AMC 10 · 2009 · #10
Grade 8 geometry-2dTriangle ABC has a right angle at B. Point D is the foot of the altitude from B, AD=3, and DC=4. What is the area of △ABC?
Pick an answer.
AMC 10 2009 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: Right triangle ABC has its right angle at B. The foot of the altitude from B to the hypotenuse AC is called D. The two pieces of the hypotenuse are AD = 3 and DC = 4. Find the area of triangle ABC.
Givens: Triangle ABC has a right angle at B.; D is the foot of the altitude drawn from B to AC.; AD = 3 and DC = 4.
Unknowns: The area of triangle ABC.
Understand
Restated: Right triangle ABC has its right angle at B. The foot of the altitude from B to the hypotenuse AC is called D. The two pieces of the hypotenuse are AD = 3 and DC = 4. Find the area of triangle ABC.
Givens: Triangle ABC has a right angle at B.; D is the foot of the altitude drawn from B to AC.; AD = 3 and DC = 4.
Plan
Primary tool: #7 Identify Subproblems
Secondary: #1 Draw a Diagram, #4 Introduce a Variable
The area needs a base and a matching height. AC is an easy base since AC = AD + DC = 7, so the real subproblem is the height BD. The altitude from the right angle cuts the big right triangle into two smaller right triangles that are copies of the whole, and that similarity pins down BD. Solve that piece, then combine.
Execute — Answer: B
8.G.A.4 Step 1 Split with the altitude
- The altitude BD drops from the right angle onto the hypotenuse AC.
- It cuts triangle ABC into two smaller right triangles, ABD and BCD.
- In triangle ABD the acute angle at A plus angle ABD make 90 degrees; in the whole triangle the angle at A plus the angle at C also make 90 degrees.
- So angle ABD equals angle C, and both small triangles share a right angle.
- That makes triangle ABD similar to triangle BCD.
💡 An altitude from a right angle always makes two mini-triangles shaped exactly like the original.
8.G.A.4 Step 2 Similar triangles give BD squared
- In similar triangles matching sides are in the same ratio.
- Line up triangle ABD with triangle BCD so that the short leg AD in the first matches the leg BD in the second, and the leg BD in the first matches the leg DC in the second.
- That gives AD over BD equals BD over DC.
- Cross-multiplying, BD times BD equals AD times DC, so BD squared is 3 times 4, which is 12.
💡 The altitude to the hypotenuse is the geometric mean of the two pieces it lands between.
8.EE.A.2 Step 3 Take the square root
- BD is a length, so it is the positive square root of 12.
- Since 12 is 4 times 3, the square root of 12 is the square root of 4 times the square root of 3, which is 2 times the square root of 3.
💡 Pull the perfect-square factor out from under the root to simplify it.
6.G.A.1 Step 4 Compute the area
- Use AC as the base and BD as the height, because BD is perpendicular to AC.
- The base is AC = AD + DC = 3 + 4 = 7, and the height is BD = 2 times the square root of 3.
- Area is half the base times the height, so it is one half times 7 times 2 root 3, which is 7 root 3.
- The answer is (B).
💡 With the altitude as height and the full hypotenuse as base, the area is just half their product.
8.G.A.4 The altitude BD drops from the right angle onto the hypotenuse AC. It cuts trian 8.G.A.4 In similar triangles matching sides are in the same ratio. Line up triangle ABD 8.EE.A.2 BD is a length, so it is the positive square root of 12. Since 12 is 4 times 3, 6.G.A.1 Use AC as the base and BD as the height, because BD is perpendicular to AC. The Review
Reasonableness: The height BD = 2 root 3 is about 3.46, and the base AC is 7, so the area is about one half times 7 times 3.46, roughly 12.1. The answer 7 root 3 is about 12.12, which matches. Checking the choices: 4 root 3 (about 6.9) and 14 root 3 (about 24.2) and 42 are too far off, and 21 would need a height near 6, which is impossible since BD sits inside a triangle only 7 wide. Only 7 root 3 fits.
Alternative: Use the Pythagorean theorem twice with the leg relations. Each leg squared equals its own hypotenuse piece times the whole hypotenuse: AB squared = AD times AC = 3 times 7 = 21, and BC squared = DC times AC = 4 times 7 = 28. Then AB = root 21 and BC = 2 root 7, and the area is one half times root 21 times 2 root 7 = root 147 = 7 root 3, the same answer.
CCSS standards used (min grade 8)
8.G.A.4Understand that a two-dimensional figure is similar to another using transformations (Recognizing that the altitude splits the right triangle into two smaller triangles similar to each other, then reading off the proportion AD/BD = BD/DC to get BD squared = 12.)8.EE.A.2Use square root and cube root symbols to represent solutions (Taking the square root of 12 and simplifying it to 2 root 3.)6.G.A.1Find area of triangles, special quadrilaterals, and polygons by composing (Combining base AC = 7 and height BD = 2 root 3 in the one-half-base-times-height formula.)
⭐ When a right angle drops a height onto the far side, that height is the geometric mean of the two pieces it lands between.
⭐ When a right angle drops a height onto the far side, that height is the geometric mean of the two pieces it lands between.
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