Question

e. Find the length of sides of the triangle whose vertices are A (7,4), B (-8, 6) Ć(1, -3).​

Answer 1

Answer:

Length of sides of triangle are: AB = 15.13, BC = 12.72, AC = 9.21

Step-by-step explanation:

We need to find the length of sides of the triangle whose vertices are A (7,4), B (-8, 6) C(1, -3).​

We have three sides of triangle AB, BC and AC

The length of side can be calculated using distance formula:

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

Now finding lengths of sides AB, BC and AC

i) Length of side AB

We have A (7,4), B (-8, 6)

and $x_1=7, y_1=4, x_2=-8, y_2=6$

Putting values in formula and finding length

$Length \ of \ side \ AB \ =\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\Length \ of \ side \ AB \ =\sqrt{(-8-7)^2+(6-4)^2}\\Length \ of \ side \ AB \ =\sqrt{(-15)^2+(2)^2}\\Length \ of \ side \ AB \ =\sqrt{225+4}\\Length \ of \ side \ AB \ =\sqrt{229}\\Length \ of \ side \ AB \ =15.13$

So, Length of side AB is 15.13

ii) Length of side BC

We have B (-8, 6) and C(1, -3)

and $x_1=-8, y_1=6, x_2=1, y_2=-3$

Putting values in formula and finding length

$Length \ of \ side \ BC \ =\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\Length \ of \ side \ BC \ =\sqrt{(1-(-8))^2+(-3-6)^2}\\Length \ of \ side \ BC \ =\sqrt{(1+8)^2+(-9)^2}\\Length \ of \ side \ BC \ =\sqrt{(9)^2+(-9)^2}\\Length \ of \ side \ BC \ =\sqrt{81+81}\\Length \ of \ side \ BC \ =\sqrt{162}\\Length \ of \ side \ BC \ =12.72$

So, Length of side BC is 12.72

iii) Length of side AC

We have A (7,4)and C(1, -3)

and $x_1=7, y_1=4, x_2=1, y_2=-3$

Putting values in formula and finding length

$Length \ of \ side \ AC \ =\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\Length \ of \ side \ AC \ =\sqrt{(1-7)^2+(-3-4)^2}\\Length \ of \ side \ AC \ =\sqrt{(-6)^2+(-7)^2}\\Length \ of \ side \ AC \ =\sqrt{36+49}\\Length \ of \ side \ AC \ =\sqrt{85}\\Length \ of \ side \ AC \ =9.21$

So, Length of side AC is 9.21

So, length of sides of triangle are: AB = 15.13, BC = 12.72, AC = 9.21