If BD = 8 cm & DC=6 cm and radius of the circle is 4 cm find the perimeter of ∆ABC
Answers
Given :- BD = 8 cm & DC=6 cm and radius of the circle is 4 cm .
To Find :- The perimeter of ∆ABC = ?
Solution :-
since tangents from same external points to the circle are equal we get,
- EB = BD = 8 cm
- DC = CF = 6 cm
- AE = AF = let x cm .
- BC = BD + DC = 8 + 6 = 14 cm
- AB = BE + EA = (8 + x) cm
- AC = AF + FC = (6 + x) cm
so,
→ semi perimeter of ∆ABC = (AB + BC + AC)/2 = (14 + 8 + x + 6 + x) / 2 = (14 + x) cm .
now,
→ Area of ∆ABC = Area ∆(AOB) + Area ∆(BOC) + Area ∆(AOC)
using :-
- Area of ∆ = √s(s - a)(s - b)(s - c) = (1/2) * Base * perpendicular height .
comparing area of ∆ABC now, we get,
→ √[(14 + x)(14 + x - 14)(14 + x - 8 - x)(14 - x - 6 - x) = (1/2)[4*14 + 4*(8 + x) + 4(6 + x)]
→ √[(14 + x) * x * 6 * 8] = 2(14 + 8 + x + 6 + x)
→ 4√3x(14 + x) = (56 + 4x)
→ √3x(14 + x) = (14 + x)
squaring both sides,
→ 3x(14 + x) = (14 + x)²
→ 3x = (14 + x)
→ 3x - x = 14
→ 2x = 14
→ x = 7 cm .
therefore,
→ Perimeter of ∆ABC = 2 * semi perimeter = 2 * (14 + x) = 2 * (14 + 7) = 2 * 21 = 42 cm (Ans.)
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