Build a triangular cross-sectional area

Triangles on a parabolic floor

A solid stands on the region between the x-axis and the parabola \(y = 4x - x^2\). Its cross sections are equilateral triangles perpendicular to the x-axis, and an equilateral triangle has three equal sides. At \(x = 1\), the orange base on the floor is 3 long.

The area of a triangle is half of the base times the height. The height is the upright orange segment, at a right angle to the base.

Here the height is \(\tfrac{\sqrt{3}}{2}\cdot 3\), about 2.6, so the area is \(\tfrac{1}{2}\cdot 3\cdot\tfrac{3\sqrt{3}}{2}\), about 3.90.

Drag the slice to \(x = 2\). The base is 4, the height is \(2\sqrt{3}\), and the area is \(4\sqrt{3}\). Both lengths change with \(x\), so the area changes too.

Half of the base times the height

The orange base is \(w\), the upright orange segment is the height \(h\), and the area of the teal triangle is \(\tfrac{1}{2} w h\).

Press the buttons. An isosceles right triangle has a right angle between two equal sides, called the legs, and its longest side is the hypotenuse. Either the hypotenuse or a leg can lie on the base.

For each shape the height is a fixed multiple of \(w\): \(\tfrac{\sqrt{3}}{2}w\), \(\tfrac{1}{2}w\) or \(w\). The area is then a fixed number times \(w^2\), and the shape sets that number.

Notation

1. The area of a triangle is half of the base times the height, where the height is measured at a right angle to the base.

   \[\cb{A} = \frac{1}{2}\,\cy{w}\,\cy{h}\]

2. If a leg lies on the base, the other leg stands upright and is also \(w\) long, so the height is \(w\).

   \[\cb{A} = \frac{1}{2}\,w\cdot w = \frac{1}{2}\,w^2\]

3. If the hypotenuse lies on the base, the height splits the triangle into two smaller isosceles right triangles. Each has legs \(h\) and \(\tfrac{w}{2}\), so \(h = \tfrac{w}{2}\).

   \[\cb{A} = \frac{1}{2}\,w\cdot\frac{w}{2} = \frac{1}{4}\,w^2\]

4. An equilateral triangle with side \(w\) on the base has height \(\tfrac{\sqrt{3}}{2}w\). The height splits it into two right triangles with hypotenuse \(w\) and a leg \(\tfrac{w}{2}\), so \(h^2 = w^2 - \tfrac{w^2}{4}\).

   \[\cb{A} = \frac{1}{2}\,w\cdot\frac{\sqrt{3}}{2}\,w = \frac{\sqrt{3}}{4}\,w^2\]

5. If the problem gives the height, as a number or as a multiple of \(w\), use it. Then substitute \(w(x)\), so that \(A\) depends on \(x\) only.

When it fails

The formula needs the height at a right angle to the base, and a slanted side is not that height. Press the button to draw a slanted side of the equilateral triangle in red.

The red side is as long as the base \(w\), but the orange height is only about \(0.87w\), because the side leans. Using the side as the height makes the area too large.

Example 1: a right triangle on the hypotenuse

The base is the region between the x-axis and \(y = x^2\), for \(0 \le x \le 3\). The cross sections are isosceles right triangles perpendicular to the x-axis, with the hypotenuse on the base. Find \(A(x)\).

1. The region runs from the x-axis up to the curve, so the hypotenuse is \(x^2\) long.

   \[\cy{w(x)} = x^2\]

2. With the hypotenuse on the base, the height is half of the hypotenuse.

   \[\cy{h(x)} = \frac{x^2}{2}\]

3. Multiply half of the base by the height.

   \[\cb{A(x)} = \frac{1}{2}\cdot x^2\cdot\frac{x^2}{2} = \frac{x^4}{4}\]

4. As a check, take \(x = 2\). The base is 4 and the height is 2, so the area is \(\tfrac12\cdot 4\cdot 2 = 4\). The formula gives \(\tfrac{16}{4} = 4\).

Example 2: equilateral triangles on a disk

The base is the disk \(x^2 + y^2 \le 4\). The cross sections are equilateral triangles perpendicular to the x-axis. Find \(A(x)\).

1. At \(x\), the circle gives \(y = \pm\sqrt{4 - x^2}\), so the base runs from \(-\sqrt{4 - x^2}\) up to \(\sqrt{4 - x^2}\). Subtract the lower value from the upper value.

   \[\cy{w(x)} = 2\sqrt{4 - x^2}\]

2. A tempting wrong move: Take the height equal to the side.

   \[\cr{A(x) = \frac{1}{2}\,w\cdot w = 2\left(4 - x^2\right)}\]

   At \(x = 0\) the side is 4, so this makes the triangle 4 tall. The teal equilateral triangle is only \(2\sqrt{3}\) tall, about 3.46, so the red triangle is taller than the teal one.

3. The height of an equilateral triangle is \(\tfrac{\sqrt{3}}{2}\) times the side.

   \[\begin{aligned} \cy{h(x)} &= \frac{\sqrt{3}}{2}\cdot 2\sqrt{4 - x^2} \\ &= \sqrt{3}\sqrt{4 - x^2} \end{aligned}\]

4. Multiply half of the base by the height.

   \[\begin{array}{l} \cb{A(x)} = \frac{1}{2}\,w\,h \\ \quad = \frac{1}{2}\cdot 2\sqrt{4 - x^2}\cdot\sqrt{3}\sqrt{4 - x^2} \\ \quad = \sqrt{3}\left(4 - x^2\right) \end{array}\]

Practice

The base is the region between the x-axis and \(y = 3 + x^2\), for \(0 \le x \le 2\). The cross sections are isosceles right triangles perpendicular to the x-axis, with a leg on the base. Find \(A(x)\).

One leg lies on the base, and the other leg stands upright at its end, as in the teal triangle. The two legs are equal.

Questions

The base lies between the x-axis and the line \(y = 5 - x\). The cross sections are perpendicular to the x-axis, and each question is about the slice at \(x = 1\), where the base is \(5 - 1 = 4\) long.

1. The cross section is an equilateral triangle. What is its area?

   • \(4\sqrt{3}\)
   • 8
   • \(8\sqrt{3}\)
   • \(2\sqrt{3}\)
   Answer

   \(4\sqrt{3}\)

2. The cross section is an isosceles right triangle with its hypotenuse on the base. What is its area?

   • 4
   • 8
   • 2
   Answer

   4

3. A cross section is a triangle with base \(3x\) and a height twice the base. What is \(A(x)\)?

   • \(9x^2\)
   • \(18x^2\)
   • \(\tfrac{9}{2}x^2\)
   Answer

   \(9x^2\)

Summary

You write the height of the triangle from its base, and you multiply half of the base by the height.

The shape of the triangle fixes the height as a multiple of the base, so the area is a constant times the base squared.

Use this for every triangular cross section before you set up the volume integral.

This is one of four skills in topic 8.8 of Which & Why: AP Calculus.