The sides AB, BC and AC of
AABC have lengths 7, 8 and
9 units, respectively. If D is
foot of perpendicular from 'A'
to BC, then the lengths of BD
and AD are respectively
equal to
OBD = 6, AD = 2
O
BD = 2, AD = 315
O
BD = 4, AD = 6
None
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Given: The sides AB, BC and AC of triangle ABC have lengths 7, 8 and
9 units, D is foot of perpendicular from 'A' to BC.
To find: the lengths of BD and AD are ?
Solution:
- As we have given that AD is perpendicular to BC, so we get two right triangles which are ABD and ADC.
- Now these triangles have the same height AD.
- Lets consider BD = x, So, DC will be 8-x.
- So now applying pythagoras theorem in both the triangles, we get:
BD² + AD² = AB²
x² + AD² = 7²
DC² + AD² = AC²
(8-x)² + AD² = 9²
- Now equating AD from both the equations we get:
49 - x² = 81 - (8-x)²
49 - x² = 81 - 64 - x² + 16x
cancelling x² from both sides:
49+64-81 = 16x
113-81 = 16x
32 = 16x
x = 2 units.
So, BD = 2 units and DC = 8-x = 8-2= 6 units.
Answer:
The the lengths of BD and AD are 2 units and 6 units.
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