On a Right angled triangle, number of theorem are derived. One such theorem is Mid point Theorem, with below example theorem is very well explained.
Topic : Right angle Triangle and Mid Point Theorem
Problem : Given angle C is a right angle, E is a midpoint of AC, F is the midpoint of BC, AF = √41, BE = 2√26, Find AB
Solution :
From the figure,
In ∆ACF
AC2 + CF2 = AF2 (by Pythagoras Theorem)
AC2 + (CB/2)2 = (√41)2 (as CF is half of CB)
AC2 + CB2/4 = 41
Now in ∆ECB
EC2 + CB2 = EB2
(AC/2)2 + CB2 = (2√26)2 (as EC is half of AC)
AC2/4 + CB2 = 104
Now adding both the equations, we get
AC2 + CB2/4 + AC2/4 + CB2 = 41 + 104
(1+1/4)AC2 + (1+1/4)CB2 = 145
5AC2/4 + 5CB2/4 = 145
5/4(AC2 + CB2) = 145
Now, in ∆ACB, AC2 + CB2 = AB2
So we get, 5/4 AB2 = 145
AB2 = 145 * 4/5
AB2 = 116
AB = √116
AB = 2√29
Hope the above elaborated explanation will help you to understand mid point theorem and help you to solve similar kind of problems.
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