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  • The area of an isosceles triangle having base 2 cm and the length of one of the equal sides 4 cm, is (A) sqrt{15}cm^{2} (B) sqrt{frac{15}{2}}cm^{2} (C) 2sqrt{15}cm^{2} (D) 4sqrt{15}cm^{2}

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

(A) \sqrt{15}cm^{2}

(B) \sqrt{\frac{15}{2}}cm^{2}

(C) 2\sqrt{15}cm^{2}

(D) 4\sqrt{15}cm^{2}

Answers (1)

[A]

                                  

We\; know\; that,We\; know\; that,semi-perimeter

S= \frac{a+b+c}{2}

S= \frac{2+4+4}{2}

S= \frac{10}{2}= 2cm

Using Heron’s formula area of \DeltaABC = \sqrt{S\left ( S-a \right )\left ( S-b \right )\left ( S-c \right )}

= \sqrt{5\left ( 5-2 \right )\left ( 5-4 \right )\left ( 5-4 \right )}

= \sqrt{5\times 3\times 1\times 1}

= \sqrt{15}cm^{2}

Hence, area of given triangle is \sqrt{15}cm^{2}.

Hence option (A) is correct.

Posted by

infoexpert21

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