Difference between revisions of "2007 AMC 12A Problems/Problem 6"

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[[Image:2007_AMC12A-6.png]]
 
[[Image:2007_AMC12A-6.png]]
  
We angle chase, and find out that:
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We angle chase and find out that:
* <math>DAC=\frac{180-140}{2} = 20</math>
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* <math>\angle DAC=\frac{180-140}{2} = 20</math>
* <math>BAC=\frac{180-40}{2} = 70</math>
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* <math>\angle BAC=\frac{180-40}{2} = 70</math>
* <math>BAD=BAC-DAC=50\ \mathrm{(D)}</math>
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* <math>\angle BAD=\angle BAC- \angle DAC=50\ \mathrm{(D)}</math>
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~minor edits by mobius247
  
 
==Solution 2==
 
==Solution 2==
[[File:Mihir_Borkar_Solution_2007_AMC_10A_Problem_6.png]]
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[[File:Mihir_Borkar_Solution_2007_AMC_10A_Problem_6_p_2.png]]
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 +
Since triangle <math>ABC</math> is isosceles we know that angle <math>\angle BAC = \angle BCA</math>.
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Also since triangle <math>ADC</math> is isosceles we know that <math>\angle DAC = \angle DCA</math>.
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This implies that  <math>\angle BAD = \angle BCD</math>.
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Then the sum of the interior angles of quadrilateral <math>ABCD</math> is  <math>40 + 220 + 2\angle BAD = 360</math>.
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Solving the equation, we get <math>\angle BAD = 50</math>.
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Therefore the answer is <math>\mathrm{(D)}</math>.
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==Solution 3==
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Using the previous drawing we know that ACD and DAC are both equal to 20 (40/2 or (180-40)/2). We also know that BAC and BCA are both 70 or (180-40)/2. Thus BDA is 70-20 or 50. Easy. -RealityWrites
  
 
==See also==
 
==See also==

Latest revision as of 23:20, 29 October 2022

The following problem is from both the 2007 AMC 12A #6 and 2007 AMC 10A #8, so both problems redirect to this page.

Problem

Triangles $ABC$ and $ADC$ are isosceles with $AB=BC$ and $AD=DC$. Point $D$ is inside triangle $ABC$, angle $ABC$ measures 40 degrees, and angle $ADC$ measures 140 degrees. What is the degree measure of angle $BAD$?

$\mathrm{(A)}\ 20\qquad \mathrm{(B)}\ 30\qquad \mathrm{(C)}\ 40\qquad \mathrm{(D)}\ 50\qquad \mathrm{(E)}\ 60$

Solution 1

2007 AMC12A-6.png

We angle chase and find out that:

  • $\angle DAC=\frac{180-140}{2} = 20$
  • $\angle BAC=\frac{180-40}{2} = 70$
  • $\angle BAD=\angle BAC- \angle DAC=50\ \mathrm{(D)}$

~minor edits by mobius247

Solution 2

Mihir Borkar Solution 2007 AMC 10A Problem 6 p 2.png

Since triangle $ABC$ is isosceles we know that angle $\angle BAC = \angle BCA$.

Also since triangle $ADC$ is isosceles we know that $\angle DAC = \angle DCA$.

This implies that $\angle BAD = \angle BCD$.

Then the sum of the interior angles of quadrilateral $ABCD$ is $40 + 220 + 2\angle BAD = 360$.

Solving the equation, we get $\angle BAD = 50$.

Therefore the answer is $\mathrm{(D)}$.

Solution 3

Using the previous drawing we know that ACD and DAC are both equal to 20 (40/2 or (180-40)/2). We also know that BAC and BCA are both 70 or (180-40)/2. Thus BDA is 70-20 or 50. Easy. -RealityWrites

See also

2007 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions
2007 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 9
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png