Question

in a triangle ABC a b is equal to 15 CM BC is equal to 13 cm and ac is equal to 14 centimetre find the area of triangle ABC and hence its altitude on AC

Answer 1

Answer:

hope this answer helps you

Answer 2

Answer:

$Area\: of \: the \: triangle = 84 \: cm^{2}\\Altitude \:on \: AC = 12 \: cm$

Step-by-step explanation:

In ∆ABC , AB = c = 15 cm,

BC = a = 13 cm ,

AC = b = 14 cm,

$s =\frac{a+b+c}{2}\\=\frac{13+14+15}{2}\\=\frac{42}{2}\\=21$

$s-a = 21-13= 8,\\s-b=21-14=7,\\s-c=21-15=6$

By Heron's Formula:

$i)Area\: of \: triangle \\(A)=\sqrt{s(s-a)(s-b)(s-c)}\\=\sqrt{21\times 8\times 7\times 6}\\=\sqrt{3\times 7\times 2^{3}\times 7\times 2\times 3}$

$=3\times 7\times 4\\=84\:cm^{2}$

$ii)Given \:AC = 14 \: cm,\\Let \: the \: corresponding \: altitude = h$

$\frac{1}{2} \times base \times altitude=Area(A)$

$\implies \frac{1}{2}\times 14 \times h = 84$

$\implies 7 \times h = 84$

$\implies h = \frac{84}{7}\\=12\: cm$

Therefore,

$Area\: of \: the \: triangle = 84 \: cm^{2}\\Altitude \:on \: AC = 12 \: cm$

•••♪