Question

find the area of triangles formed by the vertices (1,2) (5,7) and (5,-3)​

Answer 1

Given: Vertices of a triangle =

A(1,2) ; B(5,7) ; C(5,-3)

To find the area of the traingle,

To find the area of the traingle, We have to find the length of each side.

=>Length of side A = Distance between point A and B

=>Length of side B = Distance between point B and C

=>Length of side C = Distance between point C and A

Coordinates of A = $\rm x_1 = 1, \: y_1 = 2$

Coordinates of B = $\rm x_2 = 5, \: y_2 = 7$

Coordinates of C = $\rm x_3 = 5, \: y_3 = -3$

Using Distance formula,

Distance between two points $\rm A(x_1, y_1)$ and $\rm B(x_2, y_2)$ is,

$\rm \implies \sqrt{(x_2 - x_1)^2 + (y_2 -y_1)^2}$

$\texttt{\green{First, We will Find the length of AB,}}$

A = $x_1 = 1, \: y_1 = 2$

B = $x_2 = 5, \: y_2 = 7$

$\rm \therefore \sqrt{(5-1)^2 + (7-2)^2} \\ \rm \implies \sqrt{4^2 + 5^2} \\ \rm \implies \sqrt{16 + 25} \\ \rm \implies \sqrt{41} \\ \rm \implies \bf \red{6.4cm} \\ \bold{\blue{\texttt{Hence, AB = 6.4cm}}}$

$\texttt{\green{Similarly, We will find the Length of BC}}$

B = $x_1 = 5, \: y_1 = 7$

C = $x_2 = 5, \: y_2 = -3$

$\therefore \rm \sqrt{(5-5)^2 + (-3 -7)^2} \\ \rm \implies \sqrt{0^2 + (-10)^2} \\ \rm \implies \sqrt{0 + 100} \\ \rm \implies \sqrt{100} \\ \rm \implies \bf \red{10cm} \\ \bold{\blue{\texttt{Hence, BC = 10cm}}}$

$\texttt{\green{Also, we will find the length of CA}}$

C = $x_1 = 5, \: y_1 = -3$

A = $x_2 = 1, \: y_2 = 2$

$\therefore \rm \sqrt{(1-5)^2 + (2-(-3))^2} \\ \rm \implies \sqrt{-4^2 + 5^2} \\ \rm \implies \sqrt{-16 + 25} \\ \rm \implies \sqrt{9} \\ \rm \implies \bf \red{3cm} \\ \bold{\blue{\texttt{Hence, CA = 3cm}}}$

Hence, we have lengths of the traingle ABC = 3cm, 6.4cm, 10cm

Now,

Using heron's formula for area of traingle,

$\boxed{\bold{\purple{\texttt{Area of Δ = \sqrt{s(s-a)(s-b)(s-c)} }}}}$

Where, $\bold{\red{\texttt{ s = semi perimeter = \frac{a + b + c}{2} }}}$

And, $\bold{\red{\texttt{a, b, and c are sides of the triangle}}}$

So, here value of s,

$\rm \implies \frac{3 + 6.4 + 10}{2} \\ \rm \implies \frac{19.4}{2} \\ \rm \implies \red{9.7}$

s = 9.7 cm

Putting the values in the formula,

$\therefore \rm \sqrt{9.7(9.7-3)(9.7-6.4)(9.7-10)} \\ \implies \rm \sqrt{9.7 \times 6.7 \times 3.3 \times 0.3} \\ \implies \rm \sqrt{64.34} \\ \implies \rm \bold{\red{8.02cm^2}} \: \: (approx.)$

On, rounding off,

Hence, the area of the ΔABC of the vertices shown is $\rm \bold{8cm^2}$