solve this question correct please
Answers
Step-by-step explanation:
Given AB, BC and AC are tangents to the circle at E, D and F.
BD = 30 cm and DC = 7 cm and ∠BAC = 90°
Recall that tangents drawn from an exterior point to a circle are equal in length
Hence, BE = BD = 30 cm
Also FC = DC = 7 cm
Let AE = AF = x → (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
AC = 7 + x = 7 + 5 = 12 cm
Let ‘O’ be the centre of the circle and ‘r’ the radius of the circle.
Join point O, F; points O, D and points O, E.
From the figure,
Area of (ΔABC) = Area (ΔAOB) + Area (ΔBOC) + Area (ΔAOC)
⇒
1
2
x AC x AB =
1
2
x AB x OE +
1
2
x BC x OD +
1
2
x AC x OC
⇒
AC x AB = (AB x OE) + (BC x OD) + (AC x OC)
⇒
12 x 35 = (35 x r) + (37 x r) + (12 x r)
⇒
12 x 35 = 84r
∴ r = 5
Thus the radius of the circle is 5cm
Step-by-step explanation:
Step-by-step explanation:
Given AB, BC and AC are tangents to the circle at E, D and F.
BD = 30 cm and DC = 7 cm and ∠BAC = 90°
Recall that tangents drawn from an exterior point to a circle are equal in length
Hence, BE = BD = 30 cm
Also FC = DC = 7 cm
Let AE = AF = x → (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
AC = 7 + x = 7 + 5 = 12 cm
Let ‘O’ be the centre of the circle and ‘r’ the radius of the circle.
Join point O, F; points O, D and points O, E.
From the figure,
Area of (ΔABC) = Area (ΔAOB) + Area (ΔBOC) + Area (ΔAOC)
AC x AB = (AB x OE) + (BC x OD) + (AC x OC)
⇒
12 x 35 = (35 x r) + (37 x r) + (12 x r)
⇒
12 x 35 = 84r
∴ r = 5
Thus the radius of the circle is 5cm