Question

OR
In the given figure, BC, BA and AC are tangents to the circle
touching the circle at D, E and F respectively. If BD = 30 cm,
CD = 7 cm and ZA = 90°, find the radius of the circle.​

Answer 1

Radius is 5 cm.

Step-by-step explanation:

Here, BD = 30 cm & CD = 7 cm

Theorems that can be implied :

1.) Length of two tangents are equal when drawn from same external point  

2.) Pythagoras Theorem - which states that -

$(Hypotenuse)^2 = (Perpendicular)^2 + (Base)^2$

Explanation

i) AB, BC, and AC are tangents to the circle at E, D, and F.

BD = 30 cm with DC = 7 cm further ∠BAC = 90°

As per the theorem,

BE = 30 cm = BD,

Also FC = 7 cm = DC

Let AE = x = AF…. (1)

Then AB = BE + AE = (30 + x)

AC = AF + FC = (7 + x)

BC = BD + DC = 30 + 7 = 37 cm

• Consider right Δ ABC, by Pythagoras theorem we have

$BC^2 = AB^2 + AC^2$

⇒ $(37)^2 = (30 + x)^2 + (7 + x)^2$

⇒ 1369 = 900 + 60x + $x^2$ + 49 + 14x + $x^2$

⇒ $2x^2$+ 74x + 949 – 1369 = 0

⇒ $2x^2$+ 74x – 420 = 0

⇒ $x^2$ + 37x – 210 = 0

⇒ $x^2$ + 42x – 5x – 210 = 0

⇒ x (x + 42) – 5 (x + 42) = 0

⇒ (x – 5) (x + 42) = 0

⇒ (x – 5) = 0 or (x + 42) = 0

⇒ x = 5 or x = – 42

⇒ x = 5 [Since x cannot be negative]

∴ AF = 5 cm [From (1)]

Therefore AB =30 +x = 30 + 5 = 35 cm

ii)AC = 7 + x = 7 + 5 = 12 cm

Let ‘O’ be the centre of the circle and ‘r’ the radius of the circle.

Join two of the points O, F; points O, D and points O, E.

From the figure,

$\frac {1}{2} \times AC \times AB = (\frac {1}{2} \times AB \times OE) + (\frac {1}{2} \times BC \times OD) + ( \frac {1}{2} \times AC \times OC)$

⇒ $AC \times AB = AB \times OE + BC \times OD + AC \times OC$

⇒ $12 \times 35 = 35 \times r + 37 \times r + 12 \times r$

⇒ 420 = 84 r

∴ r = 5

Thus the radius of the circle is 5 cm.