Playing with Constructions · Lesson 6 of 8
Constructing Rectangles from Angles and Diagonals
“Locate rectangle vertices by combining angle directions with given distances.”
• Plan rectangles whose diagonals split corner angles into specified parts. • Use perpendiculars and angle rays to locate missing vertices. • Construct a rectangle from one side length and one diagonal length. • Explain why an intersection satisfies both construction conditions. • Compare perpendicular and compass-transfer methods for completing a rectangle.
Given information tells us where to look
A rectangle can be constructed even when its two side lengths are not both supplied. An angle can tell us a diagonal direction, and a diagonal length can tell us how far away a vertex lies. Neither condition works by itself. The construction succeeds when we combine it with the rectangle right angles and find a point satisfying both requirements.
Always start with a rough figure. Label the given side, diagonal, or angle before deciding a step sequence. Ask which segment can be drawn immediately and which point still has an unknown position. A line or ray can narrow that unknown position to a direction; a circle can narrow it to a distance. An intersection then selects a position that meets both conditions.
A diagonal that makes 60° and 30°
Name the desired rectangle ABCD around its boundary, with A lower left, B lower right, C upper right, and D upper left. We want diagonal AC to make 60° with AB at A and 30° with AD. Since the whole corner is 90°, specifying one part specifies the other. No side length is imposed, so we may choose AB freely.
Problem
Construct a rectangle whose diagonal divides its endpoint right angles into 60° and 30°.
- 1.Draw a convenient segment AB, for example 3 cm. At B draw a perpendicular ray to AB on the chosen interior side.
- 2.At A use a protractor to draw a ray making 60° with AB. Its intersection with the perpendicular through B is C. Now AC has the required direction and BC has the required rectangle direction.
- 3.Draw a perpendicular to AB through A. The angle between this ray and AC is 90° − 60° = 30°.
- 4.Through C draw a perpendicular to BC. Where it meets the ray through A, mark D. Join the boundary and verify the angles and opposite side lengths.
- 5.Alternatively, copy BC with a compass onto the perpendicular through A to locate D. Then join CD. Equal heights on the two perpendiculars give the same rectangle.
The perpendicular through B is not enough to choose C: it contains many possible points. The 60° ray is not enough either. Their common point supplies both conditions. The choice of AB sets the size. If you repeat the method with a longer base, the rectangle becomes larger but its angle split remains 60° and 30°.
The two completion methods express the same rectangle relationships. The perpendicular through C establishes a right angle at C and a top edge in the base direction. Copying BC to AD instead establishes equal opposite heights. In both cases, the fourth vertex lies on the perpendicular through A and the finished top side is parallel to AB.
Problem
Construct a rectangle whose diagonal makes 50° with its base and 40° with the adjoining side.
- 1.Choose and draw a base AB. At B draw the perpendicular ray on the intended side.
- 2.At A draw a 50° ray from AB. Its intersection with the ray at B locates C.
- 3.Draw the perpendicular at A. The remaining part at A is 90° − 50° = 40°.
- 4.Locate D by either completion method. Check the 50° and 40° parts at the other endpoint of AC, allowing small protractor error in the drawing.
Problem
Use the same method for 45° and 45°. What happens to the sides?
- 1.Draw AB and the perpendicular at B. Draw a 45° ray from A to meet it at C.
- 2.The diagonal lies exactly halfway between the horizontal base direction and the perpendicular direction at A. The horizontal and vertical distances to C are equal.
- 3.Therefore BC equals AB. Completing the rectangle gives four equal sides as well as four right angles: the result is a square.
- 4.This agrees with the diagonal investigation. Equal parts of the endpoint right angle occur in the square case.
A 60° angle from AB and a 60° angle from AD are different directions. Use the protractor baseline along the stated side. Check that the two corner parts total 90°; do not add a second unrelated angle outside the rectangle.
One side and a diagonal
Now suppose the given information is a side of 5 cm and a diagonal of 7 cm. Use rectangle ABCD with D lower left, C lower right, B upper right, and A upper left. We know DC = 5 cm and DB = 7 cm. Draw the known base first. The rectangle condition puts B on the perpendicular through C, but its height is not given.
Every possible point 7 cm from D lies on the circle centred at D with radius 7 cm. Therefore B must lie on that circle as well as on the perpendicular through C. Drawing both turns an uncertain search into a precise intersection. This is the same fixed-distance idea that made the first compass circle.
Problem
Construct rectangle ABCD with DC = 5 cm and DB = 7 cm.
- 1.Draw DC = 5 cm. At C construct a perpendicular ray to DC above the base and call its line l.
- 2.Set the compass to 7 cm. With D as centre draw a circle, or an arc near the upper part of l.
- 3.Mark the upper intersection of the circle and l as B. It is on the required perpendicular and is exactly 7 cm from D.
- 4.Complete A by drawing a perpendicular to DC at D and a perpendicular to BC through B. Their intersection is A. You may instead copy CB onto the perpendicular at D.
- 5.Join the sides and verify DC = AB = 5 cm, AD = BC, four right angles, and diagonal DB = 7 cm. Do not assume that BC is also 7 cm: that given length belongs to the diagonal.
Only a short arc near the expected intersection is needed. The full circle helps explain the condition, but most of it does no locating work. Keep the same centre and opening when replacing a circle by an arc. Changing either would change the distance condition rather than merely shorten the drawn curve.
On the full perpendicular line there is also an intersection below the base. It can be used to make a reflected rectangle on the other side. Our chosen upward ray selects the upper one. Choosing the interior side early keeps the construction consistent and avoids marking an unintended lower vertex.
Apply the same reasoning to new lengths
For a 4 cm side and an 8 cm diagonal, draw a 4 cm base, erect the perpendicular at its far endpoint, and use an 8 cm arc centred at the other endpoint. For a 3 cm side and a 7 cm diagonal, repeat with those new values. The method changes only the distances; the intersection still joins a direction condition to a distance condition.
Problem
Explain the construction without guessing the missing side length.
- 1.Draw DC = 4 cm and a perpendicular ray at C on the chosen side.
- 2.Draw an arc centred at D with radius 8 cm. Its intersection with the ray is B, because DB must be 8 cm.
- 3.Use perpendiculars or copy BC to locate A. Join AB and AD and check the rectangle conditions.
- 4.Measure the missing side only after construction if you want its approximate length. It was not needed to locate B.
Problem
For DC = 3 cm and DB = 7 cm, a student draws a radius-7 cm arc centred at C. What goes wrong?
- 1.The intended distance is from D to B. A circle centred at C instead gives CB = 7 cm.
- 2.On the perpendicular through C, the marked point is 7 cm from C, so the student has set the height instead of the diagonal.
- 3.Keep DC and the perpendicular, but redraw the radius-7 cm arc with centre D. Mark its upper intersection and complete the rectangle.
- 4.Checking DB after construction detects the mistake. The correct centre follows the first endpoint of the specified distance.
A useful feasibility check is to compare the diagonal with the given side. In a genuine rectangle, a diagonal crosses the width and also reaches a different height, so it is longer than either adjoining side. A diagonal shorter than the given side cannot reach the perpendicular ray. If it is exactly the side length, it reaches only the base endpoint and gives zero height, not a rectangle.
Setting both base and height to the two supplied numbers would solve a different problem. Label the diagonal on the rough figure first. The compass centre must be its known endpoint, and the opening must be the diagonal length.
Quiz
Where is C found in the 60° angle construction?
A diagonal makes 50° with the base at a rectangle corner. What angle does it make with the adjoining side?
For DC = 5 cm and DB = 7 cm, which circle locates B?
Why can a short arc replace the full circle?
How can you locate A after D, C, and B are known?
A rectangle diagonal divides its endpoint right angle into 45° and 45°. What is the rectangle?
Can a nonzero-height rectangle have a 6 cm side and a 5 cm diagonal?
Practice Problems
- Construct the 60°–30° rectangle using an arbitrary base length. Explain how C is located.
- Complete a second copy by transferring BC onto AD rather than drawing the perpendicular through C. Compare the results.
- Construct a rectangle with a 50°–40° diagonal split and check the corresponding angles at both endpoints.
- Construct the 45°–45° case and use measurements and reasoning to identify the resulting shape.
- Construct a rectangle with side 5 cm and diagonal 7 cm, keeping the circle or arc visible until checking.
- Repeat the construction for side 4 cm and diagonal 8 cm, and for side 3 cm and diagonal 7 cm.
- Explain why drawing a full circle is optional but choosing the correct centre and radius is essential.
- A student uses a 7 cm height instead of a 7 cm diagonal in the 5 cm side problem. Explain the error and how to correct it.
- Decide whether side–diagonal pairs 5 cm–4 cm, 5 cm–5 cm, and 5 cm–6 cm can give genuine rectangles. Explain using the intersection.
The ray intersections establish the chosen diagonal angle and a perpendicular side. Complementary parts total 90°. Copying BC onto AD gives the same fourth vertex as the perpendicular completion.
Key Takeaways
• Use a rough figure to distinguish sides, diagonals, and angle conditions. • An angle ray and perpendicular can jointly locate a rectangle vertex. • The two parts of an endpoint right angle add to 90°; equal parts give a square. • A circle represents a fixed distance from its centre. • A perpendicular-circle intersection satisfies direction and distance together. • An arc and full circle serve the same locating purpose when their centre and radius are unchanged.