MATH AI / UNIT 7 / SL

UNIT 7 · SL · 90 MINUTES

Geometry investigations

Use familiar mathematics to make and test decisions.

You will compare two panels, design cable anchors and choose a camera position. No new HL content is required. Your task is to decide how to combine the methods you already know.

Work with a partner, but keep your own labelled diagrams, calculations and explanations. Use your GDC in degree mode. Keep full precision until the final answer.

First 24 minutes: panels.Next 27 minutes: cable anchors.Next 20 minutes: camera placement.

Use the final 19 minutes for the independent check and corrections. Open a hint only after an attempt; open the worked answer after recording your reasoning.

8 cm13 cm53°Panel A8 cm13 cm127°Panel B

Two panels use the same pair of side lengths but different included angles.

INVESTIGATION 1 · PAIRS · 10 MINUTES

Compare two panel designs

Two triangular panels each have sides of 8 cm and 13 cm. The included angle is 53° in panel A and 127° in panel B.

Predict which panel has greater area and which needs more edging. Then calculate the area and perimeter of each.

A manufacturer requires an area of at least 40 cm² and a perimeter no greater than 34 cm. Decide which panel meets both conditions.

Explain how equal area can occur even though the panels are not congruent.

8 cm13 cm53°Panel A8 cm13 cm127°Panel B

Two panels use the same pair of side lengths but different included angles.

INVESTIGATION 1 · CONSTRUCT · 12 MINUTES

Create new designs with the same side lengths

Keep the two sides at 8 cm and 13 cm, but choose new included angles.

Create a pair of different triangles with equal areas larger than 41.5 cm². Then create a second pair with equal areas smaller than 41.5 cm².

For every triangle, give its included angle and verify the area. Compare with another pair: how many different answers could be valid?

Find the greatest possible area with these two sides. Explain geometrically why no design can exceed it.

A construction needs verification

A sketch alone does not establish that the areas satisfy the conditions. Record the calculations or a geometric argument.

For an extra challenge, find a new angle whose panel meets both of the manufacturer’s conditions from the previous slide.

INVESTIGATION 2 · PAIRS · 8 MINUTES

Test a proposed cable anchor

A straight cable is exactly 13 m long. It starts at A = (0, 0, 1). The third coordinate is height above level ground.

The anchor B must lie on the wall x = 3, with 0 ≤ y ≤ 10 and z > 1. The cable must make an angle no greater than 60° above the ground.

A designer proposes B = (3, 4, 13). Does this use the full cable? Does it satisfy the angle restriction?

Show the planar distance and vertical rise. Passing one requirement does not guarantee that the design is acceptable.

ABPlanar distance = √(3² + y²)z − 113 melevation

Elevation view: the vertical rise from A is z − 1, not z.

INVESTIGATION 2 · CONSTRUCT · 12 MINUTES

Construct two acceptable anchors

Keep A = (0,0,1), the 13 m cable and the wall x = 3. The anchor still requires 0 ≤ y ≤ 10, z > 1 and an elevation at most 60°.

Create two acceptable anchors with different heights. Verify every restriction.

Choose a vertical rise first, then calculate the required y-coordinate. Or choose y first and work backwards to the height. State coordinates exactly where practical.

ABPlanar distance = √(3² + y²)z − 113 melevation

Elevation view: the vertical rise from A is z − 1, not z.

INVESTIGATION 2 · GENERALISE · 7 MINUTES

Describe all the acceptable anchors

The same conditions apply: A = (0,0,1), cable length 13 m, wall x = 3, 0 ≤ y ≤ 10, z > 1 and elevation at most 60°.

Find the smallest possible y-coordinate and the greatest possible anchor height. Then describe every acceptable anchor using y.

Explain which condition controls each boundary. Check whether the boundary values themselves are allowed.

Explain the boundary

Moving an anchor closer to A across the ground forces a greater vertical rise if the cable length remains fixed.

The angle restriction prevents the planar distance becoming too small. The wall’s upper y-boundary limits movement in the other direction.

INVESTIGATION 3 · PAIRS · 10 MINUTES

Reconstruct the camera survey

B is 600 m due east of A. A drone is vertically above ground point C. The bearing of C from A is 040°; the bearing from B is 310°. The drone is 150 m above C.

Find AC and BC. Then locate C using coordinates with A = (0,0), east as positive x and north as positive y.

M is the midpoint of AB. Calculate MC without measuring the drawing. Keep unrounded values for the next slide.

NNABMC040°310° bearing600 m

Plan view: C is the ground projection of a drone 150 m above C. North lines are parallel.

INVESTIGATION 3 · DECIDE · 10 MINUTES

Choose a camera position

The drone is 150 m above C. From the survey, AC ≈ 385.7 m and MC = 300 m, where M is the midpoint of AB.

A camera may be placed at A or M. Its straight-line range is at most 400 m, and its viewing angle above the ground must be between 20° and 30° inclusive.

Test both positions against both conditions. Recommend a location with numerical evidence.

At your chosen location, find the full range of drone heights that still satisfies both camera limits.

NNABMC040°310° bearing600 m

Plan view: C is the ground projection of a drone 150 m above C. North lines are parallel.

INDEPENDENT CHECK · 14 MINUTES

Make two independent decisions

Use your notebook and GDC, without consulting a partner or the worked answers. Show the diagram and the first equation for each decision.

1 · Choose a panel

Two panels have sides of 10 cm and 14 cm meeting at either 40° or 140°. A design needs an area of at least 44 cm² and a perimeter no greater than 40 cm.

Test both panels and justify your recommendation.

2 · Revise a cable anchor

A cable joins A = (2,1,1) to B = (8,9,13), in metres. It may be at most 16 m long and at most 45° above the ground.

Decide whether B is acceptable. Keeping its first two coordinates unchanged, find the greatest permissible height of B.

In the next five minutes, check your work and correct the first step that needs changing. Keep your original attempt visible.

CORRECT AND EXPLAIN · 5 MINUTES

Check your decision, not just your calculation

Review the independent check. Choose one step to correct or one answer to justify more fully.

Which condition controlled the decision? How did you check that every other condition was also satisfied?

Write a short explanation identifying the diagram, mathematical method and final interpretation. A correct length does not settle an angle restriction; a sufficient area does not settle a perimeter restriction.

Finish by writing one connection that helped you solve a less familiar problem using mathematics you already knew.

Keep a useful record

Your notebook should contain the three investigations, your independent first attempt and a visible correction or verification.

Keep exact or unrounded intermediate values. Include units in the final interpretation and distinguish a vertical rise from an absolute height.

Worked answers

Hint: separate height from the third side

For area, use the perpendicular height associated with the two given sides. For perimeter, the third side must be calculated separately. Draw the two included angles to compare the heights.

Check the two decisions

Both areas are ½(8)(13)sin 53° = ½(8)(13)sin 127° ≈ 41.5 cm².

The third sides are √(233 − 208cos 53°) ≈ 10.4 cm and √(233 − 208cos 127°) ≈ 18.9 cm. The perimeters are approximately 31.4 cm and 39.9 cm.

Only panel A meets both conditions. The heights are equal, but changing the included angle changes the third side and hence the perimeter.

Compare possible constructions

One larger-area pair uses 70° and 110°: each area is approximately 48.9 cm². One smaller-area pair uses 30° and 150°: each area is 26 cm². Many other pairs are possible.

Supplementary included angles give equal perpendicular heights, and therefore equal areas for the fixed sides. The maximum area is 52 cm² when the sides are perpendicular: the full 13 cm side becomes the height to the 8 cm base.

Check the proposed anchor

The coordinate changes are (3,4,12). The planar distance is 5 m, the cable length is √(9+16+144) = 13 m, and the elevation is tan⁻¹(12/5) ≈ 67.4°.

The proposed anchor uses the full cable and lies on the allowed wall, but the elevation exceeds 60°. It is not acceptable.

Hint: encode the fixed cable length
3² + y² + (z − 1)² = 169.

The vertical rise cannot exceed 13sin 60°. Remember that the rise is measured from height 1, not from ground level.

Compare two valid anchors

Two examples are B₁ = (3, √60, 11) and B₂ = (3, √96, 9).

For B₁, the squared length is 9+60+100 = 169; y ≈ 7.75 is within the wall. The elevation is sin⁻¹(10/13) ≈ 50.3°.

For B₂, the squared length is 9+96+64 = 169; y ≈ 9.80 is within the wall. The elevation is sin⁻¹(8/13) ≈ 38.0°.

Both use the full cable and satisfy every stated restriction.

Check the complete range

At the limiting elevation of 60°, the planar distance is 13cos 60° = 6.5 m. Since the planar distance squared is 3²+y², the smallest y is √33.25 ≈ 5.77 m.

The maximum rise is 13sin 60°, so the greatest height is 1+13sin 60° ≈ 12.3 m.

B = (3, y, 1 + √(160 − y²)),
√33.25 ≤ y ≤ 10.

The positive square root keeps the anchor above A. Both endpoints are allowed because the angle and wall restrictions are inclusive.

Hint: begin with the plan view

The interior angle at A is 50° and the interior angle at B is 40°. C is therefore a right angle. Use the sine rule for AC and BC, then resolve AC into east and north components.

Check the survey

AC = 600sin 40° ≈ 385.7 m and BC = 600sin 50° ≈ 459.6 m.

C ≈ (247.906, 295.442), while M = (300,0). Thus MC = √((247.906−300)²+295.442²) = 300 m, using unrounded coordinates.

Alternatively, the midpoint of the hypotenuse of a right triangle is equally distant from its three vertices, so MC = 300 m.

Check the camera recommendation

At A, the straight-line distance is √((600sin 40°)²+150²) ≈ 414 m, and the elevation is approximately 21.3°. The angle is acceptable, but the range is not.

At M, the distance is √(300²+150²) ≈ 335 m and the elevation is tan⁻¹(150/300) ≈ 26.6°. M satisfies both limits.

At M the angle restriction gives 300tan 20° ≤ h ≤ 300tan 30°, approximately 109 m to 173 m. The range restriction gives h ≤ √(400²−300²) ≈ 265 m, so it is less restrictive. The acceptable heights are approximately 109–173 m, with boundaries included.

Check after both decisions are complete

Panel: Both areas are 70sin 40° ≈ 45.0 cm². The third sides are approximately 9.03 cm and 22.6 cm, giving perimeters of 33.0 cm and 46.6 cm. Only the 40° panel satisfies both limits.

Cable: The changes are (6,8,12); the planar distance is 10 m. The length is √244 ≈ 15.6 m, but the angle is tan⁻¹(12/10) ≈ 50.2°, so the proposed anchor fails the angle condition.

At the 45° limit the rise can be at most 10 m, so the greatest anchor height is 11 m. The cable then has length √200 ≈ 14.1 m, within its 16 m limit.