Prove that the points (3, 4), (8, -6) and (13,9) are the vertices of a right angled triangle. Find its area
Answers
Step-by-step explanation:
Given:
ΔABC is right angled at C.
P and Q are the points on AC and AB respectively such that
AP:PC=2:1=AQ:QB
i.e
AB
AQ
=
2+1
2
⇒AB=
2
3
×AQ
AC
AP
=
2+1
2
⇒AC=
2
3
×AP .......... (i)
Now, ΔABC is right angled at C
So, applying Pythagoras' theorem,
AB
2
=AC
2
+BC
2
Using (i), we get
(
2
3
×AQ)
2
=(
2
3
×AP)
2
+(BC)
2
⇒
4
9
×AQ
2
=
4
9
×AP
2
+BC
2
⇒9AQ
2
=9AP
2
+4BC
2
solution
Step-by-step explanation:
AC=√(13-3)^2+(4-9)^2
= √(10)^2+(5)^2
= √100+25
=√125 units
AB=√(3-8)^2+(4+6)^2
=√(-5)^2+(10)^2
= √25+100
=√125 units
BC=√(8-13)^2+(-6-9)^
=√(-5)^2+(-15)^2
= √25+225
= √250 units
BC^2=(√250)^2
=250
AC^2+AB^2=(√125)^2+(√125)^2
=125+125
=250
As, BC^2=AC^2+AB^2
SO, These points form a right angled triangle
Now, BC=√250 units =hypotenuse
AC=√125 units = height
AB=√125 units = base
Area of triangle= 1/2x base x height
=1/2x AB x AC
= 1/2 x √125 x √125
= 1/2 x 125
= 62.5 sq. units
so, area of triangle is 62.5 sq. units