In a triangle ABC right angled at B and BD perpendicular to AC if AD=5cm CD=6cm find BD AB
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Let AB = x, BD = y and BC = z.
From triangle ABC,
{x}^{2} + {z}^{2} = {(4 + 5)}^{2} = 81 - - (1)
From triangle ADB,
{x}^{2} = {y}^{2} + 16 - - - (2)
From triangle BDC,
{z}^{2} = {y}^{2} + 25 - - - (3)
Substituting (2) and (3) in (1),
( {y}^{2} + 16) + ( {y}^{2} + 25) = 81 \\ {y}^{2} = 20 - - - (4) \\ y = 2 \sqrt{5}
Substituting (4) in (2),
{x}^{2} = 20 + 16 \\ x = 6
Thus BD = 2√5; AB = 6
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