Question

solve it guys only answer if you know​

Answer 1

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.

___________________________

Answer 2

$\huge{\fbox{\fbox{\bigstar{\mathfrak{\red{Answer:-}}}}}}$

${\huge{Given :-}}$

AP = 6 cm

BP = 8 cm

Area of ∆ABP = 84 cm²

Radius of circle = 4cm

${\huge{Find :-}}$

The length of side AC and BC

${\Huge{Solution :-}}$

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

_______________________________

The rest two sides of the ∆ABC = 13 , 15 cm