Question

Question:-
Find the area of a triangle, two sides of which are 8cm and 11cm and the perimeter is 32cm​

Answer 1

A᭄ɴsᴡᴇʀ

A≈43.82cm²

Let a,b,c be the sides of the given triangle and 2s be its perimeter such that

a=8 cm, b=11 cm and 2s=32 cm

Now, a+b+c=2s

8+11+c=32

c=13

Therefore,

s−a=16−8=8,s−b=16−11=5,s−c=16−13=3

Hence, area of given triangle =

√s(s−a)(s−b)(s−c)

= √16×8×5×3

=8 √30cm²

Answer 2

$\huge{\underline{\mathtt{\red{Ꭺ}\red{ղ}\green{Տ}\blue{ω}\purple{Ꭼ}\orange{я᭄}}}}$

Concept:-

Here the concept of Heron's Formula and Perimeter of Triangle has been used. We see that we are given the perimeter of triangle and two of its side. Then firstly we can find the third side of the triangle. Then we can find the semi - perimeter of triangle. After that using Heron's Formula, we can find the area of triangle.

Let's do it!!

★FormulaUsed:-

$\begin{gathered}\\\;\boxed{\sf{\purple{Perimeter\;of\;Triangle\;=\;\bf{Sum\;of\;all\;sides}}}}\end{gathered}$

$\begin{gathered}\\\;\boxed{\sf{\pink{Semi-Perimeter\;of\;Triangle\;=\;\bf{\dfrac{Perimeter}{2}}}}}\end{gathered}$

$\begin{gathered}\\\;\boxed{\sf{\purple{Area\;of\;Triangle\;=\;\bf{\sqrt{s(s\:-\:a)(s\:-\:b)(s\:-\:c)}}}}}\end{gathered}$

Solution:-

Given,

Given,» First side of Triangle = a = 8 cm

Given,» First side of Triangle = a = 8 cm» Second side of Triangle = b =11 cm

Given,» First side of Triangle = a = 8 cm» Second side of Triangle = b =11 cm» Perimeter of the Triangle = 32 cm

• Given,» First side of Triangle = a = 8 cm» Second side of Triangle = b =11 cm» Perimeter of the Triangle = 32 cmLet the third side of the triangle be c
• Let the semi - perimeter of the triangle be s

~for the value of c::

We know that,

$\large\bf{\red{\begin{gathered}\\\;\sf{\rightarrow\;\;Perimeter\;of\;Triangle\;=\;\bf{Sum\;of\;all\;sides}}\end{gathered}} }$

By applying values, we get

$\begin{gathered}\\\;\sf{\rightarrow\;\;32\;=\;\bf{a\:+\:b\:+\:c}}\end{gathered}$

$\begin{gathered}\\\;\sf{\rightarrow\;\;32\;=\;\bf{8\:+\:11\:+\:c}}\end{gathered}$

$\begin{gathered}\\\;\sf{\rightarrow\;\;32\;=\;\bf{19\:+\:c}}\end{gathered}$

$\begin{gathered}\\\;\sf{\rightarrow\;\;c\;=\;\bf{32\:-\:19}}\end{gathered}$

$\begin{gathered}\\\;\bf{\rightarrow\;\;c\;=\;\bf{\orange{13\;\;cm}}}\end{gathered}$

Hence, third side of triangle = 13 cm

~ For the value of semi perimeter of triangle ::

We know that,

$\large\bf{\green{\begin{gathered}\\\;\sf{\rightarrow\;\;Semi-Perimeter\;of\;Triangle\;=\;\bf{\dfrac{Perimeter}{2}}}\end{gathered}} }$

By Applying the value we get

$\begin{gathered}\\\;\sf{\rightarrow\;\;s\;=\;\bf{\dfrac{32}{2}}}\end{gathered}$

$\begin{gathered}\\\;\bf{\rightarrow\;\;s\;=\;\bf{\red{16\;\;cm}}}\end{gathered}$

-----------------------------------------------------------

~ For the Area of the Triangle ::

We know that,

$\large\bf{\red{\begin{gathered}\\\;\sf{\Longrightarrow\;\;Area\;of\;Triangle\;=\;\bf{\sqrt{s(s\:-\:a)(s\:-\:b)(s\:-\:c)}}}\end{gathered}} }$

By applying the values, we get

$\large\bf{\green{\begin{gathered}\\\;\sf{\Longrightarrow\;\;Area\;of\;Triangle\;=\;\bf{\sqrt{(16)(16\:-\:8)(16\:-\:11)(16\:-\:13)}}}\end{gathered}} }$

$\large\bf{\red{\begin{gathered}\\\;\sf{\Longrightarrow\;\;Area\;of\;Triangle\;=\;\bf{\sqrt{(16)(8)(5)(3)}}}\end{gathered}} }$

$\large\bf{\green{\begin{gathered}\\\;\sf{\Longrightarrow\;\;Area\;of\;Triangle\;=\;\bf{\sqrt{1920}}}\end{gathered}} }$

$\large\bf{\red{\begin{gathered}\\\;\sf{\Longrightarrow\;\;Area\;of\;Triangle\;=\;\bf{\sqrt{2^{7}\:\times\:3\:\times\:5}}}\end{gathered}} }$

$\large\bf{\green{\begin{gathered}\\\;\sf{\Longrightarrow\;\;Area\;of\;Triangle\;=\;\bf{2^{3}\sqrt{2\:\times\:3\:\times\:5}}}\end{gathered}} }$

$\large\bf{\red{\begin{gathered}\\\;\sf{\Longrightarrow\;\;Area\;of\;Triangle\;=\;\bf{8\sqrt{30}}}\end{gathered}} }$

Since, √30 = 5.477

$\large\bf{\pink{\begin{gathered}\\\;\sf{\Longrightarrow\;\;Area\;of\;Triangle\;=\;\bf{8(5.477)}}\end{gathered}} }$

$\large\bf{\blue{\begin{gathered}\\\;\sf{\Longrightarrow\;\;Area\;of\;Triangle\;=\;\bf{43.817\;\;cm^{2}}}\end{gathered}} }$

Here the value is in decimals so we can take round off to two decimal points.

$\begin{gathered}\\\;\sf{\Longrightarrow\;\;Area\;of\;Triangle\;=\;\bf{43.817\;\;cm^{2}\;\approx\;43.82\;\;cm^{2}}}\end{gathered}$

$\begin{gathered}\\\;\bf{\Longrightarrow\;\;Area\;of\;Triangle\;=\;\bf{\orange{43.82\;\;cm^{2}}}}\end{gathered}$

$\begin{gathered}\\\;\underline{\boxed{\tt{Area\;of\;Triangle\;=\;\bf{\purple{43.82\;\;cm^{2}}}}}}\end{gathered}$