Question

The area of an isosceles triangle having base 2cm and the length of equal side 4cm is=?

Answer 1

Heya !!!




Base( B) = 2 cm



And,



Two equal sides (A) = 4 cm



Therefore,



Area of an isosceles triangle = 1/4 × B✓4A² -B²





=> 1/4 × 2✓ 4 × (4)² - (2)²




=> 1/4 × 2✓4 × 16 - 4




=> 1/4 × 2✓64 -4





=> 1/4 × 2✓60




=> 14 × 2× 7.74





=> 1/2 × 7.74






=> 3.87 cm².


HOPE IT WILL HELP YOU..... :)

Answer 2

We know, Area of isosceles triangle = $\frac{1}{4}b \sqrt{4a^2 - b^2}$

a = equal side 
   = 4 cm

b = base 
   = 2 cm
===================================

Area = $\frac{2}{4} \sqrt{4(4)^2 - (2)^2}$

        = $\frac{1}{2} \sqrt{4(16) - 4}$

        = $\frac{1}{2} \sqrt{64 - 4}$

        = $\frac{ \sqrt{60} }{2}$

        = $\frac{2 \sqrt{15} }{2}$

        = √15 cm²




i hope this will help you

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