AMC 10 · 2006 · #23
Grade 7 geometry-2dA triangle is partitioned into three triangles and a quadrilateral by drawing two lines from vertices to their opposite sides. The areas of the three triangles are 3, 7, and 7, as shown. What is the area of the shaded quadrilateral?
Pick an answer.
AMC 10 2006 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: A triangle is cut by two cevians (a segment from a vertex to a point on the opposite side), one drawn from each of two vertices. These two lines split the triangle into three smaller triangles and one quadrilateral. The three triangles have areas 3, 7, and 7. Find the area of the quadrilateral piece.
Givens: Two cevians are drawn, one from each of two vertices to the opposite side; The two cevians cross at a single point inside the triangle; The three triangular pieces have areas 3, 7, and 7; The two area-7 triangles lie on either side of the crossing point along one cevian; the area-3 triangle lies along the other cevian
Unknowns: The area of the shaded quadrilateral piece
Understand
Restated: A triangle is cut by two cevians (a segment from a vertex to a point on the opposite side), one drawn from each of two vertices. These two lines split the triangle into three smaller triangles and one quadrilateral. The three triangles have areas 3, 7, and 7. Find the area of the quadrilateral piece.
Givens: Two cevians are drawn, one from each of two vertices to the opposite side; The two cevians cross at a single point inside the triangle; The three triangular pieces have areas 3, 7, and 7; The two area-7 triangles lie on either side of the crossing point along one cevian; the area-3 triangle lies along the other cevian
Plan
Primary tool: #7 Identify Subproblems
Secondary: #1 Draw a Diagram, #4 Introduce a Variable
The figure is one compound shape, but every piece is a triangle sharing a side (a cevian) with its neighbors. The winning move is to break the figure into triangles that share the same height: for those, area ratio equals base ratio, so each given area quietly reports a length ratio. A clean labeled diagram fixes which region is which, and naming the one still-unknown sub-triangle turns those ratios into a single equation to solve.
Execute — Answer: D
6.G.A.1 Step 1 Label the points and pieces
- Call the triangle ABC.
- One cevian runs from A to a point D on BC; the other runs from B to a point E on AC.
- They cross at F.
- This makes four pieces: triangle AFE = 3, triangle ABF = 7, triangle BFD = 7, and the shaded quadrilateral FDCE, whose area we want.
💡 A clear map of which region is which turns a tangle of lines into a handful of simple triangles.
6.RP.A.3 Step 2 Equal areas force a midpoint
- Two triangles with the same height have areas in the same ratio as their bases.
- Triangles ABF and BFD both point up to B and have bases AF and FD lying on the line AD, so they share the height from B.
- Their areas are equal (7 and 7), so their bases must be equal: AF = FD.
- That means F is the midpoint of AD.
💡 Same height and same area can only happen if the bases match.
6.RP.A.3 Step 3 Read the split on the other cevian
- Now apply the same rule to the cevian BE.
- Triangles ABF and AFE both point to A and have bases BF and FE on the line BE, so they share the height from A.
- Their area ratio equals their base ratio, so BF : FE = 7 : 3.
💡 The areas 7 and 3 measure directly how the point F splits segment BE.
6.G.A.1 Step 4 Cut the quadrilateral and name a variable
- Draw segment CF; it splits the quadrilateral FDCE into triangles CFD and CFE.
- Let [CFE] = x.
- Since F is the midpoint of AD, triangles CFD and CFA have equal bases FD = FA and the same apex C, so [CFD] = [CFA].
- Point E lies on AC, so triangle CFA is made of AFE and CFE: [CFA] = 3 + x.
- Therefore [CFD] = 3 + x, and the whole quadrilateral is [FDCE] = [CFD] + [CFE] = (3 + x) + x = 3 + 2x.
💡 The midpoint copies the area 3+x across F, so the quadrilateral is 3 plus two copies of the mystery triangle.
7.EE.B.4 Step 5 Use the 7:3 split to solve for x
- Triangles BFC and EFC share the apex C, with bases BF and FE on the line BE, so their areas keep the 7 : 3 ratio: [BFC] : [EFC] = BF : FE = 7 : 3.
- Here [EFC] = x and [BFC] = [BFD] + [CFD] = 7 + (3 + x) = 10 + x.
- Set up and solve the equation.
💡 The same 7:3 ratio that splits BE also splits the two triangles hanging off vertex C.
7.G.B.6 Step 6 Add up the quadrilateral
- Substitute x = 15/2 into the quadrilateral's area: [FDCE] = 3 + 2x = 3 + 2 (15/2) = 3 + 15 = 18.
- The shaded quadrilateral has area 18, which is choice (D).
💡 Plug the mystery triangle back in and total the pieces.
6.G.A.1 Call the triangle ABC. One cevian runs from A to a point D on BC; the other runs 6.RP.A.3 Two triangles with the same height have areas in the same ratio as their bases. 6.RP.A.3 Now apply the same rule to the cevian BE. Triangles ABF and AFE both point to A 6.G.A.1 Draw segment CF; it splits the quadrilateral FDCE into triangles CFD and CFE. Le 7.EE.B.4 Triangles BFC and EFC share the apex C, with bases BF and FE on the line BE, so 7.G.B.6 Substitute x = 15/2 into the quadrilateral's area: [FDCE] = 3 + 2x = 3 + 2 (15/2 Review
Reasonableness: Check the pieces sum sensibly and re-derive from the two cevians. The four areas total 3 + 7 + 7 + 18 = 35. Splitting by cevian AD: [ABD] = 7 + 7 = 14 and [ACD] = 3 + 18 = 21, so BD : DC = 14 : 21 = 2 : 3. Splitting by cevian BE: [ABE] = 7 + 3 = 10 and [CBE] = 7 + 18 = 25, so AE : EC = 10 : 25 = 2 : 5. Both side-ratios are clean, and each split re-gives the quadrilateral as 21 - 3 = 18 and 25 - 7 = 18. Everything agrees, and 18 is choice (D).
Alternative: Mass points give the same answer fast. From AF = FD put equal masses at A and D; from BF : FE = 7 : 3 put mass 3 at B and 7 at E. Balancing at E (on AC) forces mass 2 at C, so BD : DC = 2 : 3 and AE : EC = 2 : 5. Then [ACD] = [ABD] (DC/BD) = 14 (3/2) = 21, and the quadrilateral is 21 - 3 = 18. A coordinate/shoelace computation of FDCE also lands on 18.
CCSS standards used (min grade 7)
6.G.A.1Find area of triangles, special quadrilaterals, and polygons by composing (Decomposing the figure into triangles and the quadrilateral, and rebuilding the quadrilateral from two sub-triangles)6.RP.A.3Use ratio and rate reasoning to solve real-world and mathematical problems (Turning 'triangles with the same height' into equal-base and base-ratio conclusions (AF = FD and BF : FE = 7 : 3))7.EE.B.4Use variables to represent quantities and construct simple equations and inequalities (Naming the unknown sub-triangle x and solving (10 + x)/x = 7/3)7.G.B.6Solve real-world problems involving area, surface area, and volume (Assembling the final quadrilateral area 3 + 2x = 18)
⭐ When two triangles share the same height, their areas line up exactly with their bases, so every area label is secretly telling you a length ratio.
⭐ When two triangles share the same height, their areas line up exactly with their bases, so every area label is secretly telling you a length ratio.
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