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Elementary Geometry For College St...

7th Edition
Alexander + 2 others
ISBN: 9781337614085

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BuyFindarrow_forward

Elementary Geometry For College St...

7th Edition
Alexander + 2 others
ISBN: 9781337614085
Textbook Problem

The center of a circle of radius 2 in. is at a distance of 10 in. from the center of a circle of radius length 3 in. To the nearest tenth of an inch, what is the approximate length of a common internal tangent? Use the hint provided in Exercise 38.

(HINT: use similar triangles to find OD and DP. Then apply the Pythagorean Theorem twice.)

To determine

To find:

To find length of a common internal tangent.

Explanation

Given that, the center of a circle of radius 2 in. is at a distance of 10 in. from the center of a circle of radius length 3 in.

That is OA¯=2,PB¯=3andOP¯=10 then AB¯ is common internal tangent.

The diagrammatic representation is given below,

Since AB¯ is common internal tangent therefore, AB¯OA¯andAB¯PB¯

Using the vertical angle theorem to get the following,

mADO=mPDBmOAD=mPBD

Therfore, mOADmPBD

If two triangles are similar, then the ration of any two corresponding segments (such as altitudes, medians, or angle bisectors) equals the ratio of any two corresponding sides.

OA¯PB¯=OD¯PD¯23=OD¯PD¯OD¯PD¯=233OD¯=2PD¯

We know that,

OP¯=OD¯+PD¯10=OD¯+PD¯PD¯=10OD¯. Since OP¯=10

Substitute this in the above equation to get the following,

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