Question

Find the area of a triangle whose sides are 34 cm and 20 cm and 42 CM hence find the length of the altitude corresponding to the shortest side ​

Answer 1

Answer:

using heron's formula

s = (20+34+42)/2 = 96/2 = 48

area = √s(s-a)(s-b)(s-c)

= √48*(48-20)(48-34)(48-42)

= √48*28*14*6

= √6*4*2*7*4*7*2*6

= √6*6*4*4*7*7*2*2

= 6*4*7*2

= 336 sqcm

again

area = 336

1/2* base*height = 336

1/2* 20*height = 336

10*height = 336

height = 33.60 cm

Answer 2

AnswEr :

Given Sides of the ∆ are 34 cm, 20 cm & 42 cm.

We've to find length of the altitude corresponding to the shortest side.

$\underline{\bf{\dag} \:\mathfrak{By \; using \: Heron's \: Formula \: :}}$

$\implies\sf S = \dfrac{a + b + c}{2}$

$\sf{Here}\begin{cases}\sf{S = semiperimeter} \\\sf{a = 34}\sf{b = 20}\\\sf{c = 42}\end{cases}$

$\implies\sf S = \dfrac{34 + 20 + 42}{2}$

$\implies\sf S = \cancel\dfrac{96}{2}$

$\implies\boxed{\sf{\blue{S = 48}}}$

$\star\: \boxed{\sf{\sqrt{s(s - a) (s - b) (s - c)}}}$

$\implies\sf\sqrt{48(48 - 34) (48 - 20) (48 - 42)}$

$\implies\sf\sqrt{48 \times 14 \times 28 \times 6}$

$\implies\sf\sqrt{3 \times 16 \times 2 \times 7 \times 4 \times 7 \times 3 \times 2}$

$\implies\sf 4 \times 3 \times 7 \times 2 \times 2$

$\small\implies{\underline{\boxed{\sf{\: 336\:}}}}$

Here, we get area of the ∆.

$\rule{300}{2}$

Now, we'll calculate altitude.

$\dag\: \small{\underline{\boxed{\sf{Area_{triangle} = \dfrac{1}{2} \Big(Altitude \times Base \Big)}}}}$

$\implies\sf 336 = \dfrac{1}{\cancel{2}} \times \cancel{20} \times Altitude$

$\implies\sf \dfrac{336}{10} = Altitude$

$\huge\implies{\underline{\boxed{\sf{\: 33.6\: \: cm}}}}$

Hence, Altitude of the ∆ is 33.6 cm & area of the ∆ is 336 cm².