Find the area of the Triangle formed by the points (0,0) (4,0) and (4,3)
Answers
5ab + 8ab +17cd +69cd
Step-by-step explanation:
5ab + 8ab +17cd +69cd
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.