Thread: to find the lengths of the sides of triangle

A triangle ABC circumscribes a circle of radius 4 cm. AB, BC and CA are the tangents of the given circle. D is the point of contact such that AD = 6 cm and DB = 8 cm. Find the lengths of other sides viz. BC and CA. This problem has come under Tangent to the Circle. I do not know how to go about?

Kindly enlighten me.

Advance Thanks.

Aranga

Hi Aranga,

I've got some thoughts on how to find the length of AB.

1. draw a circle and select three dots on in as A, B, and C;
2. DB=8cm, since the radius is 4cm, then DB=8=2x4 must be a diameter;
3. then triangle BAD is a right triangle with angle-BAD being 90 degree;
4. now in the right triangle BAD, you know lengths of two sides AD and BD, guess you could figure out how to find the length of the third side, i.e., AB.

*about the lengths of AC and BC, I find the provided constraints sort of insufficient; Is there any info that may be provided in the question statement?

No other information other than the ones I have provided. Hence, even I could not think beyond that. probably I have to use some properties pertaining to tangent / radius of the circle. Further, AB = AD + DB = 6 + 8 = 14 cm.

angles A and B are defined by the givens.The third angle follows to give the info to solve for the lengths of BC and AC

I did n't get you. How to go about. I do n't know. could you please elaborate further.

construct the figure.Draw a circle of radius 4 cm center o. Draw a tangent to circle perpendicular to a radius.From point of tangencymark off AD =6 cm and DB = 8 cm.FRom A draw another tangentmeeting circle at E. Do the samefrom B.Tangents to a circle from a point outside the circle are equal in length.Look at the right triangles which can beformed to the three sides of triangle and that by trig you can find angles A and B. Angle C = 180-A-B. There is enough info to calculate BC and AC

Originally Posted by arangu1508

No other information other than the ones I have provided. Hence, even I could not think beyond that. probably I have to use some properties pertaining to tangent / radius of the circle. Further, AB = AD + DB = 6 + 8 = 14 cm.

Hi Aranga,

Since triangle BAD is a right triangle and angle-BAD is the right angle, it should be:
AD^2 + AB^2 = BD^2 (not AB=AD+BD as you suggested).

Hope it's helpful and welcome further discussions.

Originally Posted by kylehk

Hi Aranga,

Since triangle BAD is a right triangle and angle-BAD is the right angle, it should be:
AD^2 + AB^2 = BD^2 (not AB=AD+BD as you suggested).

Hope it's helpful and welcome further discussions.

Hello Arangu.
What say you?

Originally Posted by bjhopper

Hello Arangu.
What say you?

Just saw you're suggesting that AB = AD + DB = 6 + 8 = 14 cm. You may consider more about it.

Since is a right triangle and is the right angle, we have . So you could solve the length of AB in this equation instead of AD+BD.

Originally Posted by ibdutt

WOW, excellent!!!

Here's another solution that uses trigonometry.

tan 1/2A =4/6 1/2A =m1 deg
tan1/2B =4/8 1/2B =m2 deg
measure 1 /2 C = 90--m1- m2
length of tangents from C =7