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Heron’s Formula — Maths STD 9 — Question

Gujarat BoardEnglish MediumSTD 9MathsHeron’s Formula1 Mark
Question
Write True or False and justify your answer: The area of the isosceles triangle is $\frac{5}{4}\sqrt{11}\text{cm}^2,$ if the perimeter is $11\ cm$ and the base is $5\ cm.$
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Answer

Let equal of an isosceles triangle be b. $\therefore$ Perimeter of a triangle, $2\text{s}=\text{b}+\text{b}+5$
$\big[\because\ 2\text{s}=\text{a}+\text{b}+\text{c}\big]$
$\therefore\ 11=2\text{b}+5$
$\Rightarrow2\text{b}=11-5$
$\Rightarrow2\text{b}=6$
$\Rightarrow\text{b}=\frac{6}{2}=3\text{cm}$ We know that,
area of an isosceles triangle,
$=\frac{\text{a}}{4}\sqrt{4\text{b}^2-\text{a}^2}$
Here, sides of triangle are $a = 5cm$ and $b = 3cm$
$\therefore$ Area of an isosceles triangle $=\frac{5\sqrt{4(3)^2-(5)^2}}{4}$
$=\frac{5\sqrt{4\times9-25}}{4}$
$=5\frac{\sqrt{36-25}}{4}$
$=\frac{5\sqrt{11}}{4}\text{cm}^2$

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