Q)  A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC are of lengths 10 cm and 8 cm respectively. Find the lengths of the sides AB and AC, if it is given that area ▲ ABC = 90 cm².

Ans: 

Step 1: Let’s join Point A, B, C with center O and also consider that circle touches AB at point E and AC at point F.

Step 2: Here, BD = BE = 10 cm (tangents on a circle from same point)

Similarly, CD = CF = 8 cm (given)

and ∵ length if AE or AF is not given, let’s consider AE = A F  = x

Step 3: Now, Side AB = AE + BE = 10 + x

and AC = AF + CF = 8 + x

and BC = BD + CD = 10 + 8 = 18 cm

Step 4: Next, Area of Δ ABC = Area Δ AOB + Area Δ BOC + Area Δ AOC 

∵ Area of Δ ABC = 90 cm²                  (given)

∴ 90 = OE x AB + OD x BC + OF x AC

Since OE = OD = OF = 4 cm                         (radii of circle)

∴ 90 = x ( 4 x AB + 4 x BC + 4 x AC)

∴ 90 = x 4 (AB + BC + AC)

∴ 90 = 2 [(10 + x) + 18 + (8 + x)]                         ……. from equation (i)

∴ 45 = 10 + x + 18 + 8 + x = 36 + 2 x

∴  2 x = 45 – 36 = 9

∴  x = = 4.5 cm

Step 5: ∵ AB = 10 + x

∴ AB = 10 + 4.5 = 14.5 cm

and ∵ AC = 8 + x

∴ AC = 8 + 4.5 = 12.5 cm

Therefore, lenth of AB is 14.5 cm and AC is 12.5 cm.

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