Question

Draw a triangle with vertices o(0,0)A(3,0)B(3,4).classify the triangle and also find its area.

Answer 1

hey mate

here is ur answer

$\Huge\orange{\boxed{\boxed{\underline{ANSWER}}}}$

given points are

O(0,0),A(3,0),B(3,4)

$\bf to \: find \: the \: area \: of \: triangle$

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

$distance \: between \: two \: points =$

$\bf \underline {\sqrt {(x_2-x_1)^2+(y_2-y_1)^2}}$

$a = \sqrt{(3 - 0)^{2} + (4 - 0) ^{2} } = \sqrt{25} = 5 \: units$

$b = \sqrt{(3 - 0)^{2} + (0) } = 3 \: units$

$c = \sqrt{ {(3 - 3)}^{2} + {(4 - 0)}^{2} } = \sqrt{16} = 4 \: units$

these form Pythagorean triplet

so,the points form

$\Huge \bf Right \: angled \: triangle$

let us find the value of s

$s = \frac{a + b + c}{2} = \frac{12}{2} = 6$

now the area of triangle

$= \sqrt{6(6 - 5)(6 - 3)(6 - 4)} \\ = \sqrt{6 \times 1 \times 3 \times 2} \\ = \sqrt{36} \\ = 6 \: square \: units$

$area \: of \: triangle =6\: sq.units$

hope the answer helps

Answer 2

Step-by-step explanation:

Area of triangle is 6 square units