Exercise a

Chapter 8, Section 8-1, Page 345
enVision Geometry 1st Edition by Al Cuoco
ISBN: 9780328931583

Solution

Verified

Using Triangle Angle-Sum Theorem on ABC\triangle ABC, we have:

mA+mB+mBCA=180m\angle A+ m\angle B+ m\angle BCA=180

45+mB+90=18045+ m\angle B+90=180

mB=45m\angle B=45

Because mA=mBm\angle A= m\angle B, then AB\angle A\cong\angle B so by the Converse of the Isosceles Triangle Theorem, ACBC\overline{AC}\cong\overline{BC}. Hence, AC=BC=52AC=BC=5\sqrt{2}. So, the area of ABC\triangle ABC is:

AABC=12(AC)(BC)A_{ABC}=\dfrac{1}{2}(AC)(BC)

AABC=12(52)(52)A_{ABC}=\dfrac{1}{2}(5\sqrt{2})(5\sqrt{2})

AABC=25 sq. units\color{#c34632}{A_{ABC}=25\text{ sq. units}}

By AA Similarity Theorem, ABCACD\triangle ABC\sim\triangle ACD so mCDA=90m\angle CDA=90 and AD=CDAD=CD. Using Pythagorean Theorem,

AC2=AD2+CD2AC^2=AD^2+CD^2

(52)2=AD2+AD2(5\sqrt{2})^2=AD^2+AD^2

50=2AD250=2AD^2

25=AD225=AD^2

5=AD=CD5=AD=CD

So, the area of ADC\triangle ADC is:

AADC=12(AD)(CD)A_{ADC}=\dfrac{1}{2}(AD)(CD)

AADC=12(5)(5)A_{ADC}=\dfrac{1}{2}(5)(5)

AADC=12.5 sq. units\color{#c34632}{A_{ADC}=12.5\text{ sq. units}}

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