Question

Find the area of an isosceles triangle having the base 2cm and length of one of the equal sides 4cm

Answer 1

$\pink \bigstar\sf\large \red{\underbrace\frak{ \blue{ \underline{ \purple{Given }}}} }$

• Isosceles triangle having the base = 2cm and
• Length of the one equal sides = 4cm

$\pink \bigstar\sf\large \red{\underbrace\frak{ \blue{ \underline{ \purple{To \: find }}}} }$

• Area of isosceles triangle

$\pink \bigstar\sf\large \red{\underbrace\frak{ \blue{ \underline{ \purple{need \: formula }}}} }$

• Pythagoras Theorem
• $\sf \: area \: of \: tringle \: = \frac{1}{2}(base \: \times height)$
• Alternate method

$\pink \bigstar\sf\large \red{\underbrace\frak{ \blue{ \underline{ \purple{Solution }}}} }$

Let ABC be an isosceles triangle in which

• AB = AC = 4cm ( in isosceles triangle two side are equal so, AB = AC)
• BC = 2cm

In right angle triangle ADB,

$\: \sf\longmapsto \qquad \: AB^2 = AD^2+ BD^2 \qquad\qquad( By Pythagoras Theorem) \\ \\ \sf\longmapsto \qquad \: (4)^2 = AD^2+ (1)^2 \\ \\ \: \sf\longmapsto \qquad \: 16 = AD^2+ 1 \\ \\ \: \sf\longmapsto \qquad \: AD^2 = 16 - 1 \\ \\ \sf\longmapsto \qquad \: AD^2 =15 \\ \\ \sf\longmapsto \qquad \red{\: AD = \sqrt{15} \: cm }$

Taking positive square root because length is always positive

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Now we find area of triangle

Formula $\sf \: area \: of \: triangle \: = \frac{1}{2} \times (base \times height)$

Putting values in formals

$\sf \: area \: of \: triangle \: = \frac{1}{2} \times AC × BC \ \: \\ \\ \sf \: area \: of \: triangle \: = \frac{1}{ \cancel2} \times \cancel{2} \times \: \sqrt{15} \\ \\\sf \: area \: of \: triangle \: = 1 \times \sqrt{15} \\ \\ \sf \: area \: of \: triangle \: = \sqrt{15} \: cm$

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Alternate method

We know that

Area of an isosceles triangle = $\sf \: \frac{a}{4} \sqrt{4 {b}^{2} - {a}^{2} }$

Where, b the length of equal and a is the length of the base

Hence, the length of the side be b = 4cm and a = 2cm

$\sf\longmapsto Area \: of \: isosceles \: triangle = \frac{\cancel2 \sqrt{4 {(4)}^{2} - 4} }{\cancel4} \\ \\ \sf\longmapsto Area \: of \: isosceles \: triangle = \frac{ \sqrt{64 - 4} }{2} \\ \\ \sf\longmapsto Area \: of \: isosceles \: triangle = \frac{ \sqrt{60} }{2} \\ \\ \sf\longmapsto Area \: of \: isosceles \: triangle = \frac{\cancel 2 \sqrt{15} } { \cancel2} \\ \\ \sf\longmapsto Area \: of \: isosceles \: triangle = \sqrt{15} \: cm {}^{2}$

Therefore, the area isosceles triangle be √15 cm²

Answer 2

Answer:

Answer is square root 15...

Step-by-step explanation:

semiperimeter=(4+4+2)/2=5cm

root 5(5-4)(5-4)(5-2)

root 5*1*1*3

root 5*3

root 15 square cm...

Thank you