Question

Question:-

Find the area of triangle, two sides of which are 8cm and 11cm and the perimeter is 32cm (See the figure).

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Answer 1

Answer:

= 1144cm²

Step-by-step explanation:

I hope it's helpful.....

Answer 2

$\\\;\underbrace{\underline{\sf{Understanding\;the\;Concept\;:-}}}$

Here the Concept of Perimeter and Heron's Formula has been used . First we shall find the third side of the triangle using the Perimeter and then find its semi perimeter . After finding that we can use Heron's Formula to find the Area of Triangle .

Let's do it !!

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★ Formula Used :-

$\\\;\boxed{\sf{Sum\;of\;all\;Sides\;=\;\bf{Perimeter}}}$

$\\\;\boxed{\sf{Semi\:-\:Perimeter\;of\;\Delta\;=\;\bf{\dfrac{Sum\;of\;all\;sides}{2}}}}$

$\\\;\boxed{\sf{Area\;of\;\Delta\;=\;\bf{\sqrt{s(s\;-\;a)(s\;-\;b)(s\;-\;c)}}}}$

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★ Solution :-

Given,

» Perimeter of ∆ABC = 32 cm

» Length of AC = 11 cm

» Length of BC = 8 cm

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~ For the length of AB ::

We know that,

$\\\;\sf{\pink{:\rightarrow\;\;Sum\;of\;all\;Sides\;=\;\bf{Perimeter}}}$

By applying values, we get,

$\\\;\sf{:\Longrightarrow\;\;AB\;+\;BC\;+\;AC\;=\;\bf{32}}$

$\\\;\sf{:\Longrightarrow\;\;AB\;+\;8\;+\;11\;=\;\bf{32}}$

$\\\;\sf{:\Longrightarrow\;\;AB\;+\;19\;=\;\bf{32}}$

$\\\;\sf{:\Longrightarrow\;\;AB\;=\;\bf{32\;-\;19}}$

$\\\;\bf{:\Longrightarrow\;\;AB\;=\;\bf{\blue{13\;\:cm}}}$

$\\\;\underline{\boxed{\tt{Hence,\;\;length\;\;of\;\;AB\;=\;\bf{\blue{13\;\:cm}}}}}$

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★ For Area of the Triangle ::

Firstly we need to find the Semi Perimeter of Triangle that is s ::

$\\\;\sf{\pink{:\rightarrow\;\;Semi\:-\:Perimeter\;of\;\Delta,\;s\;=\;\bf{\dfrac{Sum\;of\;all\;sides}{2}}}}$

By applying values, we get

$\\\;\sf{:\rightarrow\;\;s\;=\;\bf{\dfrac{AB\;+\;BC\;+\;AC}{2}}}$

$\\\;\sf{:\rightarrow\;\;s\;=\;\bf{\dfrac{Perimeter}{2}}}$

$\\\;\sf{:\rightarrow\;\;s\;=\;\bf{\dfrac{32}{2}}}$

$\\\;\sf{:\rightarrow\;\;s\;=\;\bf{\orange{16\;\:cm}}}$

Now using the Heron's Formula we get,

$\\\;\sf{\pink{:\rightarrow\;\;Area\;of\;\Delta\;=\;\bf{\sqrt{s(s\;-\;a)(s\;-\;b)(s\;-\;c)}}}}$

• Here a = AB = 13 cm

• Here b = BC = 8 cm

• Here c = AC = 11 cm

By applying values, we get,

$\\\;\sf{:\rightarrow\;\;Area\;of\;\Delta\;=\;\bf{\sqrt{s(s\;-\;a)(s\;-\;b)(s\;-\;c)}}}$

$\\\;\sf{:\rightarrow\;\;Area\;of\;\Delta\;=\;\bf{\sqrt{(16)(16\;-\;13)(16\;-\;8)(16\;-\;11)}}}$

$\\\;\sf{:\rightarrow\;\;Area\;of\;\Delta\;=\;\bf{\sqrt{1920}}}$

$\\\;\sf{:\rightarrow\;\;Area\;of\;\Delta\;=\;\bf{\purple{43.82\;\:cm^{2}}}}$

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

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★ More to know :-

$\\\;\sf{\leadsto\;\;Area\;of\;Square\;=\;(Side)^{2}}$

$\\\;\sf{\leadsto\;\;Area\;of\;Rectangle\;=\;Length\:\times\:Breadth}$

$\\\;\sf{\leadsto\;\;Area\;of\;Triangle\;=\;\dfrac{1}{2}\:\times\:Base\:\times\:Height}$

$\\\;\sf{\leadsto\;\;Area\;of\;Parallelogram\;=\;Base\:\times\:Height}$

$\\\;\sf{\leadsto\;\;Area\;of\;Trapezium\;=\;\dfrac{1}{2}\:\times\:(Sum\:of\;||^{e}\:sides)\:\times\:Height}$