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7.1 Under Construction
Supplement for Mathematics Vision Project (MVP) curriculum · Secondary Math I, Module 7, Section 7.1
Independent supplementary instruction; not affiliated with or endorsed by Mathematics Vision Project.
Human review completed .
Use circles to copy lengths, locate exact intersections, and construct shapes whose sides and angles meet the required conditions.
1. In class: A Develop Understanding Task
This section explains some of the ideas you explored in class through the guided exercises. Use it alongside the “A Develop Understanding Task” portion of Section 7.1 in your MVP packet.
Main task: printed pp. 1–3 / PDF pp. 3–5.
What each tool guarantees
An unmarked straightedge draws a straight line through two points or extends a segment. A compass draws a circle with a fixed center and radius, and transfers a length without measuring it numerically. An arc is part of a circle. You can draw just the arcs needed to find an intersection.
Every point on a circle is the same distance—the radius—from its center. If you keep the compass opening unchanged, circles drawn at different centers have the same radius. This is how a construction guarantees equal lengths.
A construction uses these facts to locate points. A sketch that only looks right does not establish equal lengths or right angles. Leave useful compass arcs visible so someone can follow your work. “Congruent segments” means segments of equal length; the symbol is \(\cong\).
Main 1–2 — Construct a rhombus
A rhombus is a quadrilateral with four equal side lengths. The packet starts with segment \(BA\) and a ray from \(B\). Use that segment as the compass opening.
Main 1 — Locate C
Original problem: Locate \(C\) on the given ray so that \(\overline{BC}\cong\overline{BA}\), and explain why.
Worked solution: Place the compass point at \(B\), open it to \(A\), and draw an arc crossing the ray. Label that intersection \(C\). Both \(A\) and \(C\) are on the circle centered at \(B\), so \(BA=BC\).
Main 2 — Locate D
Original problem: Locate \(D\) so that \(\overline{CD}\cong\overline{DA}\cong\overline{AB}\), and explain why.
- Keep the opening \(AB\). Draw arcs centered at \(A\) and \(C\).
- These circles meet at \(B\) and at another point. Label the other intersection \(D\).
- Use the straightedge to draw \(CD\) and \(DA\), completing \(ABCD\).
Why it works: The arcs give \(AD=AB\) and \(CD=AB\). Together with \(BC=AB\), all four sides are equal.
Main 3–4 — Construct a perpendicular bisector
A midpoint divides a segment into two equal lengths. To bisect a segment is to divide it at its midpoint. A perpendicular bisector is a line that passes through that midpoint and meets the segment at a right angle, \(90^\circ\).
Main 3 — Fold R onto S
Original problem: Fold so that \(R\) is reflected onto \(S\). What do you know about the crease?
Answer: The crease is the reflection line, so it crosses \(RS\) at its midpoint and is perpendicular to \(RS\). Thus it is the perpendicular bisector.
Main 4 — Find the same line with a compass
Original problem: Locate at least two points on the crease line with a compass, then draw the line and describe the strategy.
- Open the compass enough that equal circles centered at \(R\) and \(S\) intersect at two points. The radius must be greater than half of RS; using the full length \(RS\) is a simple choice.
- Without changing that opening, draw intersecting arcs above and below the segment, first from \(R\), then from \(S\).
- Label the intersections \(P\) and \(Q\). Draw the line through them. Its intersection with \(RS\) is \(M\), the midpoint.
Why it works: Each intersection is equally far from \(R\) and \(S\). Reflecting across the crease that swaps \(R\) and \(S\) swaps the two equal circles. Each of their intersections stays fixed: one is on each side of \(RS\). Both intersections therefore lie on that crease, the perpendicular bisector.
If the arcs do not cross, increase the compass opening and redraw both pairs with that same opening. If the radius is exactly half of RS, there is only one intersection, which cannot determine the whole line.
Main 5 — Complete a square with side RM
A square has four equal sides and four right angles. The packet uses \(RM\) as its side, where \(M\) is the midpoint of \(RS\). Its side is therefore half of \(RS\).
Main 5 — Use the constructed right angle
Original problem: Finish a square using \(RM\) as one side and the perpendicular line through \(M\).
- Set the compass opening to \(RM\). Center it at \(M\) and mark \(T\) on the perpendicular line, on either side of \(RS\). Then \(MT=RM\).
- With the same opening, draw arcs centered at \(R\) and \(T\). Their intersections are \(M\) and another point \(U\).
- Join \(T\) to \(U\) and \(U\) to \(R\). The square is \(RMTU\).
Why it works: The construction gives four equal sides and a right angle at \(M\). A rhombus is a parallelogram, so its opposite angles are equal and adjacent angles add to \(180^\circ\). A rhombus with one right angle consequently has four right angles.
2. Ready — Circles and midpoints
Ready 1–2: printed p. 4 / PDF p. 6.
Concentric circles share a center but have different radii. For Problem 1, copy each opening from center \(A\) to center \(B\), then compare the intersections of matching-radius circles. Use the perpendicular-bisector construction to interpret what you see. Try both problems before checking.
Check your work — Ready 1–2
1. The intersections of corresponding equal-radius circles lie on the perpendicular bisector of \(AB\), because each is equally distant from \(A\) and \(B\). Connecting an upper and lower intersection gives that line; where it meets \(AB\) is the midpoint. Choose radii large enough for intersections. Construction not shown.
2. Fold the paper so that \(A\) lands exactly on \(B\). The crease bisects \(AB\) at right angles, so its intersection with the segment is the midpoint.
3. Set — Construction methods
Set 3: printed p. 4 / PDF p. 6; Set 4–6: printed p. 5 / PDF p. 7; Set 7–8: printed p. 6 / PDF p. 8.
Bisect an angle — Set 3
An angle bisector is a ray from an angle’s vertex that divides the angle into two equal angles. This divides the angle opening; it does not depend on the drawn sides being the same length.
- Center the compass at the vertex \(B\). Draw one arc crossing rays \(BA\) and \(BC\) at new points \(Q\) and \(P\). Then \(BQ=BP\).
- With the same opening, draw arcs centered at \(P\) and \(Q\). Mark their intersection \(X\) inside the angle, other than \(B\).
- Draw ray \(BX\). It bisects \(\angle ABC\).
Why it works: \(B\) and \(X\) are each equally far from \(P\) and \(Q\), so line \(BX\) is their perpendicular bisector. Reflecting across it swaps \(P\) and \(Q\), and therefore swaps the two angle sides while fixing \(B\). The two resulting angles are equal.
Paper folding: Fold so the two rays lie on top of one another, keeping the vertex fixed. The crease inside the angle is its bisector.
Copy a segment — Set 4
Draw a ray starting at a new point \(D'\). Set the compass opening to the original length \(DE\). Center the compass at \(D'\) and mark the ray’s intersection with the arc as \(E'\). The copied segment \(D'E'\) has length \(DE\), because it is a radius drawn with that opening.
Copy an angle — Set 5
You must transfer both the radius and the distance between the arc’s two intersection points.
- Label the original vertex \(V\). Draw an arc centered at \(V\) crossing its rays at \(P\) and \(Q\).
- Draw a new ray from a new vertex \(V'\). With the same opening, draw an arc centered at \(V'\) crossing the ray at \(P'\).
- Now change the opening to the distance \(PQ\). Center the compass at \(P'\) and mark its intersection with the new arc as \(Q'\), on the chosen side of the new ray.
- Draw ray \(V'Q'\). The copied angle is \(\angle P'V'Q'\).
Why it works: Both arcs use the same radius, and the straight segments \(PQ\) and \(P'Q'\) have the same length. On equal-radius circles, equal distances between the two arc points give equal angle openings at the centers. Copying only the radius would leave the second ray undetermined.
Construct a square on a whole given side — Set 7
To use \(CD\) as a side, you need a perpendicular at an endpoint, rather than at the midpoint of \(CD\).
- Extend \(CD\) beyond \(D\). Set the opening to \(DC\) and mark \(E\) on the extension so that \(DE=DC\). Then \(D\) is the midpoint of \(CE\).
- Construct the perpendicular bisector of \(CE\). It passes through \(D\).
- On this perpendicular, mark \(F\) with \(DF=CD\).
- With opening \(CD\), draw arcs centered at \(C\) and \(F\). Label their intersection other than \(D\) as \(G\).
- Join \(F\) to \(G\) and \(G\) to \(C\). Square \(CDFG\) has the entire given segment as a side.
Find the center of an equilateral triangle — Set 8
An equilateral triangle has three equal sides. Construct perpendicular bisectors of two of its sides. Their intersection \(O\) is equally far from all three vertices. For an equilateral triangle it is the center of rotation: turns of \(120^\circ\) and \(240^\circ\) carry the triangle onto itself. The third perpendicular bisector also passes through \(O\), providing a useful check.
Use these methods and the rhombus construction to attempt Set 3–8 on your packet before opening the checks.
Check your work — Set 3–8 construction checks
Completed packet constructions not shown. The diagrams above illustrate the methods. Your answers must include the requested constructions on the packet figures.
3. Both the compass construction and the fold should give the same ray from \(B\), splitting \(\angle ABC\) into two equal angles. Use the angle-bisector steps; keep the vertex fixed when folding.
4. Label the copy \(D'E'\), with \(D'E'=DE\). Its orientation and position can differ from the original. Use the segment-copy steps.
5. The new angle must have the same opening as the original. Retain the equal-radius arcs and the transferred distance between arc points as evidence. Use the angle-copy steps.
6. Center an arc at \(A\) with radius \(AB\) to mark \(C\) on the given ray from \(A\). With that same radius, draw arcs from \(B\) and \(C\); label their intersection other than \(A\) as \(D\). Join \(BD\) and \(DC\). In vertex order \(A,B,D,C\), all four sides equal \(AB\). Notice that this problem’s starting angle is at A, whereas the main task’s starting angle is at B.
7. Use the endpoint-perpendicular method. Check \(CD=DF=FG=GC\) and all four right angles. A square using only half of \(CD\) would not satisfy this problem.
8. The answer is the intersection of the side perpendicular bisectors, inside the triangle. Check that the compass distances from that point to all three vertices agree.
4. Go — Review systems of equations
Go 9–14: printed p. 6 / PDF p. 8.
A system’s solution is an ordered pair \((x,y)\) that satisfies both equations. For a refresher on finding an intersection by setting two output rules equal, see Section 3.6 Ready.
Substitution: Replace a variable with its equal expression from the other equation. Elimination: Add or subtract equations to cancel one variable; multiply an entire equation first if needed. Find the remaining variable, then check both original equations.
Go 9 — Substitution reminder
Original problem:
\[\begin{cases}x=11+y\\2x+y=19\end{cases}\]Worked solution: Substitute \(11+y\) for \(x\) in the second equation:
\[2(11+y)+y=19\quad\Rightarrow\quad3y=-3\quad\Rightarrow\quad y=-1.\]Then \(x=11+(-1)=10\). Check: \(10=11-1\) and \(2(10)-1=19\). Thus \((x,y)=(10,-1)\).
Try the remaining systems before checking. The key includes short algebraic checks of the method.
Check your work — Go 9–14
9. \((10,-1)\). See the visible worked solution.
10. \((9,5)\). From \(x-3y=-6\), use \(x=3y-6\). Then
\[-4(3y-6)+9y=9\Rightarrow -3y=-15\Rightarrow y=5,\qquad x=9.\]11. \(\left(8,\frac32\right)\). Subtract the second equation from the first:
\[(x+2y)-(x-4y)=11-2\Rightarrow6y=9\Rightarrow y=\frac32.\]Then \(x=11-2\left(\frac32\right)=8\).
12. \((0,1)\). Equate the two expressions for \(y\):
\[-x+1=2x+1\Rightarrow3x=0\Rightarrow x=0,\qquad y=1.\]13. \((3,1)\). Substitute the first rule into the second equation:
\[-3x+(-2x+7)=-8\Rightarrow-5x=-15\Rightarrow x=3,\qquad y=1.\]14. \((11,37)\). Double the first equation and add it to the second:
\[\begin{aligned}8x-2y&=14\\-6x+2y&=8\\\hline 2x&=22.\end{aligned}\]Thus \(x=11\) and \(y=4(11)-7=37\). Check: \(44-37=7\), and \(-66+74=8\).
Additional learning resources
- Khan Academy — Geometric constructions: perpendicular bisector. Demonstrates the equal-radius arc construction used in Main 4 and Set 8.
- Khan Academy — Geometric constructions: angle bisector. Demonstrates the compass-and-straightedge method for Set 3.
Teacher/parent review of the source packet
No serious mathematical errors were found in Section 7.1. Keep two distinctions explicit: Main 5 uses RM as a square’s side, while Set 7 uses the entire given CD; Set 6 starts its rhombus angle at A, while Main 1–2 starts at B.
Coverage — Section 7.1
| Packet component | Support and remaining student work |
|---|---|
| Main 1–2 · pp. 1–2 | Tool definitions, rhombus steps, radius reasoning, and illustrative diagram. Students construct on the packet. |
| Main 3–5 · pp. 2–3 | Fold explanation, perpendicular-bisector and square procedures, checked illustrative diagrams. |
| Ready 1–2 · p. 4 | Concentric-circle definition, linked bisector explanation, observation and folding key; packet construction omitted. |
| Set 3–5 · pp. 4–5 | Angle bisection with compass and paper, copying segments and angles, and checking criteria; completed packet constructions omitted. |
| Set 6–8 · pp. 5–6 | Rhombus adaptation, perpendicular at an endpoint for a whole-side square, and equilateral-triangle rotation center; completed packet constructions omitted. |
| Go 9–14 · p. 6 | Linked earlier intersection instruction, substitution/elimination reminders, worked Problem 9, and all six checked solutions. Algebraic methods selected from the packet’s allowed choices. |
Source prompts and notation: Mathematics Vision Project, Secondary Mathematics I, Module 7 student packet, Section 7.1, printed pp. 1–6 / PDF pp. 3–8, licensed under CC BY 4.0. Explanations, solutions, and diagrams created for this independent supplement.
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