Question

the vertices of triangle are (2,1), (5,2), (4,4). The lengths of the perpendiculars from these vertices on opposite sides are​

Answer 1

First find length of sides of triangle

$Formula :d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}d=$

$AB=\sqrt{(5-2)^2+(2-1)^2}AB=$

AB=3.162AB=3.162

$BC=\sqrt{(4-5)^2+(4-2)^2}BC=$

$AC=\sqrt{(4-2)^2+(4-1)^2}AC=$

Now to find area of triangle

a = 3.162

b = 2.236

c =3.605

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

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

$s=\frac{3.162+2.236+3.605}{2}s=$

$Area=\sqrt{4.5015(4.5015-3.162)(4.5015-2.236)(4.5015-3.605)}Area= </p><p>4.5015(4.5015−3.162)(4.5015−2.236)(4.5015−3.605)$

Area=3.4995Area=3.4995

The base corresponding to point A is BC

So, To find length of the perpendicular from the first vertex to the opposite side

$Area of triangle = \frac{1}{2} \times Base \times height$

$3.4995=\frac{1}{2} \times 2.236 \times Height.$

Hence the length of the perpendicular from the first vertex to the opposite side is 3.13 units