AMC 8 · 2015 · #6

Easy mode Grade 8

Problem

Picture a triangle $ABC$ . Two of its sides have the same length: $AB = 29$ and $BC = 29$ . The third side is $AC = 42$ .

What is the area of this triangle?

$\textbf{(A) }100\qquad\textbf{(B) }420\qquad\textbf{(C) }500\qquad\textbf{(D) }609\qquad \textbf{(E) }701$

Pick an answer.

(A)

100

(B)

420

(C)

500

(D)

609

(E)

701

View mode:

Toolkit + CCSS Solution

Understand

Restated: Triangle $ABC$ is isosceles with $AB = BC = 29$ and base $AC = 42$. Find its area.

Givens: $AB = BC = 29$ (the two equal legs); $AC = 42$ (the base); Answer choices: (A) $100$, (B) $420$, (C) $500$, (D) $609$, (E) $701$

Unknowns: The area of $\triangle ABC$

Understand

Restated: Triangle $ABC$ is isosceles with $AB = BC = 29$ and base $AC = 42$. Find its area.

Givens: $AB = BC = 29$ (the two equal legs); $AC = 42$ (the base); Answer choices: (A) $100$, (B) $420$, (C) $500$, (D) $609$, (E) $701$

Plan

Primary tool: #12 Draw a Picture

Secondary: #13 Use Symmetry

We have three side lengths but no height, and the area formula needs a base and a height. Tool #12 (Draw a Picture) makes the isosceles triangle visible so we can add an altitude from $B$ to the base $AC$. Tool #13 (Use Symmetry) is the key insight: because $AB = BC$, the altitude from $B$ is also the perpendicular bisector of $AC$, so it splits the base exactly in half ($21$ and $21$) and creates two congruent right triangles. From there, the Pythagorean theorem gives the height in one line, and the area formula finishes the job.

Execute — Answer: B

#13 Use Symmetry 4.G.A.3 Step 1

Draw $\triangle ABC$ with $AB = BC = 29$ and base $AC = 42$.
Drop an altitude from the apex $B$ to the base, meeting $AC$ at point $D$.
Because the triangle is isosceles, this altitude is the axis of symmetry, so $D$ is the midpoint of $AC$ and $AD = DC = 21$.

$$AD = DC = \dfrac{AC}{2} = \dfrac{42}{2} = 21$$

💡 Recognizing the line of symmetry in an isosceles triangle is a Grade 4 idea about symmetric figures.

#12 Draw a Picture 8.G.B.7 Step 2

Look at right triangle $\triangle ABD$.
It has hypotenuse $AB = 29$, one leg $AD = 21$, and the other leg is the height $h = BD$ that we want.
Apply the Pythagorean theorem.

$$AD^2 + BD^2 = AB^2 \;\Rightarrow\; 21^2 + h^2 = 29^2$$

💡 The altitude split the isosceles triangle into two right triangles — a Grade 8 "apply the Pythagorean theorem" setup.

#12 Draw a Picture 8.EE.A.2 Step 3

Solve for $h$.
Compute the two squares and subtract.

$$h^2 = 29^2 - 21^2 = 841 - 441 = 400 \;\Rightarrow\; h = \sqrt{400} = 20$$

💡 Taking the positive square root of $400$ is the Grade 8 "square roots of small perfect squares" skill — and $20\text{-}21\text{-}29$ is a well-known Pythagorean triple.

#12 Draw a Picture 6.G.A.1 Step 4

Use the area formula with base $AC = 42$ and height $BD = 20$.

$$\text{Area} = \tfrac{1}{2} \times 42 \times 20 = 21 \times 20 = 420 \;\Rightarrow\; \textbf{(B)}$$

💡 Applying $\tfrac{1}{2} \times b \times h$ to a triangle is the Grade 6 area formula.

[1] #13 4.G.A.3 Draw $\triangle ABC$ with $AB = BC = 29$ and base $AC = 42$. Drop an altitude fr

[2] #12 8.G.B.7 Look at right triangle $\triangle ABD$. It has hypotenuse $AB = 29$, one leg $AD

[3] #12 8.EE.A.2 Solve for $h$. Compute the two squares and subtract.

[4] #12 6.G.A.1 Use the area formula with base $AC = 42$ and height $BD = 20$.

Review

Reasonableness: Sanity-check with the bounding rectangle: a $42 \times 20$ rectangle has area $840$, and a triangle that fits inside it covers exactly half, giving $420$. That matches our answer. Also, $20\text{-}21\text{-}29$ is a standard Pythagorean triple ($20^2 + 21^2 = 400 + 441 = 841 = 29^2$), so the height is exact, not an approximation.

Alternative: Tool #17 (Use a Formula) — Heron's formula. The semi-perimeter is $s = \tfrac{29 + 29 + 42}{2} = 50$, so $\text{Area} = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{50 \cdot 21 \cdot 21 \cdot 8} = \sqrt{176400} = 420$. Same answer, but Heron's formula is overkill once you see the symmetry trick.

CCSS standards used (min grade 8)

4.G.A.3 Recognize a line of symmetry for a two-dimensional figure (Seeing that the altitude from $B$ in the isosceles triangle is the line of symmetry, so it bisects $AC$ into two segments of length $21$.)
6.G.A.1 Find the area of triangles and other polygons (Applying $\text{Area} = \tfrac{1}{2} \times \text{base} \times \text{height} = \tfrac{1}{2} \times 42 \times 20 = 420$.)
8.G.B.7 Apply the Pythagorean Theorem to determine unknown side lengths in right triangles (Using $21^2 + h^2 = 29^2$ in the right triangle $\triangle ABD$ to find the altitude.)
8.EE.A.2 Use square root symbols to represent solutions, evaluate square roots of small perfect squares (Computing $h = \sqrt{400} = 20$ to get the height from $h^2 = 400$.)

⭐ The symmetry of an isosceles triangle turns this into a Grade 8 Pythagorean theorem problem with the $20\text{-}21\text{-}29$ triple — then a Grade 6 area formula finishes it.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Review

CCSS standards used (min grade 8)

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