AMC 8 · 2007 · #14

Grade 8 geometry-2d

Archetype Arithmetic Expression Decomposition └ Sub-archetype Weighted Group Aggregation

area-trianglespythagorean-theoremisosceles-triangle identify-subproblems ↑ Prerequisites: area-trianglespythagorean-theorem

📏 Short solution 💡 2 insights

Problem

The base of isosceles $\triangle ABC$ is $24$ and its area is $60$ . What is the length of one
of the congruent sides?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2007 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.

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Toolkit + CCSS Solution

Understand

Restated: Isosceles $\triangle ABC$ has base $24$ and area $60$. Find the length of one of the two congruent sides.

Givens: $\triangle ABC$ is isosceles, so the two non-base sides have the same length; Base length is $24$; Area is $60$; Answer choices: (A) $5$, (B) $8$, (C) $13$, (D) $14$, (E) $18$

Unknowns: The length of one congruent side

Understand

Restated: Isosceles $\triangle ABC$ has base $24$ and area $60$. Find the length of one of the two congruent sides.

Givens: $\triangle ABC$ is isosceles, so the two non-base sides have the same length; Base length is $24$; Area is $60$; Answer choices: (A) $5$, (B) $8$, (C) $13$, (D) $14$, (E) $18$

Plan

Primary tool: #1 Draw a Diagram

Secondary: #5 Look for a Pattern

Tool #1 (Draw a Diagram) is the natural opening: sketch the isosceles triangle and drop the altitude from the apex to the base. Because the triangle is isosceles, that altitude bisects the base, splitting the figure into two congruent right triangles. The picture turns the problem into a Pythagorean Theorem exercise where the congruent side is the hypotenuse. Tool #5 (Look for a Pattern) is the finisher: once the two legs are $5$ and $12$, recognizing the classic $5$-$12$-$13$ Pythagorean triple gives the hypotenuse instantly — no square root needed.

Execute — Answer: C

#1 Draw a Diagram 4.G.A.1 Step 1

Sketch $\triangle ABC$ with base $BC = 24$ and apex $A$.
Drop the altitude from $A$ to the base; call its foot $M$.
In an isosceles triangle, the altitude to the base also bisects the base, so $BM = MC = 12$ and $\angle AMB = 90^\circ$.
This produces two congruent right triangles, $\triangle AMB$ and $\triangle AMC$.

$$BM = MC = \tfrac{24}{2} = 12, \quad \angle AMB = 90^\circ$$

💡 Adding the altitude is the Grade 4 move of drawing a perpendicular line to expose a right angle hidden in the figure.

#1 Draw a Diagram 6.G.A.1 Step 2

Use the area to find the altitude $AM$.
The area formula $\text{Area} = \tfrac{1}{2} \times \text{base} \times \text{height}$ gives a single linear equation for $AM$.

$$60 = \tfrac{1}{2} \times 24 \times AM \;\Rightarrow\; 60 = 12 \cdot AM \;\Rightarrow\; AM = 5$$

💡 The Grade 6 triangle-area formula reads backwards: given area and base, the height is forced.

#5 Look for a Pattern 8.G.B.7 Step 3

Now $\triangle AMB$ is a right triangle with legs $AM = 5$ and $BM = 12$, and hypotenuse $AB$ — exactly one of the congruent sides we want.
Spot the pattern: $5$ and $12$ are the legs of the famous $5$-$12$-$13$ Pythagorean triple, so the hypotenuse is $13$ without any square-root work.

$$AB^2 = 5^2 + 12^2 = 25 + 144 = 169 \;\Rightarrow\; AB = 13 \;\Rightarrow\; \textbf{(C)}$$

💡 Recognizing the $5$-$12$-$13$ triple is a Grade 8 Pythagorean Theorem shortcut. The pattern saves the square root.

[1] #1 4.G.A.1 Sketch $\triangle ABC$ with base $BC = 24$ and apex $A$. Drop the altitude from

[2] #1 6.G.A.1 Use the area to find the altitude $AM$. The area formula $\text{Area} = \tfrac{1

[3] #5 8.G.B.7 Now $\triangle AMB$ is a right triangle with legs $AM = 5$ and $BM = 12$, and hy

Review

Reasonableness: Check the triangle inequality: sides $13$, $13$, $24$ satisfy $13 + 13 = 26 > 24$, so the triangle exists (barely — it is long and flat). Recompute the area to confirm: with base $24$ and altitude $5$, area $= \tfrac{1}{2} \cdot 24 \cdot 5 = 60$, which matches. Among the choices, $5$ is the altitude (a distractor), $8$ is half the perimeter minus the base (a distractor), $14$ and $18$ would force altitudes of $\sqrt{14^2 - 12^2} = \sqrt{52}$ and $\sqrt{18^2 - 12^2} = \sqrt{180}$, neither matching $5$. Only $13$ fits.

Alternative: Tool #6 (Guess & Check) on the answer choices works without algebra. For each choice $s$, the altitude must satisfy $s^2 = 12^2 + h^2$, so $h = \sqrt{s^2 - 144}$, and the area must equal $\tfrac{1}{2} \cdot 24 \cdot h = 60$, i.e. $h = 5$. Test $s = 13$: $h = \sqrt{169 - 144} = \sqrt{25} = 5$. Match — answer (C). Same result, slower path because each choice has to be tried.

CCSS standards used (min grade 8)

4.G.A.1 Draw and identify points, lines, line segments, rays, angles, and perpendicular lines (Adding the altitude from the apex to the base and marking the right angle at its foot, which exposes two congruent right triangles inside the isosceles triangle.)
6.G.A.1 Find the area of right triangles, other triangles, and special quadrilaterals (Using $\text{Area} = \tfrac{1}{2} \cdot \text{base} \cdot \text{height}$ with area $60$ and base $24$ to solve for the altitude $AM = 5$.)
8.G.B.7 Apply the Pythagorean Theorem to determine unknown side lengths in right triangles in real-world and mathematical problems (Computing the congruent side as the hypotenuse of the $5$-$12$-right triangle: $5^2 + 12^2 = 13^2$.)

⭐ Drop the altitude in an isosceles triangle and it bisects the base — the picture hands you a right triangle, and the $5$-$12$-$13$ Pythagorean triple finishes the problem.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 8)

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