Question

If line segment BC is considered the base of triangle ABC, what is the corresponding height of the triangle?

Answer 1

The answer is 1.6 on e2020

Answer 2

Answer:

1.6 units

Step-by-step explanation:

Coordinates of A : (-1,1)

Coordinates of B : (3,2)

Coordinates of C :( -1,-1)

Now to find the length of the sides of the triangle we will use distance formula :

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

To find length of AB

A = $(x_1,y_1)=(-1,1)$

B= $(x_2,y_2)=(3,2)$

Now substitute the values in the formula:

$d =\sqrt{(3-(-1))^2+(2-1)^2}$

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

$d =\sqrt{16+1}$

$d =\sqrt{17}$

$d =4.12$

Now to find length of BC

B= $(x_1,y_1)=(3,2)$

C = $(x_2,y_2)=(-1,-1)$

Now substitute the values in the formula:

$d =\sqrt{(-1-3)^2+(-1-2)^2}$

$d =\sqrt{(-4)^2+(-3)^2}$

$d =\sqrt{16+9}$

$d =\sqrt{25}$

$d =5$

Now to find length of AC

A = $(x_1,y_1)=(-1,1)$

C = $(x_2,y_2)=(-1,-1)$

Now substitute the values in the formula:

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

$d =\sqrt{(0)^2+(-2)^2}$

$d =\sqrt{4}$

$d =2$

Thus the sides of triangle are of length 4.12, 5 and 2

Now to find area of triangle we will use heron's formula :

To calculate the area of given triangle we will use the heron's formula :

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

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

a,b,c are the side lengths of triangle  

a = 4.12

b=5

c=2

Substitute the values

Now substitute the values :

$s = \frac{4.12+5+2}{2}$

$s =5.56$

$Area = \sqrt{5.56(5.56-4.12)(5.56-5)(5.56-2)}$

$Area = 3.99$

Now to find the height of altitude corresponding to side BC

So, formula of area of triangle $=\frac{1}{2} \times Base \times Height$

Since Area = 3.99 square units

So,$3.99 =\frac{1}{2} \times BC \times Height$

So,$3.99 =\frac{1}{2} \times 5\times Height$

$3.99 =2.5 \times Height$

$\frac{3.99}{2.5}=Height$

$1.596=Height$

Thus the altitude corresponding to the BC is of length 1.596 unit  ≈ 1.6 units

Thus Option D is correct.