Connected ideas

Circle, trigonometry and area

Keep a chord fixed and move a triangle vertex.

Predict, test and explain

Step 1 of 3

Predict

Keep radius and chord fixed; move P along its major arc. Must triangle APB keep the same area?

\text{area}=\tfrac12 AB\times h\approx 28.38

Scroll the diagram sideways to see every label.

Fixed angle

\angle APB=40^\circ

Height

h\approx 8.83

\text{area}\approx 28.38

A fixed chord gives a fixed angle in this segment. The perpendicular height changes as P moves, so the area changes. The height may meet an extension of the chord. The largest area is at the top of the circle.

Check the working

Quantity	Value
Radius	5
Chord AB	$\approx 6.4279$
Angle at P	40°
Perpendicular height	$\approx 8.8302$
Area	$\approx 28.3798$

Explore and compare

Can you break the rule?

Triangles on the same chord, with their third vertex on the same arc, have equal areas.

Always, sometimes or never? Test it in the simulation.

Hint

Keep the circle and chord fixed. Move the third vertex and compare the perpendicular heights.

Check the reasoning

Sometimes. The subtended angle stays fixed, but the perpendicular height can change. Equal heights give equal areas; different heights do not.

$A=\tfrac12 bh$

One counterexample disproves “always”. Examples alone do not prove it.

Keep the circle and chord fixed. Predict whether the angle and triangle area both stay the same.

Predict

Try it in the simulation. Change one thing at a time.

Hint

Choose Angle and area. Set radius 5 and central angle 80°. Save values at arc position 25, then move P to 50.

Explain

Notes and saved values stay in this tab. They disappear when you reload.

Hints, self-check and connections

Try a self-check

Fixed values, separate from the diagram controls.

Two triangles share an 8 cm chord as their base. Their third vertices lie on the same major arc of the same circle, with perpendicular heights to the chord of 3 cm and 5 cm. Must their areas be equal? Calculate them.

$b=8\,\text{cm},\quad h_1=3\,\text{cm},\quad h_2=5\,\text{cm}$

Hint

Each area is half the chord length times its perpendicular height.

Reveal answer

The areas are 12 cm² and 20 cm². The chord subtends equal angles at these vertices, but different heights give different areas.

$A_1=\tfrac12\times8\times3=12\,\text{cm}^2,\quad A_2=\tfrac12\times8\times5=20\,\text{cm}^2$

Build an explanation

Hint 1

A fixed chord AB subtends the same angle at each point C on the same major arc, away from A and B. This angle is half the minor angle AOB at the centre.

Hint 2

Join the centre O to the chord's midpoint M. OM is perpendicular to AB and AM = AB/2, giving a right triangle for Pythagoras and trigonometry.

Hint 3

For an inscribed triangle, a/sin A = 2R, where R is the circle's radius. Area = 1/2 × base × perpendicular height, so the area can change even when the angle stays fixed.

Worked example

This example uses fixed values, separate from the diagram controls.

A chord AB is 6 cm long in a circle of radius 5 cm. C lies on the major arc AB. Find sin ∠ACB exactly, then find the triangle's area when C is at the point of the circle furthest from AB.

Let O be the centre and M the midpoint of AB. Then AM = 3 cm and OA = 5 cm.
Right triangle OMA gives OM = √(5² − 3²) = 4 cm.
∠ACB is half the minor central angle AOB, so it equals ∠AOM. Therefore sin ∠ACB = AM/OA = 3/5.
This also checks the extended sine rule: AB/sin ∠ACB = 6/(3/5) = 10 cm = 2R.
At the furthest point, the perpendicular height is R + OM = 5 + 4 = 9 cm. The area is 1/2 × 6 × 9 = 27 cm².

Answer: sin ∠ACB = 3/5; area = 27 cm² at the furthest point.

Connect this idea

Watch out: Equal angles do not imply equal triangle areas. As C moves along the same major arc, ∠ACB stays fixed but the perpendicular height and area change.