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Math
GeometryElementary Geometry for College Students6th Edition

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Chapter 11.CR, Problem 14CR
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Elementary Geometry for College St...

6th Edition
Daniel C. Alexander + 1 other
ISBN: 9781285195698

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Elementary Geometry for College St...

6th Edition
Daniel C. Alexander + 1 other
ISBN: 9781285195698
Textbook Problem
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In Exercises 13 to 17, use the Law of S i n e s or the Law of C o s i n e s to solve each problem. Angle measures should be found to the nearest degree; distances should be found to the nearest tenth of a unit.

Two sides of a parallelogram are 50 cm and 70 cm long. Find the length of the shorter diagonal if a larger angle of the parallelogram measures 105 ° .

To determine

To find:

The length of the shorter diagonal of a parallelogram.

Explanation

Procedure used:

In any triangle ABC such that

AB=c,

BC=a,

CA=b,

m∠A=α,

m∠B=β, and

m∠C=γ.

The Law of cosines is given by

a2=b2+c2-2bccos⁡α

b2=c2+a2-2cacos⁡β

c2=a2+b2-2abcos⁡γ

The Law of sines is given by

asin⁡α=bsin⁡β=csin⁡γ

Calculation:

Given:

Length of the sides of the parallelogram are 50 cm and 70 cm.

The larger angle =105°

Let ABCD be the parallelogram with diagonals AC and BD, which intersect at ‘O’.

From the above figure we have

AD=BC=70 cm

CD=AB=50 cm

m∠ADC=105°

The diagonal BD is shorter than the diagonal AC.

To find the length of the shorter diagonal BD, it is necessary to find the angle m∠BAD.

By the property of parallelogram, consecutive angles of any parallelogram is supplementary

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Sect-11.1 P-1ESect-11.1 P-2ESect-11.1 P-3ESect-11.1 P-4ESect-11.1 P-5ESect-11.1 P-6ESect-11.1 P-7ESect-11.1 P-8ESect-11.1 P-9ESect-11.1 P-10E
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