Question

Find the area of triangle whose side are 91m 98 m and 105m.also find the height of corresponding to the longest side

Answer 1

$\textbf{Answer:}$

We will use Heron's Formula here,

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

S=semi perimeter

$s = \frac{91 + 98 + 105}{2} = 147$

$area = \sqrt{147(56)(49)(42)}$

=4116m²

Height corresponding to longest side

▶️Area of the triangle-$\boxed{1/2 x base x height }$

4116m²=1/2 x105 x height

h=4116x2/105

$\boxed{h=78.4m}$

$\textbf{Hope it helps you! }$

Answer 2

$\large \bf{ \mathfrak{ HELLO \: \: FRIENDS!! }}$

$\large{ \mathfrak{Here \: is \: y our \: answer↓}}$


⬇⬇⬇⬇⬇⬇⬇⬇⬇⬇⬇⬇⬇⬇



$\huge \bf \sf{Given:-)}$


↪➡ Sides of triangle are 91m , 98m and 105m.


$\huge \bf \sf{To \: Find:-)}$

↪➡ Area of triangle.

↪➡ And height of triangle correspondence to the longest side.


$\bf{By \: using \: Heron's \: formula:-)}$

$\boxed{ = > \sqrt{s(s - a)(s - b)(s - c)} .}$

→ Here a = 91m, b = 98m , and c = 105m.


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


$\bf = > s = \frac{91 + 98 + 105}{2} .$


$\bf = > s = \frac{294}{2} .$


$\huge \boxed{ = > s = 147.}$


$\bf = > area \: of \: triangle = \sqrt{s(s - a)(s - b)(s - c)} .$


$\bf = \sqrt{147(147 - 91)(147 - 98)(147 - 105).}$


$\bf = \sqrt{147 \times 56 \times 49 \times 42} .$


$\bf = \sqrt{3 \times 7 \times 7 \times 2 \times 2 \times 2 \times 7 \times 7 \times 2 \times 3 \times 7} .$


$\bf = 3 \times 2 \times 2 \times 7 \times 7 \times 7.$



$\huge \bf \boxed { = 4116 {m}^{2} .}$

▶ We know that :-)

$\boxed{area \: of \: triangle = \frac{1}{2} \: base \times height.}$


$\bf = > 4116 = \frac{1}{2} \times 105 \times h.$
[ note:- longest base is 105 m.]


$\bf = > h = \frac{4116 \times 2}{105} .$


$\huge \bf \boxed{ = > h = 78.4m.}$


✅✅ Hence, all are founded ✔✔.



$\huge \boxed{ \mathfrak{THANKS}}$



$\huge \bf \underline{ \mathfrak{Hope \: it \: is \: helpful \: for \: you}}$