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

\sqrt{15} c m^{2}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\sqrt{\frac{15}{2}} c m^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2 \sqrt{15} c m^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

4 \sqrt{15} c m^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $\sqrt{15} c m^{2}$
$s = \frac{4 + 4 + 2}{2} = 5$

Area of the triangle
$Δ = \sqrt{s (s - a) (s - b) (s - c)} = \sqrt{5 (5 - 4) (5 - 4) (5 - 2)} = \sqrt{15} c m^{2}$

Suggest Corrections

Similar questions

If AD is the median of a triangle ABC and area of triangle $A D C = 15 c m^{2}$ , then $\frac{a r (Δ A B C)}{15} =$ ____.

Find the area of an equilateral triangle whose area is equal to the triangle of sides 12 cm, 8 cm and 16 cm.

Find the area of an isosceles triangle, whose unequal side is 12 cm and the perimeter of triangle, is 60 cm.

Join BYJU'S Learning Program