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

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

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 24: / Line 24: @@
 This implies that  <math>\angle BAD = \angle BCD</math>.
-Then the sum of the angles in quadrilateral <math>ABCD</math> is  <math>40 + 220 + 2\angle BAD = 360</math>.
+Then the sum of the interior angles of quadrilateral <math>ABCD</math> is  <math>40 + 220 + 2\angle BAD = 360</math>.
-Solving the equation we get <math>\angle BAD = 50</math>.
+Solving the equation, we get <math>\angle BAD = 50</math>.
-Therefore the answer is '''(D)'''.
+Therefore the answer is <math>\mathrm{(D)}</math>.
-<math>\mathrm{D}</math>
+==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
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