The origins of algebra are closely connected to geometry. Problems that connect questions about length, area, and volume lead to interesting equations. And early algebraic equalities were viewed as relationships among geometric quantities.

**Problem.**

A square is inscribed in a triangle as shown in the picture below.

The base of the triangle has length b, and its height has length h. Find the length x of the side of the square. (The triangle doesn't have to be equilateral.)

**A modern solution:**

The area of the small triangle on the top is x*(h-x)/2

The area of the square is x^{2}.

The area of the triangle that is obtained by sliding together the two smaller triangles standing on the base is (b-x)*x/2.

Therefore,

b*h/2 = x*(h-x)/2 + x^{2} + (b-x)*x/2

b*h = x*(h-x) + 2*x^{2} + (b-x)*x

= x*h - x^{2} + 2*x^{2} + b*x - x^{2}

= x*h + b*x

= x*(h + b)

So

x = b*h/(b + h).

**A second problem.**

In some triangles (e.g., those with three angles that measure 90 degrees or less), three different squares may be inscribed, one for each of the three sides. Do the three squares have the same area?

The answer is, not always; the area of the inscribed square is a function of the length of the side on which it sits. Below we derive the formula.

But Area_{1} + Area_{2} + Area_{3} = A. Therefore,

(x^{2}/b^{2})*A + ((b-x)^{2}/b^{2})*A + x^{2} = A

x^{2} + (b-x)^{ 2} + x^{2}*b^{2}/A = b^{2}

x^{2} + b^{2} - 2*b*x + x^{2} + x^{2}*b^{2}/A = b^{2}

(2 + b^{2}/A)*x^{2} - 2*b*x = 0

See also http://www.hawaii.edu/suremath/jsquareInTriangle.html

**A third problem. **

Compute x = b*h/(b+h), where b = length of base on which the square will sit, and h = height of triangle from that base.

Using an index card, mark length x on one of its sides. Run the unmarked edge of the index card along the base until the mark x cuts the edge of the triangle. Draw length x perpendicular to the base b, as in picture 2.

Turn the card 90 degrees, and along the top of the card draw a segment of length x parallel to the base (picture 3).

Again using the index card, finish the square by drawing another segment of length x perpendicular to the base (picture 4).

Last Modified: July 26, 2004