2017 AMC 10 B

Complete problem set with solutions and individual problem pages

Problem 22 Medium

The diameter AB of a circle of radius 2 is extended to a point D outside the circle so that BD=3. Point E is chosen so that ED=5 and line ED is perpendicular to line AD. Segment AE intersects the circle at a point C between A and E. What is the area of \triangle ABC? (2017 AMC 10B Problem, Question#22)

  • A.

    \frac{120}{37}

  • B.

    \frac{140}{39}

  • C.

    \frac{145}{39}

  • D.

    \frac{140}{37}

  • E.

    \frac{120}{31}

Answer:D

Solution 1

Notice that ADE and ABC are right triangles.

Then AE=\sqrt{7^{2}+5^{2}}=\sqrt{74},\sin DAE=\sin ABC=\dfrac{5}{\sqrt{74}}=\sin BAE=\dfrac{BC}{4},so BC=\dfrac{20}{\sqrt{74}}. We also find that AC=\dfrac{28}{\sqrt{74}}, and thus the area of ABC is \dfrac{\dfrac{20}{\sqrt{74}}\cdot\dfrac{28}{\sqrt{74}}}{2}=\dfrac{\dfrac{560}{74}}{2}=\dfrac{140}{37}.

Solution 2

We note that \triangle ACB\sim\triangle ADE by AA similarity. Also, since the area of \triangle ADE=\dfrac{7\cdot5}{2}=\dfrac{35}{2} and AE=\sqrt {74},\dfrac{\left[ ABC\right]}{\left[ ADE\right]}=\dfrac{\left[ ABC\right]}{\dfrac{35}{2}}=\left( \dfrac{4}{\sqrt{74}}\right)^{2} , so the area of \triangle ABC=\frac {140}{37}.

Solution 3

As stated before, note that \triangle ACB,\triangle ADE. By similarity, we note that \dfrac{\overline{AC}}{\overline{BC}} is equivalent to \frac 75. We set \overline{AC} to 7x  and \overline{BC} to 5x. By the Pythagorean Theorem, \left( 7x\right)^{2}+\left( 5x\right)^{2}=4^{2}. Combining, 49x^{2}+25x^{2}=16. We can add and divide to get x^2=\frac 8{37}. We square root and rearrange to get x=\dfrac{2\sqrt{74}}{37}. We know that the legs of the triangle are 7x and 5x. Mulitplying x by 7 and 5 eventually gives us \dfrac{14\sqrt{74}}{37}\dfrac{10\sqrt{74}}{37}. We divide this by 2, since \frac 12bh is the formula for a triangle. This gives us \frac {140}{37}.

Solution 4

Let's call the center of the circle that segment AB is the diameter of,O. Note that \triangle ODE is an isosceles right triangle. Solving for side OE, using the Pythagorean theorem, we find it to be 5\sqrt 2. Calling the point where segment OE intersects circle O, the point I, segment IE would be 5\sqrt 2-2. Also, noting that \triangle ADE is a right triangle, we solve for side AE, using the Pythagorean Theorem, and get \sqrt {74}. Using Power of Point on point E, we can solve for CE. We can subtract CE from AE to find AC and then solve for CB using Pythagorean theorem once more (AE)(CE). = (Diameter of circle O+IE ),\left( IE\right)\rightarrow\sqrt{74}\left( CE\right)=\left( 5\sqrt{2}+2\right)\left( 5\sqrt{2}-2\right)\Rightarrow CE=\dfrac{23\sqrt{74}}{37},AC=AE-CE\rightarrow AC=\sqrt{74}-\dfrac{23\sqrt{74}}{37}\Rightarrow AC=\dfrac{14\sqrt{74}}{37},Now to solve for CB:AB^{2}-AC^{2}=CB^{2}\rightarrow4^{2}+\dfrac{14\sqrt{74}^{2}}{37}=CB^{2}\Rightarrow CB=\dfrac{10\sqrt{74}}{37} ,

Note that \triangle ABC is a right triangle because the hypotenuse is the diameter of the circle. Solving for area using the bases AC and BC, we get the area of triangle ABC to be \frac {140}{37}.

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