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Question

In the given figure, a triangle $$ABC$$ is drawn to circumscribe a circle of radius 3 cm such that the segments BD and DC into which BC is divided by the point of contact D, are of lengths 6 cm and 9 cm respectively. If the area of $$\Delta ABC = 54 cm^2$$ then find the lengths of sides AB and AC.
1713958_dab3c99c4bf24147bc5bdf25f68920bd.png


Solution

In the given figure, $$\Delta ABC$$ circumscribed the circle with centre O.

Radius $$OD = 3 cm$$

$$BD = 6 cm, DC = 9 cm$$

Area of $$\Delta ABC = 54 cm^2$$

To find :  Length of $$AB$$ and $$AC$$.

$$AF$$ and $$EA$$ are tangents to the circle at point A.

Let $$AF = EA = x$$

$$BD$$ and $$BF$$ are tangents to the circle at point B.

$$BD = BF = 6 cm$$

$$CD$$ and $$CE$$ are tangents to the circle at point C.

$$CD = CE = 9 cm$$

Now, new sides of the triangle are:

$$AB = AF + FB = x + 6 cm$$
$$AC = AE + EC = x + 9 cm$$
$$BC = BD + DC = 6 + 9 = 15 cm$$

Now, using Heron's formula:

Area of triangle $$ABC = \sqrt{s(s - a) (s - b) (s-c)}$$

Where $$S = \dfrac{a + b + c}{2}$$

$$S = 1/2(x + 6 + x + 9+15) = x + 15$$

Area of $$ABC = \sqrt{(x + 15) (x + 15 - (x + 6)) (x + 15 - (x - 9)) (x + 15 - 15)}$$ 

Or

$$54 = \sqrt{(x + 15) (9)(6) (x)}$$

Squaring both sides, we have
$$54^2 = 54 x (x + 15)$$
$$x^2 + 15x - 54 = 0$$

Solve this quadratic equation and find the value of x.
$$x^2 + 18x - 3x - 54 = 0$$
$$x(x + 18) - 3(x + 18) = 0$$
$$(x - 3) (x + 18) = 0$$

Either $$x = 3$$ or $$x = - 18$$

But x cannot be negative.

So, $$x = 3$$

Answer :-

$$AB = x + 6 = 3 + 6 = 9 cm$$

$$AC = x + 9 = 3 + 9 = 12 cm$$

1540967_1713958_ans_5393ed5ecdc3467aa5d7f382767b9f08.png

Mathematics
RS Agarwal
Standard X

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