find the area of triangle formed by the points (2,3) (6,3) (2,6) by using Heron's formula
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241
Let A=(2,3),B=(6,3),C=(2,6)
AB=√(6-2)²+(3-3)²
AB=√(4)²+0
AB=√16=4
BC=√(2-6)²+(6-3)²
BC=√(-4)²+(3)²
BC=√16+9
BC=√25=5
AC=√(2-2)²+(6-3)²
AC=√0+3²
AC=√9=3
Let
AB=c=4
BC=a=5
AC=b=3
Heron's Formula is given by
A=√s(s-a)(s-b)(s-c)
where s=a+b+c/2
s=4+5+3/2
s=12/2
s=6
s-a=6-5=1
s-b=6-3=3
s-c=6-4=2
Area of a triangle, ΔABC
=√6(2)(1)(3)
=√36
=6 square units
hope it helps you
AB=√(6-2)²+(3-3)²
AB=√(4)²+0
AB=√16=4
BC=√(2-6)²+(6-3)²
BC=√(-4)²+(3)²
BC=√16+9
BC=√25=5
AC=√(2-2)²+(6-3)²
AC=√0+3²
AC=√9=3
Let
AB=c=4
BC=a=5
AC=b=3
Heron's Formula is given by
A=√s(s-a)(s-b)(s-c)
where s=a+b+c/2
s=4+5+3/2
s=12/2
s=6
s-a=6-5=1
s-b=6-3=3
s-c=6-4=2
Area of a triangle, ΔABC
=√6(2)(1)(3)
=√36
=6 square units
hope it helps you
mysticd:
Plz , change AB=c, BC= a, CA= b
Units are not given
Remove cm, cm^2
Write square units in the last line
thank u and u r correct
:)
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