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Answers
Given :-
- AP = 6cm.
- BP = 8cm.
- Area of ∆ = 84cm².
- inradius of circle = 4cm.
To Find :-
- Length of side AC and BC . ?
Formula used :-
- Tangents From External Point to the circle are Equal.
- Area of Triangle = (Semi-Perimeter of ∆) * (inradius).
Solution :-
Let us Assume That, Length of QC is xcm.
Than,
By Tangent External Points we Have :-
→ AP = AR = 6cm.
→ PB = BQ = 8cm.
→ QC = CR = x cm.
Or, we can say That,
→ AB = 14cm.
→ AC = (6+x)cm.
→ BC = (8+x) cm.
So,
→ Semi-Perimeter of ∆ = (14 + 6 + x + 8 + x) / 2
→ s = (28 + 2x)/2
→ s = (14+x) cm.
Now, Putting both Values in above told formula, we get,
→ s * inradius = Area of ∆
→ (14+x) * 4 = 84
Dividing both sides by 4,
→ (14+x) = 21
→ x = 21 - 14
→ x = 7cm.
Hence ,
→ AC = (6+x) = 6 + 7 = 13cm.
→ BC = (8+x) = 8 + 7 = 15cm.
Hence, Rest two Sides of ∆ABC are 13 & 15 cm.
___________________________
AP = 6 cm
BP = 8 cm
Area of ∆ABP = 84 cm²
Radius of circle = 4cm
The length of side AC and BC
Let the length of QC be x cm
AP = AR = 6cm
PB = BQ = 8 cm
QC = CR = x cm
_______________________________
⇝ AB = 14 cm
⇝AC =( 6+x ) cm
⇝ BC = ( 8+x ) cm
_______________________________
So,
⇝ half-perimeter of ∆ =
( 14 + 6 + x + 8 + x ) / 2
⇝ H-P = ( 28 +2x )/2
⇝ H-P = ( 14 + x ) cm
_______________________________
⇝ H × inradius = Area of ∆
⇝ ( 14 + x) = 21
⇝ x = 21 - 14
⇝x = 7 cm
_______________________________
Hence,
⇝ AC = ( 6 + x ) = 6 + 7 = 13 cm
⇝ BC = ( 8 + x ) = 8 + 7 = 15 cm
_______________________________