Find the area of triangle whose coordinates are (1,2),(3,4) and (5,10)
Answers
A=(1,2)
B=(3,4)
C=(5,10)
We will use distance formula to find the sides of triangle
d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}d=
(x
2
−x
1
)
2
+(y
2
−y
1
)
2
\begin{lgathered}A=(x_1,y_1)=(1,2)\\B=(x_2,y_2)=(3,4)\end{lgathered}
A=(x
1
,y
1
)=(1,2)
B=(x
2
,y
2
)=(3,4)
Substitute the values
AB=\sqrt{(3-1)^2+(4-2)^2}AB=
(3−1)
2
+(4−2)
2
AB=\sqrt{(2)^2+(2)^2}AB=
(2)
2
+(2)
2
AB=\sqrt{4+4}AB=
4+4
AB=\sqrt{8}AB=
8
AB=2.82AB=2.82
\begin{lgathered}B=(x_1,y_1)=(3,4)\\C=(x_2,y_2)=(5,10)\end{lgathered}
B=(x
1
,y
1
)=(3,4)
C=(x
2
,y
2
)=(5,10)
Substitute the values
BC=\sqrt{(5-3)^2+(10-4)^2}BC=
(5−3)
2
+(10−4)
2
BC=\sqrt{(2)^2+(6)^2}BC=
(2)
2
+(6)
2
BC=\sqrt{4+36}BC=
4+36
BC=\sqrt{40}BC=
40
BC=6.32BC=6.32
\begin{lgathered}A=(x_1,y_1)=(1,2)\\C=(x_2,y_2)=(5,10)\end{lgathered}
A=(x
1
,y
1
)=(1,2)
C=(x
2
,y
2
)=(5,10)
Substitute the values
AC=\sqrt{(5-1)^2+(10-2)^2}AC=
(5−1)
2
+(10−2)
2
AC=\sqrt{(4)^2+(8)^2}AC=
(4)
2
+(8)
2
AC=\sqrt{16+64}AC=
16+64
AC=\sqrt{80}AC=
80
AC=8.94AC=8.94
Now to find the area we will use heron's formula:
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=
2
a+b+c
a=2.82
b=6.32
c=8.94
Substitute the values
s=\frac{2.82+6.32+8.94}{2}s=
2
2.82+6.32+8.94
s=9.04s=9.04
Area=\sqrt{9.04(9.04-2.82)(9.04-6.32)(9.04-8.94)}Area=
9.04(9.04−2.82)(9.04−6.32)(9.04−8.94)
Area=3.910 units^2Area=3.910units
2
Hence the area of the triangle is 3.910 sq.units.
Answer:
The answer is 4 sq units
Step-by-step explanation:
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