Question

The area of an isosceles triangle having base 2 cm and length of the one of its two equal sides
is 4 cm.​

Answer 1

Answer:

I'm solving this, [by Pythagoras theorem].

In right angled △ADB,

AB² = AD² + BD²

(4)² = AD²+ 1

AD² = 16 - 1 = 15.

∴ AD

$= \: \sqrt{15cm}$

[taking positive square root because length is always positive]

∴ Area of △ABC = 1/2 × BC × AD. [∵ Area of traingle = 1/2 × base × height].

= 1/2 × 2 × √15 = 15cm²

HOPE MY ANSWER HELPFUL TO YOU ❣️❣️

Answer 2

Step-by-step explanation:

$For \: an \: isoseles \: ∆ABC-$

$Base(BC)=b=2cm$

$equal \: sides(AB = AC)=a= 4cm$

$Area \: of \: an \: isoseles \: ∆ABC-$

$=>\frac{b}{4}√(4a²-b²)$

$=>\frac{2}{4}√4(4)²-(2)²$

$=>\frac{1}{2}√(4³ - 4)$

$=>\frac{1}{2}√4(4²-1)$

$=>\frac{1}{2}*2*√15$

$=> √15 cm²$