AMC 10 · 2005 · #10
Grade 8 geometry-2dIn △ABC, we have AC=BC=7 and AB=2. Suppose that D is a point on line AB such that B lies between A and D and CD=8. What is BD?
Pick an answer.
AMC 10 2005 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: Triangle $ABC$ has $AC=BC=7$ and $AB=2$, so it is isosceles with the two equal sides meeting at $C$. A point $D$ sits on the line through $A$ and $B$, positioned so that $B$ is between $A$ and $D$, and the distance $CD=8$. Find the length $BD$.
Givens: $AC=BC=7$, so triangle $ABC$ is isosceles with apex $C$; The base $AB=2$; $D$ lies on line $AB$ with $B$ between $A$ and $D$ (so $D$ is past $B$, away from $A$); $CD=8$; Answer choices: (A) $3$, (B) $2\sqrt{3}$, (C) $4$, (D) $5$, (E) $4\sqrt{2}$
Unknowns: The length $BD$, the distance from $B$ to $D$ along the line
Understand
Restated: Triangle $ABC$ has $AC=BC=7$ and $AB=2$, so it is isosceles with the two equal sides meeting at $C$. A point $D$ sits on the line through $A$ and $B$, positioned so that $B$ is between $A$ and $D$, and the distance $CD=8$. Find the length $BD$.
Givens: $AC=BC=7$, so triangle $ABC$ is isosceles with apex $C$; The base $AB=2$; $D$ lies on line $AB$ with $B$ between $A$ and $D$ (so $D$ is past $B$, away from $A$); $CD=8$; Answer choices: (A) $3$, (B) $2\sqrt{3}$, (C) $4$, (D) $5$, (E) $4\sqrt{2}$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #7 Identify Subproblems, #4 Introduce a Variable
Tool #1 (Draw a Diagram) turns the words into a picture: an isosceles triangle standing on base $AB$ with $D$ marked farther along the line. The key move the picture reveals is dropping a perpendicular from the apex $C$ straight down to the line, which splits everything into right triangles. Tool #7 (Identify Subproblems) then handles two right triangles in turn — one to find the height, one that contains $CD$. Tool #4 (Introduce a Variable) names $BD$ so the second right triangle becomes an equation we can solve.
Execute — Answer: A
4.G.A.3 Step 1 Drop a perpendicular from C
- Draw the height from $C$ straight down to line $AB$, meeting it at point $M$.
- Since $AC=BC$, the triangle is symmetric across this height, so $M$ is the midpoint of $AB$.
- That gives $AM=MB=\tfrac{1}{2}\cdot AB=1$.
- The point $D$ is farther out along the line past $B$.
💡 An isosceles triangle folds onto itself along the line from its apex to the middle of the base, so that line cuts the base exactly in half.
8.G.B.7 Step 2 Find the height CM
- Now look at right triangle $CMB$: it has the right angle at $M$, one leg $MB=1$, and hypotenuse $CB=7$.
- By the Pythagorean theorem, $CM^2 = CB^2 - MB^2 = 7^2 - 1^2 = 49 - 1 = 48$, so the height is $CM=\sqrt{48}=4\sqrt{3}$.
💡 The height and half the base are the two legs of a right triangle whose hypotenuse is the known slanted side.
8.G.B.7 Step 3 Set up the right triangle with CD
- Let $BD=x$.
- Since $D$ is past $B$, its distance from the midpoint $M$ is $MD = MB + BD = 1 + x$.
- The segment $CD$ is the hypotenuse of right triangle $CMD$ (right angle at $M$), with legs $CM=4\sqrt{3}$ and $MD=1+x$.
- The Pythagorean theorem gives $CM^2 + MD^2 = CD^2$, that is $48 + (1+x)^2 = 8^2 = 64$.
💡 The same height $CM$ still stands over the line, so $C$, $M$, and $D$ form another right triangle sharing that height.
8.EE.A.2 Step 4 Solve for BD
- Subtract $48$ from both sides: $(1+x)^2 = 16$.
- Taking the positive square root (a length can't be negative), $1+x = 4$, so $x = 3$.
- Therefore $BD=3$, which is choice (A).
💡 Undo the squaring by taking a square root, keeping only the positive value because it measures a distance.
4.G.A.3 Draw the height from $C$ straight down to line $AB$, meeting it at point $M$. Si 8.G.B.7 Now look at right triangle $CMB$: it has the right angle at $M$, one leg $MB=1$, 8.G.B.7 Let $BD=x$. Since $D$ is past $B$, its distance from the midpoint $M$ is $MD = M 8.EE.A.2 Subtract $48$ from both sides: $(1+x)^2 = 16$. Taking the positive square root ( Review
Reasonableness: Plug back in: with $BD=3$, the far leg is $MD = 1+3 = 4$, and the height is $CM=4\sqrt{3}\approx 6.93$. Then $CD = \sqrt{4^2 + (4\sqrt{3})^2} = \sqrt{16+48} = \sqrt{64} = 8$, exactly the given value. The answer is also sensible in size: $CD=8$ is a bit longer than the height $4\sqrt{3}\approx 6.93$, so $D$ must sit a little way past $M$ — a leg of $4$ fits. The trap choice (D) $5$ is the length $AD = AB + BD = 2+3$, the distance from $A$ rather than from $B$; the question asks for $BD$.
Alternative: Use the Law of Cosines without splitting the triangle. In triangle $ABC$, $\cos(\angle ABC) = \frac{BA^2 + BC^2 - AC^2}{2\cdot BA\cdot BC} = \frac{4+49-49}{28} = \frac{1}{7}$. Since $D$ is on the opposite side of $B$ along the line, $\angle CBD = 180^\circ - \angle ABC$, so $\cos(\angle CBD) = -\tfrac{1}{7}$. In triangle $CBD$: $CD^2 = CB^2 + BD^2 - 2\cdot CB\cdot BD\cos(\angle CBD)$, giving $64 = 49 + BD^2 + 2\cdot BD$. So $BD^2 + 2BD - 15 = 0$, which factors as $(BD-3)(BD+5)=0$, and the positive root is $BD=3$.
CCSS standards used (min grade 8)
4.G.A.3Recognize a line of symmetry for a two-dimensional figure (Using the symmetry of the isosceles triangle to conclude the perpendicular from $C$ hits the midpoint of $AB$, so $MB=1$.)8.G.B.7Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Finding the height $CM=4\sqrt{3}$ in right triangle $CMB$, and setting up $48+(1+x)^2=64$ from right triangle $CMD$.)8.EE.A.2Use square root and cube root symbols to represent solutions (Evaluating $\sqrt{48}=4\sqrt{3}$ and solving $(1+x)^2=16$ by taking the positive square root to get $BD=3$.)
⭐ Drop a straight height from the tip of an isosceles triangle to split it into right triangles, then let the Pythagorean theorem turn the known lengths into an equation for the one you want.
⭐ Drop a straight height from the tip of an isosceles triangle to split it into right triangles, then let the Pythagorean theorem turn the known lengths into an equation for the one you want.
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