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

Latest revision as of 17:28, 10 December 2024

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:

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

~minor edits by mobius247

Solution 2

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

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 1: / Line 1: @@
 {{duplicate|[[2007 AMC 12A Problems|2007 AMC 12A #6]] and [[2007 AMC 10A Problems/Problem 8|2007 AMC 10A #8]]}}
 ==Problem==
-Triangles <math>ABC</math> and <math>ADC</math> are [[isosceles]] with <math>AB=BC</math> and <math>AD=DC</math>. Point <math>D</math> is inside triangle <math>ABC</math>, angle <math>ABC</math> measures 40 degrees, and angle <math>ADC</math> measures 140 degrees. What is the degree measure of angle <math>BAD</math>?
+Triangles <math>ABC</math> and <math>ADC</math> are [[isosceles]] with <math>AB=BC</math> and <math>AD=DC</math>. Point <math>D</math> is inside triangle <math>ABC</math>, angle <math>ABC</math> measures <math>40</math> degrees, and angle <math>ADC</math> measures <math>140</math> degrees. What is the degree measure of angle <math>BAD</math>?
 <math>\mathrm{(A)}\ 20\qquad \mathrm{(B)}\ 30\qquad \mathrm{(C)}\ 40\qquad \mathrm{(D)}\ 50\qquad \mathrm{(E)}\ 60</math>

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