Monday, December 23, 2019

If the sides of a triangle are 5 cm, 4 cm and $\sqrt{41}$ cm, then the area then he take the area of triangle is

A20 $cm^2$

B$(5+4+\sqrt{41}) cm^2$

C$\frac{(5+4+\sqrt{41})}{3} cm^2$

D$10 cm^2$

Your solution

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