14.2 The Area of a Triangle

Take a rectangle ABCD and mark a point X somewhere on the top edge. Look at \triangle XDC, its base is the bottom edge DC and its apex is X up on top. Drop a perpendicular (an altitude) from X straight down to DC. That altitude carves the rectangle into pieces which reveal \triangle XDC to be precisely half the rectangle, wherever X happens to lie on the top edge.

So if the rectangle is 6 wide and 4 tall, then

\text{Area}(\triangle XDC) = \tfrac12 \times 6 \times 4 = 12 \text{ square units.}

Building the formula

To pin down a triangle’s area, we box it inside a rectangle whose base matches the triangle’s base and whose height matches the triangle’s height. The triangle is then exactly half of that rectangle:

\boxed{\;\textbf{Area of a triangle} = \tfrac12 \times \text{base} \times \text{height}.\;}

Here the height (altitude) is the perpendicular distance from a vertex to the opposite side (the base).

base height

Does it work for every triangle?

What if a triangle leans so far over that you cannot fit a rectangle on its base? Imagine \triangle ABC where the foot D of the altitude from A lands outside the segment BC. Then read \triangle ABC as the difference of two right-angled triangles, \triangle ADC and \triangle ADB, each of which does sit snugly in a rectangle:

\begin{aligned} \text{Area}(\triangle ABC) &= \tfrac12 \times h \times DC - \tfrac12 \times h \times DB \\ &= \tfrac12 \times h \times (DC - DB) \\ &= \tfrac12 \times h \times BC. \end{aligned}

The identical formula returns. So \tfrac12 \times \text{base} \times \text{height} is valid for every triangle, acute, right or obtuse.

Three heights, one area A triangle has three sides, so it offers three base–height pairs. Whichever side you treat as the base, \tfrac12 \times \text{base} \times \text{height} yields the same area. This unlocks many shortcuts: if you know the area one way, you can recover an unknown height the other way.

Worked Example

In \triangle ABC, the altitude AX to side BC is 4 and BC = 6. The altitude BY is drawn to side AC, where AC = 5. Find BY.

First find the area using base BC:

\text{Area}(\triangle ABC) = \tfrac12 \times AX \times BC = \tfrac12 \times 4 \times 6 = 12 \text{ sq. units.}

The same area, now using base AC:

\text{Area} = \tfrac12 \times BY \times AC = \tfrac12 \times BY \times 5 = \tfrac{5}{2}\,BY.

So \tfrac{5}{2}\,BY = 12, giving

BY = \frac{24}{5} = 4.8 \text{ units.}

Worked Example

A triangular flag has base 20\text{ cm} and the perpendicular height to that base is 14\text{ cm}. How much cloth does one flag use? If a string of 9 such flags is to be stitched, what is the total cloth area?

One flag:

\text{Area} = \tfrac12 \times 20 \times 14 = \tfrac12 \times 280 = 140 \text{ cm}^2.

Nine flags:

9 \times 140 = 1260 \text{ cm}^2.

A triangle split by a median

Draw both diagonals of a rectangle; they cross at the centre O and split the rectangle into four triangles. Do they all have equal area? They are not all congruent, but pick any two adjacent ones (say with bases OD and OB along a diagonal): they share the same altitude, and OB = OD because the diagonals of a rectangle bisect each other. Equal base, equal height, so equal area. All four triangles match in area.

This points to a broader fact:

The Median Splits Equally In any triangle, the segment from a vertex to the midpoint of the opposite side (a median) splits the triangle into two triangles of equal area, they have the same base length and the same height.

Triangles on the same base, between parallel lines

Let line \ell run parallel to BC. Consider every triangle with base BC whose apex slides along \ell. Since \ell keeps a fixed distance from BC, every such triangle has the same height, and therefore the same area! The apex may roam, yet the area holds firm.

These triangles do differ in perimeter, though. The one with the least perimeter has its apex on the perpendicular bisector of BC. The slick way to see this: treat \ell as a mirror, reflect C to C' on the far side, and notice that the trip B \to A \to C has the same length as B \to A \to C'. The shortest such trip is the straight segment BC', so the best apex A sits on BC', which works out to be exactly the perpendicular bisector of BC.

Figure it Out: Triangles

Practice

  1. Find the areas of these triangles: (i) base 5\text{ cm}, height 4\text{ cm}; (ii) base 4.5\text{ cm}, height 6\text{ cm}; (iii) base 6\text{ cm}, height 5\text{ cm}.
  2. In a triangle, the altitude to a side of length 10 units is 6 units. Find the altitude drawn to a different side of length 12 units.
  3. \triangle PQR is isosceles with PE \perp QR, and E is the midpoint of QR. The area of \triangle PER is 18 sq. units. Find the area of \triangle PQR.
  4. [Śulba-Sūtra] Describe a way to recast a rectangle into a triangle of equal area.
  5. [Śulba-Sūtra] Describe a way to recast a triangle into a rectangle of equal area.
  6. ABCD, BCEF and BFGH are three identical squares set side by side. (i) If the area of the red region is 64 sq. units, what is the area of the blue region? (ii) If the blue and red regions together total 200 sq. units, what is the area of each square?
  7. M and N are the midpoints of XY and XZ. What fraction of the area of \triangle XYZ is the area of \triangle XMN? (Hint: join NY.)
  8. Lakshmi walks from her cottage to a canal to collect water, then on to her storage drum. Trace the shortest route cottage \to canal \to drum.
    1. \tfrac12 \times 5 \times 4 = \mathbf{10\text{ cm}^2}. (ii) \tfrac12 \times 4.5 \times 6 = \mathbf{13.5\text{ cm}^2}. (iii) \tfrac12 \times 6 \times 5 = \mathbf{15\text{ cm}^2}.
  2. Find the area from base 10 and height 6: Area = \tfrac12 \times 10 \times 6 = 30 sq. units. Now use the other base: 30 = \tfrac12 \times 12 \times h, so h = \tfrac{60}{12} = \mathbf{5\text{ units}}.
  3. Because E is the midpoint of QR, the segment PE is a median of \triangle PQR, splitting it into two equal halves \triangle PEQ and \triangle PER. Hence Area(\triangle PQR) = 2 \times 18 = \mathbf{36\text{ sq. units}}.
  4. Draw a diagonal of the rectangle: it halves the area into a triangle. To get a triangle of the same area as the whole rectangle, keep the same base b but make the height twice the rectangle’s, that is, build a triangle of height 2h on base b, since \tfrac12 \times b \times 2h = b \times h.
  5. Reverse the idea: from a triangle of base b and height h, build a rectangle on the same base b with half the height, \tfrac{h}{2}, since b \times \tfrac{h}{2} = \tfrac12 bh. (A classic dissection cuts along the midline parallel to the base and folds the upper piece down to fill the sides.)
  6. Let each square have side s and area s^2. (i) By the symmetry of the three-square strip, the red and blue triangular regions are equal in area, so the blue region is also \mathbf{64\text{ sq. units}}. (ii) With red + blue = 200 and red = blue, each region is 100; matching this to the square geometry gives area of each square = 100\text{ sq. units} (so s = 10 units).
  7. Join NY. In \triangle XYZ, N is the midpoint of XZ, so NY is a median and \triangle XNY = \tfrac12 \triangle XYZ. In \triangle XNY, M is the midpoint of XY, so NM is a median and \triangle XMN = \tfrac12 \triangle XNY. Therefore \triangle XMN = \tfrac12 \times \tfrac12 \triangle XYZ = \mathbf{\tfrac14} of \triangle XYZ.
  8. Reflect the drum across the canal line to a point T'. The shortest route cottage \to canal \to drum is the straight segment from the cottage to T'; where it meets the canal is the spot Lakshmi should draw water. (This is the same mirror trick used for the least-perimeter triangle.)