A circle touches the side bc of triangle abc at p and the sides ab and ac produced at q and r respectively. If the sides ab bc and ac of the triangle are 13cm, 15cm,18cm respectively, show that aq= ar=23 cm
Answers
If the sides AB, BC and AC of the triangle are 13cm, 15cm,18cm respectively, then AQ = AR = 23 cm .
Step-by-step explanation:
It is given that, (as shown in the figure attached below)
ABC is a triangle with AB = 13 cm, BC = 15 cm and AC = 18 cm
A circle touches the triangle ABC at P
Side AB is produced to Q and Side AC is produced to R
We know that the lengths of the tangents drawn from an external point to a circle are equal.
Therefore, we have
AQ = AR, BQ = BP and CP = CR.
Now, we have
The Perimeter of ΔABC as,
= AB + BC + CA
= AB + (BP + PC) + (AR – CR)
= (AB + BQ) + (PC) + (AQ – PC) …….. [since AQ = AR, BQ = BP, CP = CR]
= AQ + AQ
= 2AQ
∴ AQ = AR = ½ * (Perimeter of ΔABC) = ½ * [13+15+18] = ½ * 46 = 23 cm
Hence proved that AQ = AR = 23 cm .
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