Length of Segment

In triangle ABC, AC=6 and BC=5. Point D on \overline{AB} divides it into segments of length AD=1 and DB=3. What is the length of segment CD?

Source: NCTM Mathematics Teacher 2006

SOLUTION

Using the Heron’s area formula to calculate the area of triangle ABC
Area = \sqrt{s(s-4)(s-5)(s-6)} where s=(4+5+6)/2=7.5
Area = \sqrt{7.5(7.5-4)(7.5-5)(7.5-6)}
=\sqrt{98.4375}
=9.92
CE is the altitude of triangle ABC
Area = \dfrac{1}{2}AB\times CE
9.92=(4\times EC)/2
EC=4.96
Using the Pythagorean theorem in right triangle AEC
AE^2+EC^2=AC^2
AE^2+4.96^2=6^2
AE^2=11.40
AE=3.38
DE=AE-AD
=3.38-1
=2.38
Using the Pythagorean theorem in right triangle DEC
DE^2+EC^2=CD^2
2.38^2+4.96^2=CD^2
30.27=CD^2
CD=5.5

Answer: 5.5

Alternative solution

Using the law of cosines in triangle BDC
5^2=3^2+x^2-2(3x)\text{cos}\,\theta
25=9+x^2-6x\text{cos}\,\theta\qquad (1)
Using the law of cosines in triangle ADC
6^2=1^2+x^2-2(1x)\text{cos}\,(180-\theta)
36=1+x^2-2x(\textrm{-}\text{cos}\,\theta)
36=1+x^2+2x\text{cos}\,\theta\qquad (2)
Multiply Eq. (2) by 3 and add Eq. (1)
25=9+x^2-6x\text{cos}\,\theta
108=3+3x^2+6x\text{cos}\,\theta
———————————–
133=12+4x^2
4x^2=121
x^2=30.25
x=5.5

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About mvtrinh

Retired high school math teacher.
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