Question

find the area of a triangle whose sides are 13cm ,14cm and 15cm. Also find the height of the triangle, corresponding to the longest side​

Answer 1

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Let the sides of the given triangle be a, b, c respectively.

So given,

a = 13 cm

b = 14 cm

c = 15 cm

★By using Heron's Formula

the area of triangle is,

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

Semi perimeter of a triangle = 2s

$2s = a + b + c$

$s = \frac{a + b + c}{2}$

$s = \frac{13+ 14+ 15}{2}$

$s = 21 \: cm$

$\large \bold{\underline{\sf{\blue{Therefore,}}}}$

the area of the triangle is,

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

$= \sqrt{21(21 - 13)(21 - 14)(21 - 15)}$

$= 84 \: {cm}^{2}$

hight corresponding to longest side,

$= \frac{2 \times area}{base}$

$= \frac{2 \times 84}{15}$

$= \frac{168}{15}$

$= 11.2$

so, it's area us 84 cm² and the height of the triangle, corresponding to the longest side is 11.2 cm.

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Answer 2

Answer:

A=hb/2

hb= 2 A/B

the height of the triangle is 13

Step-by-step explanation:

hope its helpful