In TriangleABC, AB: BC = 1: 2, BC: AC = 3: 5 how much is AC: AB?
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Answer:
In a triangle ABC, AB:BC is 2:3, BC:AC is 4:5, and the perimeter is 21 cm. What are the sides?
Ans :
AC = 9 cm
BC = 7.2 cm
AB = 4.8 cm
Solution :
AB:BC = 2:3 , BC:AC =4:5 , AB +BC +AC =21
AB = 2BC/3 & BC = 4AC/5
AB = (2/3)×4AC/5
AB = 8AC / 15
2. AB + BC + AC =21
8AC/15 + 4AC/5 + AC = 21
(8 + 12 + 15)AC /15 = 21 ( upon taking L.C.M)
35AC/15 = 21
AC = 21 × (15/ 35)
AC = 9 cm
3. Upon solving,
AC = 9 cm
BC = 7.2 cm
AB = 4.8 cm
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