# Area of a triangle inquiry

# The prompt

Mathematical inquiry processes: Explore; test particular cases; analyse structure; reason and prove. Conceptual field of inquiry: Area of a triangle; differentiation; iteration; cosine rule.

The prompt invites students to use the formula for the area of a non-right-angled triangle:

During classroom inquiry, students' initial questions and observations have included:

How do you work out the area of the triangle?

The triangle is not drawn to scale because it's not isosceles.

The statement must be true because the sides are getting longer.

n can't be 180 because then the angle would not exist.

You cannot work out the area of the triangle unless n = 90. Then the angle is 90 degrees and the height and base are both 90.

In the first phase of the inquiry, the teacher will introduce the formula and, if appropriate, show how it is derived from the simpler formula (half the product of base and height).

As students explore the area, they realise that the contention in the prompt is false. The greatest area is achieved when n = 131.1 (accurate to one decimal place), at which point the area starts to decrease. Students realise that the area must be a minimum at the limits of n. That is because:

As n tends to 0, the angle gets closer to 180o, and

As n tends to 180, the angle gets closer to zero degrees.

As you approach both limits, the triangle tends towards a straight line. One way to show this (and the maximum area) is to use a spreadsheet.

In the another phase of the inquiry, students have changed the parameters in the prompt to see how the value of n changes for the maximum area. For example, the side lengths could be (n - 1) and the angle (180 - 10n)o and so on.

# Proof

Shawki Dayekh, a teacher of mathematics in north London (UK), proved the maximum area occurs when n = 131.14 (accurate to 2 decimal places). He used the product rule and iteration, expressing the angle in radians (see the picture).

The value of x is 131.14 at the end of Shawki's working means an angle of 48.86o gives the maximum area.

# Alternative prompt

The prompt could be presented in a simpler form by using an acute angle (see illustration), giving students fewer cases to explore.

In this case, the area increases until n = 61.7, whereupon it starts to decrease. At n = 61.7, the angle is 28.3o and the area equals 902.4 square units (accurate to one decimal place).

# Another line of inquiry

Perimeter as n increases

After the initial phase of the inquiry, students in a year 10 class at Haverstock School (London, UK) developed another line of inquiry when they posed questions about the perimeter:

As n increases, does the perimeter increase?

What is the perimeter of the triangle when the area is at its maximum?

What is the greatest possible perimeter?

The students quickly discounted a suggestion that the greatest perimeter occurs when the angle is at its largest - that is, as it approaches 180o .

In order to find the length of the third side of the triangle, Michael Joseph, the students' teacher, took the opportunity to explain the cosine rule.

He created a spreadsheet (using radians) to show that the perimeter is at a maximum when n = 2.61308 (accurate to five decimal places) at which point the angle is (p - 2.61308) radians or 30.28o .

Michael was enthusiastic about inquiry learning,

"The inquiry was such a natural way to introduce the cosine rule at a time when students needed it to answer their own questions. That meant they got lots of practice, but with a purpose."