Use Heron's formula to find the area of a triangle of lengths $4, 5$ and $6.$

\frac{15}{4} \sqrt{7}

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\frac{13}{4} \sqrt{7}

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4 \sqrt{7}

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3 \sqrt{7}

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Solution

The correct option is B $\frac{15}{4} \sqrt{7}$
Given: Lengths of a triangle $= 4,$ $5,$ and $6$

$S = \frac{4 + 5 + 6}{2} = \frac{15}{2}$ units

Using heron's formula to find the area of the triangle:

$A = \sqrt{s (s - a) (s - b) (s - c)}$

$= \sqrt{\frac{15}{2} (\frac{15}{2} - 4) (\frac{15}{2} - 5) (\frac{15}{2} - 6)}$

$= \sqrt{\frac{15}{2} \times \frac{7}{2} \times \frac{5}{2} \times \frac{3}{2}}$

$= \sqrt{\frac{1575}{16}}$

$= \frac{15}{4} \sqrt{7}$

Hence, area of triangle is $\frac{15}{4} \sqrt{7}$ sq. units.

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Similar questions

Q. Use Heron's formula to find the area of a triangle of lengths

7, 8

and

9

Verify the following :

(i) $\frac{3}{7} \times (\frac{5}{6} + \frac{12}{13}) = (\frac{3}{7} \times \frac{5}{6}) + (\frac{3}{7} \times \frac{12}{13})$
(ii) $\frac{- 15}{4} \times (\frac{3}{7} + \frac{- 12}{5}) = (\frac{- 15}{4} \times \frac{3}{7}) + (\frac{- 15}{4} \times \frac{- 12}{5})$
(iii) $(\frac{- 8}{3} + \frac{- 13}{12}) \times \frac{5}{6} = (\frac{- 8}{3} \times \frac{5}{6}) + (\frac{- 13}{12} \times \frac{5}{6})$
(iv) $\frac{- 16}{7} \times (\frac{- 8}{9} + \frac{- 7}{6}) = (\frac{- 16}{7} \times \frac{- 8}{9}) + (\frac{- 16}{7} \times \frac{- 7}{6})$