in figure, D is a point on BC such that angle ABD=angle CAD. If AB=5cm, AD=4cm and AC=3cm, find Bc, A[triangle ACD ]/A[triangle aBCA].
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By pythagoras therom
Ab square equals to AC square + BC square so that
(5)square =(x) square +(3)square
25=(x)square +9
(x)square =25-9
(x)square =16
X=4
By pythagoras therom
Ab square equals to AC square + BC square so that
(5)square =(x) square +(3)square
25=(x)square +9
(x)square =25-9
(x)square =16
X=4
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