ABC IS AN ISOSCELES TRIANGLE WITH AB=AC=2A AND BC = A . IF AD IS PERPENDICULAR TO BC FIND THE LENGTH OF AD..
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let the perpendicular ad bisects bc
therefore length of bd=1/2 a
according to pythagoras theorem
hypotenuse square=Base square +perpendicular square
ac square=1/2 bc square +ad square
4a^2 =a^2/4+ad square
therefore ad square =4a^2- a^2/4
therefore ad=a root 15/2
therefore length of bd=1/2 a
according to pythagoras theorem
hypotenuse square=Base square +perpendicular square
ac square=1/2 bc square +ad square
4a^2 =a^2/4+ad square
therefore ad square =4a^2- a^2/4
therefore ad=a root 15/2
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In an isosceles ΔABC in which AB=AC=2a units, BC=a units
AD is the altitude. Therefore, D is the midpoint of BC
⇒BD= 2 a
We have two right triangles: ΔADB and ΔADC
By Pythagoras theorem,
AB 2 =BD 2 +AD 2
(2a) 2 =( 2 a ) 2 +AD 2
(2a) 2 = 4 a 2 +AD 2
AD 2 = 4 16a 2 −a 2 = 4 15a 2
AD= 2 a 15
Hence, the length of the altitude AD is 1/2 root 15 units.
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