Question

The sides of a triangle are 25cm,17cm,12cm,.The length of the altitude on the longest side is equal to​

Answer 1

$\huge\mathbb{SOLUTION:-}$

• First find the area of triangle which sides are given

• Sides of ∆25cm , 17cm,12 cm

By heron's formula

$\boxed{\sf{Area\:of\: triangle=\sqrt{s(s-a)(s-b)(s-c)}}}$

• Where a,b and c are the side of triangle and 's 'is semi perimeter

$\sf s=\frac{a+b+c}{2}\\ \\ \sf\implies s= \frac{25+17+12}{2}\\ \\ \sf\implies s=\cancel{\frac{54}{2}}= 27$

• Semi perimeter=27cm

Now the area of triangle

$\sf\implies Area=\sqrt{27(27-25)(27-17)(27-12)}\\ \\ \sf\implies Area=\sqrt{27\times 2\times 10\times 15}\\ \\ \sf\implies Area=\sqrt{9\times 3\times 2\times 2\times 5\times 3\times 5}\\ \\ \sf\implies Area=3\times 3\times 2\times 5\\ \\ \sf\implies Area= 9\times 10\\ \\ \sf\implies Area\:of\: triangle=90cm{}^{2}$

$\boxed{\sf{Area\:of\: triangle=90cm{}^{2}}}$

• Now we have to find the length of attitude on the longest side

• Longest side =25cm

$\boxed{\tt{Area\:of\: triangle=\frac{1}{2}\times base\times height}}$

$\sf\implies 90=\frac{1}{2}\times 25\times height\\ \\ \sf\implies 90\times 2=25\times height\\ \\ \sf\implies 180=25\times height\\ \\ \sf\implies height=\cancel{\frac{180}{25}}=7.2\\ \\ \sf\implies height=7.2cm$

$\boxed{\sf{Height\:of\: triangle\:to\: longest\: side=7.2cm}}$

Answer 2

Answer -

h = 7.2 cm

● Explaination -

Semiperimeter is given by -

$\small\textt{s = (a + b + c) / 2}$

s = (25 + 17 + 12) / 2

s = 27 cm

Area of triangle is calculated by -

Area of Triangle = √[s (s-a) (s-b) (s-c)]

Area of Triangle = √[27 (27-25) (27-17) (27-12)]

Area of Triangle = 90 cm^2

let h be Altitude of Longest side.

Area of Triangle = 1/2 × b × H

90 = 1/2 × 25 × h

h = 90 / 12.5

h = 7.2 cm

$\huge\green{\boxed{\bold{Ans.7.2cm}}}$