In triangle ABC angle ACB = 90 degree and CD perpendicular to AB provided that AD= 16cm and DB= 9 cm, find BC/AC
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AC2 + BC2 = 625
BC2 = 81 + CD2
AC2 = 256 + CD2
256+81+2CD2 = 625
CD2 = 144
CD= 12
BC2 = 81+144=225
BC=15
AC2 = 256+144=400
AC= 20
BC/AC = 15/20 = 3/4
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Answer:
Step-by-step explanation:wkt the perpendicular drawn from the vertex of the right angle to the hypotnuse divides the right triangle into 2 similar triangles and each of these is similar to the whole triangle.
ABC is similar to CBD
BA/CB=CB/BD =AC/CD by cpst
BC²=BD×AB
ABC is similar to the CAD
AB/AC= CB/AD= CA/DC by cpst
AC² =AB× CD
BC²/AC²=BD ×AB/ AB×CD
By calculating u can get the answer
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