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

Revision as of 19:04, 10 August 2014

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

We angle chase, and find out that:

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

Solution 2

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 angles in quadrilateral $ABCD$ is $40 + 220 + 2\angle BAD = 360$ .

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

Therefore the answer is (D).

2007 AMC 12A (Problems • Answer Key • Resources)
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 (Problems • Answer Key • Resources)
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.

@@ Line 25: / Line 25: @@
 Solving the equation we get <math>\angle BAD = 50</math>.
+Therefore the answer is '''(D)'''.
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Difference between revisions of "2007 AMC 12A Problems/Problem 6"

Revision as of 19:04, 10 August 2014

Contents

Problem

Solution 1

Solution 2

See also