AMC 10 · 2005 · #25
Grade 6 geometry-2dIn △ABC we have AB=25, BC=39, and AC=42. Points D and E are on AB and AC respectively, with AD=19 and AE=14. What is the ratio of the area of triangle ADE to the area of the quadrilateral BCED?
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: In triangle $ABC$ the sides are $AB=25$, $BC=39$, $AC=42$. Point $D$ sits on side $AB$ with $AD=19$, and point $E$ sits on side $AC$ with $AE=14$. Segment $DE$ cuts the triangle into a small corner triangle $ADE$ and the four-sided region $BCED$ that is left. Find the ratio of the area of triangle $ADE$ to the area of quadrilateral $BCED$.
Givens: Triangle $ABC$ with $AB=25$, $BC=39$, $AC=42$; $D$ is on $\overline{AB}$ with $AD=19$; $E$ is on $\overline{AC}$ with $AE=14$; Answer choices: (A) $\tfrac{266}{1521}$, (B) $\tfrac{19}{75}$, (C) $\tfrac13$, (D) $\tfrac{19}{56}$, (E) $1$
Unknowns: The ratio $[ADE]:[BCED]$ of the corner triangle's area to the quadrilateral's area
Understand
Restated: In triangle $ABC$ the sides are $AB=25$, $BC=39$, $AC=42$. Point $D$ sits on side $AB$ with $AD=19$, and point $E$ sits on side $AC$ with $AE=14$. Segment $DE$ cuts the triangle into a small corner triangle $ADE$ and the four-sided region $BCED$ that is left. Find the ratio of the area of triangle $ADE$ to the area of quadrilateral $BCED$.
Givens: Triangle $ABC$ with $AB=25$, $BC=39$, $AC=42$; $D$ is on $\overline{AB}$ with $AD=19$; $E$ is on $\overline{AC}$ with $AE=14$; Answer choices: (A) $\tfrac{266}{1521}$, (B) $\tfrac{19}{75}$, (C) $\tfrac13$, (D) $\tfrac{19}{56}$, (E) $1$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #1 Draw a Diagram, #16 Change Focus / Count the Complement
The quadrilateral $BCED$ has no tidy area formula, but it is just the whole triangle with the corner triangle sliced off, so the only number worth chasing is $[ADE]/[ABC]$; the quadrilateral is its complement (Tool #16). Comparing $[ADE]$ to $[ABC]$ directly is awkward because both a base and a height change at once. Tool #7 (Identify Subproblems) splits that comparison into two easy links through a middle triangle $ABE$: sliding $B$ back to $E$ changes only one dimension at a time. Each link uses the single fact that triangles with the same height have areas in the ratio of their bases, which a diagram (Tool #1) makes visible. Multiplying the two links gives $[ADE]/[ABC]$, and the complement finishes the ratio.
Execute — Answer: D
6.G.A.1 Step 1 See the quadrilateral as a leftover
- Draw triangle $ABC$ and mark $D$ on $AB$, $E$ on $AC$, then join $DE$.
- The segment $DE$ chops off the little triangle $ADE$ at corner $A$; everything else — the region with corners $B$, $C$, $E$, $D$ — is the quadrilateral.
- So the quadrilateral is the whole triangle minus the corner triangle.
- That means every quantity we need is controlled by one ratio: how big $ADE$ is compared to the whole triangle $ABC$.
💡 Cut a corner off a triangle and what remains is simply the whole minus that corner.
6.G.A.1 Step 2 Route through a middle triangle
- Going from $ADE$ straight to $ABC$ moves the base from $AD$ to $AB$ and the far vertex from $E$ to $C$ at the same time, so nothing stays fixed to compare.
- Fix that by stepping through the in-between triangle $ABE$, which shares base line $AB$ with $ADE$ and shares base line $AC$ with $ABC$.
- Then $[ADE]/[ABC]$ becomes a product of two comparisons, each of which changes only one thing.
💡 A hard comparison becomes two easy ones if you pass through a shape that shares a side with each.
6.RP.A.3 Step 3 First link: same height from E
- Look at triangles $ADE$ and $ABE$.
- Both have their far vertex at $E$, and their bases $AD$ and $AB$ lie along the same line $AB$.
- So both triangles have the exact same height — the perpendicular distance from $E$ down to line $AB$.
- When two triangles share a height, the one with the longer base has proportionally more area, so their areas are in the ratio of their bases.
💡 Same height means area just tracks base length.
6.RP.A.3 Step 4 Second link: same height from B
- Now look at triangles $ABE$ and $ABC$.
- Both have their far vertex at $B$, and their bases $AE$ and $AC$ lie along the same line $AC$.
- Again the two triangles share a height — the distance from $B$ to line $AC$ — so their areas are in the ratio of the bases $AE$ to $AC$.
- Here $AE=14$ and $AC=42$, and $14/42$ simplifies to $1/3$.
💡 The same shared-height trick works on the other pair of triangles.
5.NF.B.4 Step 5 Multiply the two links
- Multiply the two ratios to collapse the chain and land on the number we wanted, the size of the corner triangle relative to the whole.
- Notice the side $BC=39$ never appeared — the shape of the triangle does not matter, only how far along each side the points $D$ and $E$ sit.
💡 Two side fractions multiply because area scales with each direction independently.
6.NS.A.1 Step 6 Take the complement and divide
- The corner triangle is $\tfrac{19}{75}$ of the whole, so the leftover quadrilateral is the rest of the whole: $1-\tfrac{19}{75}=\tfrac{56}{75}$.
- The ratio asked for is the corner triangle over the quadrilateral.
- Dividing the two shares, the common $\tfrac{1}{75}$ cancels and leaves $19$ over $56$.
- That is answer choice (D).
💡 The quadrilateral is whatever fraction of the whole the triangle is not.
6.G.A.1 Draw triangle $ABC$ and mark $D$ on $AB$, $E$ on $AC$, then join $DE$. The segme 6.G.A.1 Going from $ADE$ straight to $ABC$ moves the base from $AD$ to $AB$ and the far 6.RP.A.3 Look at triangles $ADE$ and $ABE$. Both have their far vertex at $E$, and their 6.RP.A.3 Now look at triangles $ABE$ and $ABC$. Both have their far vertex at $B$, and th 5.NF.B.4 Multiply the two ratios to collapse the chain and land on the number we wanted, 6.NS.A.1 The corner triangle is $\tfrac{19}{75}$ of the whole, so the leftover quadrilate Review
Reasonableness: The corner triangle is $\tfrac{19}{75}\approx 0.25$ of the whole triangle, so the quadrilateral holds the other $\approx 0.75$ — the leftover really should be the bigger piece, and the answer $\tfrac{19}{56}\approx 0.34$ correctly shows the triangle is well under half the quadrilateral. It is also a useful check to reject the near-miss choices: (B) $\tfrac{19}{75}$ is $[ADE]/[ABC]$, the ratio to the whole triangle rather than to the quadrilateral, and (A) $\tfrac{266}{1521}=\tfrac{19\cdot14}{39\cdot39}$ comes from wrongly dividing by $BC$; only (D) compares the two regions the problem actually names.
Alternative: Use the sine area formula, which shares angle $A$ in both triangles. Then $[ADE]=\tfrac12\,AD\cdot AE\sin A$ and $[ABC]=\tfrac12\,AB\cdot AC\sin A$, so the common $\tfrac12\sin A$ cancels and $[ADE]/[ABC]=\dfrac{AD\cdot AE}{AB\cdot AC}=\dfrac{19\cdot14}{25\cdot42}=\dfrac{19}{75}$, exactly the same value; then $[ADE]/[BCED]=\tfrac{19}{56}$ as before.
CCSS standards used (min grade 6)
6.G.A.1Find the area of triangles and other figures by composing and decomposing shapes (Seeing quadrilateral $BCED$ as triangle $ABC$ with corner triangle $ADE$ removed, and using area $=\tfrac12\,\text{base}\times\text{height}$ to know that triangles of equal height compare by their bases.)6.RP.A.3Use ratio and rate reasoning to solve mathematical problems (Turning each equal-height pair of triangles into a base ratio: $[ADE]/[ABE]=AD/AB=19/25$ and $[ABE]/[ABC]=AE/AC=1/3$.)5.NF.B.4Multiply a fraction by a fraction (Multiplying the two links $\tfrac{19}{25}\cdot\tfrac13=\tfrac{19}{75}$ to get $[ADE]/[ABC]$.)6.NS.A.1Interpret and compute quotients of fractions (Taking the complement $1-\tfrac{19}{75}=\tfrac{56}{75}$ and dividing $\tfrac{19/75}{56/75}=\tfrac{19}{56}$.)
⭐ Triangles with the same height compare by their bases, so slide through the middle triangle $ABE$ to get $[ADE]/[ABC]=\tfrac{19}{25}\cdot\tfrac13=\tfrac{19}{75}$, then the leftover quadrilateral is $\tfrac{56}{75}$, giving $\tfrac{19}{56}$.
⭐ Triangles with the same height compare by their bases, so slide through the middle triangle $ABE$ to get $[ADE]/[ABC]=\tfrac{19}{25}\cdot\tfrac13=\tfrac{19}{75}$, then the leftover quadrilateral is $\tfrac{56}{75}$, giving $\tfrac{19}{56}$.
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