Show that the following points A(8, 2), B(5, −3) and C(0, 0) are the vertices of an isosceles triangle.
Answers
Answers
➡️Vertices of the triangle are A(-5,6), B(3,0), C(9, 8)
✔️Distance between two points=
=√(x^2 - x^1)^2+(y^2 - y^1)^2
AB = √[(3-(-5))^2 + (0-6 )^2] = √64 + 36
= √100
= 10
BC = √[(9-3)^2 + (8-0)^2]
= √36+64
=√100
= 10
AC = √[(9-(-5)^2 + (8-6)^2] = √196 + 4
= √200
= 10√2
AB = BC
Therefore, ΔABC is an isosceles triangle.
(AB)^2 + (BC)^2 = (AC)^2
(10)^2 + (10)^2
= 200 = 10√2 = AC^2
So, the triangle satisfies the pythagoras theorem and hence it is a right angled triangle.
Area of the isosceles right angled triangle
= 1/ 2 × base × height
= 1 /2 *10*10 = 50 sq units.
Hope this helps you.
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Answer:
yes it is an isoceles triangle
Step-by-step explanation:
because AB=BC=10 units
AC is 10 root 2 units
therefore, two equal sides and an unequal side
therefore, it is isoceles triangle