Playing with Constructions · Lesson 7 of 8
Equal Distances and the House Construction
“Use intersecting circles to find points at two equal distances and build a complete geometric design.”
• Locate a point at a specified equal distance from two given points. • Explain the role of circle and arc intersections. • Plan and construct the square body, roof, curved boundary, and doorway of a house. • Adapt a construction to new lengths while checking its conditions. • Construct an equal-sided quadrilateral that is not a square.
One distance gives a circle; two give intersections
Suppose B and C are fixed points and we want A to be 5 cm from both. The first condition, AB = 5 cm, puts A somewhere on a circle centred at B. The second, AC = 5 cm, puts A somewhere on a circle centred at C. A suitable point must belong to both circles at the same time.
If the circles cross, each crossing point satisfies both distance conditions. The compass point does not need to be moved repeatedly in a guessing search. Draw the two distance curves and choose an intersection according to the design. For a roof above BC, choose the upper intersection. The lower one satisfies the distances too but lies in the wrong part of the picture.
At equal distances from the specified points. A is equidistant from B and C when AB = AC.
Problem
B and C are 5 cm apart. Find A above BC with AB = AC = 5 cm.
- 1.Set the compass to 5 cm. Draw a circle centred at B; every point on it is 5 cm from B.
- 2.With the same opening draw a circle centred at C; every point on it is 5 cm from C.
- 3.Mark their upper intersection as A. Since it is on both circles, AB = AC = 5 cm.
- 4.Join A to B and A to C. These segments and BC have the same length, which checks the intended roof geometry.
Planning the house from its parts
The house design has a square body, two sloping roof sides, a curved line between the body top corners, and a doorway. The body sides and the two sloping roof sides are each 5 cm. The doorway is 1 cm wide and 2 cm high. Its smaller dimensions do not change the requirement for the main body and roof segments.
Use B for the body top-left corner, C for the top-right, D for the lower-left, and E for the lower-right. The square is BCED in boundary order. Its width BC is 5 cm even though the final drawing replaces the straight top edge with a curved line. Draw BC lightly as a supporting segment so that the roof endpoints and square body are accurately positioned.
Problem
Construct the straight parts of the 5 cm house before locating the roof vertex.
- 1.Draw DE = 5 cm. At D and E draw perpendiculars and mark B and C 5 cm above the base. A light BC closes the planned square.
- 2.For a centred doorway, the remaining base length is 5 − 1 = 4 cm. Leave 2 cm at each end to locate the doorway vertical sides.
- 3.Draw the two doorway sides 2 cm high and connect their top ends with a 1 cm horizontal segment.
- 4.Keep DE as a faint construction line through the doorway opening. Darken only the two outer base pieces and the final doorway boundary. The complete body width remains 5 cm.
Next locate A with the two radius-5 cm circles centred at B and C. Use their upper intersection and join AB and AC. Before adding the curved roof line, check that both roof sides are 5 cm. A vertex guessed above the centre may produce a symmetric-looking roof while still having the wrong side lengths.
The curve uses the roof vertex as its centre
The two equal roof lengths give exactly the opening needed for the curved boundary. AB and AC are both 5 cm, so a circle centred at A with radius 5 cm passes through both B and C. Draw the short arc between those two points on the body side of the supporting segment BC. It bends slightly into the top of the body.
Problem
After locating A, explain how to draw the curved line from B to C.
- 1.Put the compass point at A and set the pencil tip at B. The opening equals AB = 5 cm.
- 2.Turn the pencil along the short arc towards C without changing the opening.
- 3.Because AC is also 5 cm, the pencil reaches C with the same radius. Both endpoints lie on the circle centred at A.
- 4.Keep the short downward-bending arc as the final line. The unused parts of the circle are not needed in the finished house.
The construction now connects two uses of circles. Circles centred at B and C locate A. A circle centred at A then supplies the final curved line through B and C. The centres are different because the tasks are different. Confusing them can make an arc that fails to reach one endpoint or gives the wrong curve.
The compass first locates A using centres B and C. It later draws the roof curve using centre A. The equal length permits the same opening, but the centre must change with the condition you are applying.
Use arcs when the intersection is enough
Drawing two complete circles makes the reasoning visible, but the upper intersection can also be found with two short arcs. Keep the same 5 cm opening, draw an arc above BC from B, and draw another from C across it. Their intersection satisfies the same two distances as the full-circle intersection.
A short arc must extend far enough to meet the other arc. If they do not meet, check the opening and the approximate area where the intersection should be; do not move the centre to force a crossing. A planned arc construction is more efficient because it draws only the parts needed for locating the point.
Problem
Construct a house whose body sides and two sloping roof sides are 7 cm.
- 1.Construct the square body with width and height 7 cm. Label its top corners B and C.
- 2.Draw arcs of radius 7 cm centred at B and C. Their upper intersection is A, so AB = AC = 7 cm.
- 3.Join AB and AC. Draw the short arc from B to C centred at A with the same 7 cm opening.
- 4.Choose doorway dimensions that fit. For example, a 2 cm wide, 3 cm high centred doorway leaves (7 − 2) ÷ 2 = 2.5 cm at each end. Check those separate doorway conditions.
The larger doorway example is a design choice, not a claim that every detail was enlarged by one common factor. If you want a uniformly enlarged picture, every length, including doorway dimensions, must change in the same proportion. If only the main 7 cm side conditions are required, you may select a fitting doorway independently.
An equal-sided figure need not be a square
The square definition requires equal sides and right angles. We can use the same circle-intersection method to construct four equal sides with non-right corners. Start with two equal segments from A meeting at a non-right angle. Then locate a new point at that same distance from their other endpoints.
Problem
Draw a quadrilateral whose sides are all 5 cm but which is not a square.
- 1.Draw AB = 5 cm and AC = 5 cm meeting at A at 60°, rather than 90°. B and C are the two other endpoints.
- 2.Draw radius-5 cm arcs centred at B and C. A is already one intersection, since AB = AC = 5 cm. Select the other intersection and call it D.
- 3.Join BD and DC. In boundary order the quadrilateral is A-B-D-C. Its sides AB, BD, DC, and CA all equal 5 cm.
- 4.The angle at A is 60°, so the figure is not a square. It is an equal-sided quadrilateral called a rhombus; no further rhombus properties are needed for this construction.
Refining earlier artwork
Return to the Person neckline and the two eye outlines. Their arcs need centres that are equally distant from two chosen endpoints. You can replace a vague guess by comparing distances with a compass or by locating an equal-distance centre with intersecting arcs. Place upper and lower eye centres symmetrically so the resulting curves still match.
For a shallow arc, choose equal-distance compass centres farther from the endpoint span than the semicircle centre. For the two-wave design, repeat the same span and centre offset on the other half, with the placement reversed to bend the other way. This revisits the original artwork with a clearer method: choose positions through conditions, then draw the final curve from the located centre.
Quiz
A is 5 cm from both B and C. Where must A lie?
Which intersection is chosen for a roof above BC?
After A is located, which centre draws the house curve through B and C?
A 5 cm wide body has a centred doorway 1 cm wide. What is each side gap?
Why does the roof arc with centre A and radius 5 cm reach C?
In the equal-sided quadrilateral construction, A is already one circle intersection. Which point should complete the figure?
Four equal sides and a 60° corner identify which conclusion?
Practice Problems
- Construct the 5 cm house, including its 1 cm by 2 cm centred doorway. Retain supporting lines until checking.
- Write the two distance conditions used to locate the roof vertex and explain why the intersection satisfies both.
- Construct a second roof using only short intersecting arcs and compare it with the full-circle version.
- Explain why the roof arc is centred at A rather than at B or C.
- Construct a 7 cm house with a centred 2 cm by 3 cm doorway. Calculate the end gaps before drawing.
- Construct two equal 5 cm segments at 60°, then use the other circle intersection to complete a four-equal-sided nonsquare.
- Explain why choosing A again as the fourth point would not produce a genuine quadrilateral.
- Refine a Person neckline, two shallow waves, and matching Eyes using equal-distance centres. Describe how you checked each pair of endpoints.
The 5 cm body gives 2 cm gaps around its 1 cm doorway. The 7 cm body with a 2 cm doorway gives 2.5 cm gaps. In either case the roof vertex comes from equal-radius intersections at the body top corners.
Key Takeaways
• An equidistant point satisfies equal-distance conditions from the specified points. • A circle intersection satisfies both of its defining distance rules. • Choose the intersection that also fits the required position in the design. • Full circles and short arcs can locate the same point. • The house uses circles to locate a vertex and another circle to draw its final curved boundary. • Four equal sides do not guarantee a square: the right-angle condition matters.