Assume that $ AD $ is the geometric mean of $ BD $ and $ DC $ in $\triangle ABC$ shown in the accompanying drawing.The diagram features a right triangle $ \triangle ABC $ with altitude $ AD $ from vertex $ A $ to the hypotenuse $ BC $. Point $ D $ marks where the altitude meets $ BC $.(a) Suppose $ BD = 3 $ and $ DC = 6 $.Write a proportion involving $ AD $ that can be used to solve for $ AD $.\[\frac{3}{AD} = \frac{AD}{6}\] This proportion sets up the relationship based on the property of geometric means in right triangles.

Name: Assume that $ AD $ is the geometric mean of $ BD $ and $ DC $ i
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Description: Assume that $ AD $ is the geometric mean of $ BD $ and $ DC $ in $\triangle ABC$ shown in the accompanying drawing.The diagram features a right triangle $ \triangle ABC $ with altitude $ AD $ from vertex $ A $ to the hypotenuse \( BC \

in ∆ABD and ∆CAD ∠ADB=∠CDA ∵ ∠ADB=∠CDA=90∘∠ABD=∠CAD∠DAB=∠ACD so,…

Answered: Assume that AD is the geometric mean of…

Math

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Assume that AD is the geometric mean of BD and DC in AABC shown in the accompanying drawing. (a) Suppose BD = 3 and DC = 6. Write a proportion involving AD that can be used to solve for AD. AD

Elementary Geometry For College Students, 7e

7th Edition

ISBN:9781337614085

Author:Alexander, Daniel C.; Koeberlein, Geralyn M.

Publisher:Alexander, Daniel C.; Koeberlein, Geralyn M.

ChapterP: Preliminary Concepts

SectionP.CT: Test

Problem 1CT

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Concept explainers

2-Proportion Test

In a 2-proportion test, two separate proportions are measured at the same time, and their similarity is evaluated. It is also known as a 2-proportion z-test, and a comparison of two different proportions is done in this method.

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Question

Assume that $ AD $ is the geometric mean of $ BD $ and $ DC $ in $\triangle ABC$ shown in the accompanying drawing.

The diagram features a right triangle $ \triangle ABC $ with altitude $ AD $ from vertex $ A $ to the hypotenuse $ BC $. Point $ D $ marks where the altitude meets $ BC $.

(a) Suppose $ BD = 3 $ and $ DC = 6 $.

Write a proportion involving $ AD $ that can be used to solve for $ AD $.

\[
\frac{3}{AD} = \frac{AD}{6}
\]

This proportion sets up the relationship based on the property of geometric means in right triangles.

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