Question

Find the area of the Triangle formed by the points (0,0) (4,0) and (4,3)​

Answer 1

5ab + 8ab +17cd +69cd

Step-by-step explanation:

5ab + 8ab +17cd +69cd

Answer 2

Step-by-step explanation:

6 sq. units.  

Step-by-step explanation:

A=(0,0)

B = (4,0)

C = (4,3)

Now find the sides of triangle AB,BC,AC

To find AB use distance formula :

d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}d=(x2−x1)2+(y2−y1)2

(x_1,y_1)=(0,0)(x1,y1)=(0,0)

(x_2,y_2)=(4,0)(x2,y2)=(4,0)

Substitute the values in the formula :

AB=\sqrt{(4-0)^2+(0-0)^2}AB=(4−0)2+(0−0)2

AB=\sqrt{(4)^2}AB=(4)2

AB=\sqrt{16}AB=16

AB=4AB=4

To Find BC

(x_1,y_1)=(4,0)(x1,y1)=(4,0)

(x_2,y_2)=(4,3)(x2,y2)=(4,3)

Substitute the values in the formula :

BC=\sqrt{(4-4)^2+(3-0)^2}BC=(4−4)2+(3−0)2

BC=\sqrt{(0)^2+(3)^2}BC=(0)2+(3)2

BC=\sqrt{9}BC=9

BC=3BC=3

To Find AC

(x_1,y_1)=(0,0)(x1,y1)=(0,0)

(x_2,y_2)=(4,3)(x2,y2)=(4,3)

Substitute the values in the formula :

AC=\sqrt{(4-0)^2+(3-0)^2}AC=(4−0)2+(3−0)2

AC=\sqrt{(4)^2+(3)^2}AC=(4)2+(3)2

AC=\sqrt{16+9}AC=16+9

AC=\sqrt{25}AC=25

AC=5AC=5

So, sides of triangle :

a =4

b=3

c=5

Now to find area:

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

Where s = \frac{a+b+c}{2}s=2a+b+c  

a,b,c are the side lengths of triangle  

Now substitute the values :  

s = \frac{4+3+5}{2}s=24+3+5  

s =6s=6  

Area = \sqrt{6(6-4)(6-3)(6-5)}Area=6(6−4)(6−3)(6−5)  

Area = 6Area=6  

Hence the area of the given triangle is 6 sq. units.