Question

the sides of a triangle are 35,54 and 61 CM.the length of its longest altitude is. ​

Answer 1

answer: Let a = 35 cm, b = 54 cm, c = 61 cm and h the longest altitude of the triangle. Semi-perimeter of the given triangle = 35 + 54 + 61 2 cm = cm = 75 cm. = 420 5 c m 2 . The longest altitude is on the smallest side in length.

Answer 2

Answer:

Given :

• ➞ The sides of a triangle are 35cm ,54cm and 61cm.

$\begin{gathered}\end{gathered}$

To Find :

• ➞ Length of longest altitude

$\begin{gathered}\end{gathered}$

Using Formulas :

${\longrightarrow{\small{\underline{\boxed{\sf{Semi \: Perimeter = \dfrac{1}{2} \bigg(a + b + c \bigg)}}}}}}$

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

${\longrightarrow{ \small{\underline{\boxed{\sf{ Area \: of \: \triangle = \dfrac{1}{2} \times b \times h}}}}}}$

• ➟ a, b, c = sides of triangle
• ➟ s = semi perimeter
• ➟ b = breadth
• ➟ h = height

$\begin{gathered}\end{gathered}$

Solution :

Finding the area of triangle by heron's formula :

${\longrightarrow \: \: {\sf{Semi \: Perimeter = \dfrac{1}{2} \bigg(a + b + c \bigg)}}}$

${\longrightarrow \: \: {\sf{Semi \: Perimeter = \dfrac{1}{2} \bigg(35 + 54 + 61 \bigg)}}}$

${\longrightarrow \: \: {\sf{Semi \: Perimeter = \dfrac{1}{2} \bigg(150 \bigg)}}}$

${\longrightarrow \: \: {\sf{Semi \: Perimeter = \dfrac{1}{2} \times 150 }}}$

${\longrightarrow \: \: {\sf{Semi \: Perimeter = \dfrac{150}{2}}}}$

${\longrightarrow \: \: {\sf{Semi \: Perimeter = \cancel{\dfrac{150}{2}}}}}$

${\longrightarrow \: \: {\sf{Semi \: Perimeter = 75 \: cm}}}$

${\bigstar \: {\underline{\boxed{\sf{\red{Semi \: Perimeter = 75 \: cm}}}}}}$

Hence, the semi perimeter of triangle is 75 cm.

$\rule{300}{1.5}$

Now,

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

${\longrightarrow \: \: {\sf{ Area \: of \: \triangle=\sqrt{75(75-35)(75 - 54)(75 - 61)}}}}$

${\longrightarrow \: \: {\sf{ Area \: of \: \triangle=\sqrt{75(40)(21)(14)}}}}$

${\longrightarrow \: \: {\sf{ Area \: of \: \triangle=\sqrt{75 \times 40\times 21 \times 14}}}}$

${\longrightarrow \: \: {\sf{ Area \: of \: \triangle=\sqrt{882000}}}}$

${\longrightarrow \: \: {\sf{ Area \: of \: \triangle \approx 939.14 \: {cm}^{2}}}}$

${\bigstar\: {\underline{\boxed{\sf{\red{ Area \: of \: \triangle\approx 939.14 \: {cm}^{2} }}}}}}$

Hence, the area of triangle is 939.14 cm².

$\rule{300}{1.5}$

Now, finding the lenght of longest altitude :

• base = 35 cm (smallest side)
• height = lenght of longest altitude

${\longrightarrow \: \: {\sf{ Area \: of \: \triangle = \dfrac{1}{2} \times b \times h}}}$

${\longrightarrow \: \: {\sf{939.14= \dfrac{1}{2} \times 35 \times h}}}$

${\longrightarrow \: \: {\sf{939.14= \dfrac{35}{2}\times h}}}$

${\longrightarrow \: \: {\sf{939.14= \cancel{\dfrac{35}{2}}\times h}}}$

${\longrightarrow \: \: {\sf{939.14=17.5\times h}}}$

${\longrightarrow \: \: {\sf{h = 939.14 \div 17.5}}}$

${\longrightarrow \: \: {\sf{h = \dfrac{939.14}{17.5}}}}$

${\longrightarrow \: \: {\sf{h = \cancel{\dfrac{939.14}{17.5}}}}}$

${\longrightarrow \: \: {\sf{h \approx 53.67 \: cm}}}$

${\bigstar \: {\underline{\boxed{\sf{\red{Lenght \: of \: altitude \approx 53.67\: cm}}}}}}$

Hence, the lenght of longest altitude is 53.67 cm.

$\begin{gathered}\end{gathered}$

Learn More :

$\begin{gathered}\begin{gathered}\boxed{\begin{minipage}{9cm}\\ \dag\quad \Large\underline{\bf Formulas\:of\:Triangle:-}\\ \\ \star\sf Triangle \:area = \dfrac{1}{2}\times b \times h\\ \\ \star\sf Triangle \: perimeter=a+b+c\\\\ \star\sf Scalene\:\triangle=\sqrt{s (s-a)(s-b)(s-c)}\\\\\star\sf Equilateral\: \triangle\:area = \dfrac{\sqrt{3}}{4}\times{side}^{2}\\\\\star\sf Equilateral \:\triangle\:perimeter = 3 \times side\\\\\star\sf Isosceles\: \triangle\:area= \dfrac{3}{4}\sqrt{{4b}^{2}-{a}^{2}}\\\\\star\sf Isosceles\:\triangle\:perimeter=a+2b\end{minipage}}\end{gathered}\end{gathered}$

${\rule{220pt}{3pt}}$