\sf{=\sqrt{15\times10\times3\times3}}$

$\sf{=\sqrt{30\times30}}$

$\sf{\therefore{Area \ of \ triangle=30 \ cm^{2}}}$

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

$\sf{\therefore{30=\frac{1}{2}\times13\times \ h}}$

$\sf{\therefore{h=\frac{30\times2}{13}}}$

$\sf{\therefore{h=4.61 \ cm(approx)}}$

$\sf\purple{\tt{\therefore{The \ length \ of \ perpendicular \ from}}}$

$\sf\purple{\tt{opposite \ vertex \ to \ the \ side \ whose \ length}}$

$\sf\purple{\tt{is \ 13 \ cm \ is \ 4.61 \ cm}}$