Equal sides of an isosceles triangle are 13 cm each and the Base is 10 cm. find the altitude from the vertex to the Base of the triangle
Answers
Step-by-step explanation:
1) Given :
ABC be an isosceles triangle whose congruent sides are AC and BC.
AC=BC=13cm
AB = 10cm.
2) We will mark all points by taking origin as A (0,0) in first Quadrant of Cartesian Co-ordinate system.
Then,
A= (0,0)
B = (10,0)
Now,
I isosceles triangle,
CE is median passing through centroid D
Therefore,
AE = BE = 5cm
=> E = (5,0)
3) In triangle ACE,
By Pythagoras Theorem,
EC^2. = AC^2 - AE^2
=> EC^2 = 13^2 - 5^2
=> EC = 12cm
=> C = (5,12)
4) Centroid of triangle ABC,
C =( ( x1 + x2 + x3)/ 3 ,(y1+y2+y3)/3 )
=> C =((0+ 10+ 5)/3 ,(0+0+12)/3
=> C = (15/3 , 12/3 )
=> C = (5, 4)
5) Required :
By distance formula,
Distance between vertex opposite the base and controid is :
CD =
\begin{gathered}cd = \sqrt{ {(5 - 5)}^{2} + {(12 - 4)}^{2} } \\ = > cd \: = 8cm\end{gathered}cd=(5−5)2+(12−4)2=>cd=8cm
Hence,
Distance between the vertex opposite the base and the controid is 8cm.
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