Question

In the given figure a circle touches the side BC of Triangle ABC at P And touches AB and AC produced at Q and R Respectively . If AQ=5cm , Find the perimeter of Triangle ABC .

Answer 1

here is your answer
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Answer 2

The solution to the problem is: Perimeter of Triangle ABC = 0 cm.

We can begin the solution by using the following theorem:

"The tangents from a point outside a circle to the circle are equal in length."

Using this theorem, we can say that AQ = AR. Let this common length be denoted by x. Then, we can write:

AQ = AR = x

Also, we know that BP is a tangent to the circle. Therefore, angle BAP = 90 degrees.

Now, we can use the Pythagorean theorem in triangle AQP to get:

$AP^2 = AQ^2 + PQ^2$

$AP^2 = x^2 + PQ^2$

Similarly, we can use the Pythagorean theorem in triangle APR to get:

$AP^2 = AR^2 + PR^2$

$AP^2 = x^2 + PR^2$

Since $AP^2$ is common to both equations, we can equate them and simplify:

$x^2 + PQ^2 = x^2 + PR^2$

$PQ^2 = PR^2$

This implies that PQ = PR. Let this length be denoted by y. Then, we can write:

PQ = PR = y

Now, we can use the fact that the circle touches BC at P to say that angle PBC = 90 degrees.

Therefore, angle ABC = 180 - 90 - 90 = 0 degrees, which means that triangle ABC is degenerate and reduces to a straight line.

Since a degenerate triangle has zero perimeter, the perimeter of triangle ABC is 0 cm.

Therefore, the solution to the problem is: Perimeter of Triangle ABC = 0 cm.

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