Question

what is the area of trainle whose side is 9 CM 12 cm and 5 cm​

Answer 1

Answer:

$20.4 {cm}^{2}$

Step-by-step explanation:

Side9

Base12

Side=5

Using the formula

A=

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

$s = a + b + c \div 2$

Solving for A

$a = 1 \div 4 \ \sqrt{ - } {a}^{4} + 2(a + b) {}^{2} + 2(ac) {}^{2} - b {}^{4} + 2(bc) {}^{2} - c {}^{4} 1 \div 4. \sqrt{ - 9 {}^{4} } + 2.(9.12) {}^{2} + 2.(9.5) {}^{2} - 12 {}^{4} + 2.(12.5) {}^{2} - 5 {}^{4} =20.39608cm {}^{2}$