Question

if each side of triangle is doubled the find the ratio of area of new traingle then formed and the given triangle​

Answer 1

AnswEr :

❍ Let a, b and c are the sides of the triangle.

We'll use Heron's formula to find out the area of the triangle.

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S E M I – P E R I M E T E R :

$\bf{\dag}\;\boxed{\sf{Semi \; Perimeter = \bigg(\dfrac{Sum\;of\;all\;sides}{2}\bigg)}}$

◗ As Per given Condition, when the sides of the triangle is doubled then Let the semi perimeter of the new Triangle formed be s'. Therefore :

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

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A R E A :

(I) Area of old triangle is :

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$\sf A' = \sqrt{s(s - a)(s - b)(s - c)}$

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(II) Calculating Area of the triangle new formed :

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$\bf{\dag}\:\boxed{\sf{Area_{\triangle} = \sqrt{s'(s' - a)(s' - b)(s' - c)}}}$

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$\longrightarrow\sf Area_{\triangle} = \sqrt{s'(s' - 2a)(s' - 2b)(s' - 2c)} \\\\\\\longrightarrow\sf Area_{\triangle} = \sqrt{2s(2s - 2a)(2s - 2b) (2s - 2c)} \\\\\\\longrightarrow\sf Area_{\triangle} = \sqrt{16s(s - a)(s - b)(s - c)} \\\\\\\longrightarrow\sf Area_{\triangle} = \sqrt{4 \times 4s(s - a) (s - b) (s - c)} \\\\\\\longrightarrow\sf Area_{\triangle} = 4\sqrt{s(s - a)(s - b)(s - c)} \\\\\\\longrightarrow{\underline{\boxed{\sf Area_{\triangle} = 4 \times A'}}}$

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$\therefore{\underline{\textsf{Hence,\; ratio\;of\;area\;of\; the\;new\;triangle\;formed\;is\; \textbf{4:1}.}}}$

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Answer 2

Step-by-step explanation:

$\huge\sf\red{Answér}$

$Semi \: perimeter \: of \: the \: triangle = \frac{a + b + c}{2}$

$Area \: of \: the \: traiangle = \: \sqrt{s(s - a)(s - b)(s - c)}$

$Sides \: of \: the \: new \: triangle \: will \: be \: 2a,2b,2c$

$Semi \: primeter \: of \: new \: traiangle \: = a + b + c = 2s$

$Area \: of \: new \: triangle \: = \sqrt{2s(2s - 2a)(2s - 2b)(2s - 2c)}$

$= 4 \sqrt{s(s - a)(s - b)(s - c) = 4 \times area \: of \: the \: triangle}$

Hence Ratio is 4:1

$\small\bf\orange{Answered\:by\: ITZHEART29}$