Chapter 15 — Areas
In earlier chapters we discussed how to find the areas of simple figures like circles and triangles. In this chapter, we learn how to find the area of more complex figures and of simple figures in complex problems.
In this chapter
- 15.1 Similar Figures
- 15.2 Same Base / Same Altitude
- 15.3 Complicated Figures
- Area Summary Table
- Exercises & Problems
Area Summary Table
| Shape | Area Formula |
|---|---|
| Triangle | $\frac{1}{2}bh$ |
| Rectangle | $lw$ |
| Parallelogram | $bh$ |
| Trapezoid | $\frac{1}{2}(b_1 + b_2)h$ |
| Circle | $\pi r^2$ |
| Circular Sector (angle $\theta$) | $\frac{\theta}{360°}\pi r^2$ |
| Regular Polygon ($n$ sides, side $s$) | $\frac{1}{4}ns^2 \cot!\left(\frac{\pi}{n}\right)$ |
| Equilateral Triangle (side $s$) | $\frac{s^2\sqrt{3}}{4}$ |
15.1 Similar Figures
Key Principle. If two figures are similar and the ratio of corresponding lengths is $k$, then the ratio of their areas is $k^2$.
This applies to any pair of similar figures. Since all circles are similar, if one circle has a radius twice as large as another, its area is $2^2 = 4$ times as large. The same principle holds for squares, equilateral triangles, regular polygons, and any other similar figures.
Example 15-1. The area of a triangle is 36. Find the area of the triangle formed by connecting the midpoints of its sides.
Solution. We first prove that any triangle is similar to the triangle formed by connecting the midpoints of its sides. In $\triangle ABC$, let $D$, $E$, $F$ be midpoints of $BC$, $CA$, $AB$ respectively. Since $E$ and $F$ are midpoints, $\frac{AE}{AC} = \frac{AF}{AB} = \frac{1}{2}$. Since $\angle EAF = \angle CAB$, we have $\triangle CAB \sim \triangle EAF$ by SAS Similarity, so $\frac{EF}{CB} = \frac{1}{2}$. Similarly $\frac{FD}{AC} = \frac{1}{2}$ and $\frac{ED}{AB} = \frac{1}{2}$, so $\triangle ABC \sim \triangle DEF$ by SSS Similarity. Thus:
\[\frac{[DEF]}{[ABC]} = \left(\frac{1}{2}\right)^2 = \frac{1}{4}\]Hence $[DEF] = \frac{[ABC]}{4} = \frac{36}{4} = \boxed{9}$.
Example 15-2. The ratio of the areas of two squares is 6. Find the ratio of the lengths of their diagonals.
Solution. All squares are similar. The ratio of areas is the square of the ratio of any corresponding lengths. Hence the ratio of the diagonal lengths is $\sqrt{6}$.
Example 15-3. In trapezoid $ABCD$, $AB | CD$ and the diagonals meet at $E$. If $AB = 4$ and $CD = 12$, show that the area of $\triangle CDE$ is 9 times the area of $\triangle ABE$.
Proof. Since $AB | CD$, we have $\angle BAE = \angle DCE$ and $\angle ABE = \angle CDE$ (alternate interior angles). By AA Similarity, $\triangle ABE \sim \triangle CDE$. Since $\frac{CD}{AB} = \frac{12}{4} = 3$, we get:
\(\frac{[CDE]}{[ABE]} = \left(\frac{CD}{AB}\right)^2 = 3^2 = 9 \qquad \square\)
15.2 Same Base / Same Altitude
Principle. If two triangles have the same altitude, the ratio of their areas equals the ratio of their bases:
\[\frac{[ABC]}{[DEF]} = \frac{BC}{EF}\]Similarly, if two triangles have the same base, the ratio of their areas equals the ratio of their altitudes.
Tip. This is most powerful when the “same base” is literally the same segment shared by two triangles, or when the triangles have their apex on the same line parallel to the base (guaranteeing equal altitudes).
Proof Sketch
Given $\triangle ABC$ and $\triangle DEF$ with equal altitudes $h_a = h_d$ to bases $BC$ and $EF$:
\[\frac{[ABC]}{[DEF]} = \frac{\frac{1}{2}(BC)(h_a)}{\frac{1}{2}(EF)(h_d)} = \frac{BC}{EF}\]Example 15-4. Show that by drawing the three medians of a triangle, we divide the triangle into six regions of equal area.
Proof. Let $G$ be the centroid of $\triangle ABC$, and let $D$, $E$, $F$ be the midpoints of $BC$, $CA$, $AB$ respectively.
Step 1. $[ACD] = \frac{[ABC]}{2}$ because $\triangle ACD$ and $\triangle ABC$ share altitude from $A$, and $\frac{CD}{CB} = \frac{1}{2}$.
Step 2. $[GCD] = \frac{[ACD]}{3}$ because $\triangle GCD$ and $\triangle ACD$ share altitude from $C$, and $\frac{GD}{AD} = \frac{1}{3}$ (the centroid divides each median in ratio $2:1$).
\[[GCD] = \frac{[ACD]}{3} = \frac{[ABC]/2}{3} = \frac{[ABC]}{6}\]Similarly each of the other 5 small triangles has area $\frac{[ABC]}{6}$. $\square$
Example 15-5. In $\triangle ABC$, $D$ is the midpoint of $AB$, $E$ is the midpoint of $DB$, and $F$ is the midpoint of $BC$. If the area of $\triangle ABC$ is 96, find the area of $\triangle AEF$. (AHSME 1976)
Solution. Since $\triangle ABF$ shares the same altitude as $\triangle ABC$ (from $C$ and from $F$ respectively — actually, $F$ is on $BC$, so $\triangle ABF$ shares altitude from vertex $A$ to… let’s reason differently).
$\triangle ABF$ has the same altitude from $A$ as $\triangle ABC$, and base $BF = \frac{1}{2}BC$, so $[ABF] = \frac{[ABC]}{2} = 48$.
Now $\triangle AEF$ shares the same altitude from $F$ as $\triangle ABF$, and base $AE = \frac{3}{4}AB$ (since $E$ is the midpoint of $DB$, so $AE = AD + DE = \frac{1}{2}AB + \frac{1}{4}AB = \frac{3}{4}AB$). Thus:
\([AEF] = \frac{3}{4}[ABF] = \frac{3}{4}(48) = \boxed{36}\)
Example 15-6. Line $\ell$ is parallel to segment $AB$. Show that for all points $X$ on $\ell$, $[ABX]$ is the same.
Proof. No matter where $X$ is on $\ell$, the altitude from $X$ to $AB$ is the same (the perpendicular distance between the parallel lines). Since $AB$ is constant, $[ABX] = \frac{1}{2}(AB)(h)$ is constant. $\square$
Example 15-7. If diagonal $AC$ of quadrilateral $ABCD$ divides diagonal $BD$ into two equal segments, prove that $[ACD] = [ACB]$. (M&IQ 1992)
Proof. Let $X$ be the intersection of the diagonals. Then $X$ is the midpoint of $BD$. Triangles $\triangle ACD$ and $\triangle ACB$ share base $AC$, so we need to show they have equal altitudes to $AC$.
Draw $BY \perp AC$ and $DZ \perp AC$. Since $DX = BX$ and $\angle DXZ = \angle BXY$ (vertical angles), triangles $\triangle DXZ$ and $\triangle BXY$ are congruent by SA for right triangles. Hence $DZ = BY$, so:
\([ACD] = \frac{1}{2}(AC)(DZ) = \frac{1}{2}(AC)(BY) = [ACB] \qquad \square\)
15.3 Complicated Figures
Sometimes the area of a figure must be found by breaking it into simpler pieces—triangles, rectangles, sectors—and adding or subtracting their areas.
Tips for complicated area problems:
- Draw radii to separate sectors and circular segments from the rest of the diagram.
- Look for right and equilateral triangles. Draw lines to isolate them.
- Draw diagonals of quadrilaterals to split them into triangles.
- Think of the problem as a puzzle: which simple areas can be added or subtracted to produce the desired region?
Example 15-8. Find the area between two concentric circles with radii 2 and 3.
Solution. The larger circle has area $9\pi$ and the smaller has area $4\pi$. The shaded annulus has area:
\(9\pi - 4\pi = \boxed{5\pi}\)
Example 15-9. Find the area of a regular octagon with side length 2.
Solution. A regular octagon can be formed by cutting isosceles right triangle corners from a square. If the octagon has side $s = 2$, each corner triangle has legs of length $\frac{s}{\sqrt{2}} = \sqrt{2}$. The side of the square is $2 + 2\sqrt{2}$, so the square has area $(2+2\sqrt{2})^2 = 12 + 8\sqrt{2}$. Each corner triangle has area $\frac{(\sqrt{2})^2}{2} = 1$, and there are 4 corners:
\(\text{Octagon area} = (12 + 8\sqrt{2}) - 4(1) = \boxed{8 + 8\sqrt{2}}\)
Example 15-10. $\triangle ABC$ is an isosceles right triangle with $AB = 2\sqrt{2}$. The midpoint of $AB$ is the center of a semicircle on $AB$, and $C$ is the center of a quarter circle through $A$ and $B$. Find the shaded area (the region inside the semicircle but outside the quarter circle, plus the region inside the quarter circle but outside the semicircle). (MAΘ 1990)
Solution. Since $AB = 2\sqrt{2}$ and $\triangle ABC$ is isosceles right, $AC = CB = 2$.
- $[ABC] = \frac{1}{2}(2)(2) = 2$
- Quarter circle $ABC$ (radius $BC = 2$): area $= \frac{\pi(2)^2}{4} = \pi$
- Semicircle on $AB$ (radius $\sqrt{2}$): area $= \frac{\pi(\sqrt{2})^2}{2} = \pi$
The shaded area $=$ (semicircle) $+$ (triangle) $-$ (quarter circle) $= \pi + 2 - \pi = \boxed{2}$.
Example 15-11. A square has side length 4, and four semicircles have the sides of the square as diameters. Find the area of the “leaves” where semicircles overlap.
Solution. Each semicircle has area $\frac{\pi(2)^2}{2} = 2\pi$. The four semicircles together have total area $4(2\pi) = 8\pi$. Since the semicircles cover the square, but each “leaf” is counted twice (once in each of two semicircles), the overlap area equals total semicircle area minus square area:
\(8\pi - 4^2 = \boxed{8\pi - 16}\)
Example 15-12. Three circles of radius 6 have centers equally spaced on a fourth circle also of radius 6. Find the total area of the three “leaf” regions common to three of the four circles. (MATHCOUNTS 1992)
Solution. Drawing line $AC$ to bisect one leaf, and radii $AB$ and $BC$ of the lowest circle: since $AB = BC = AC = 6$, $\triangle ABC$ is equilateral.
The circular segment from arc $AC$ is: (sector $ABC$) $-$ ($\triangle ABC$):
\[\text{Sector area} = \frac{60°}{360°}\pi(6)^2 = 6\pi, \qquad [\triangle ABC] = \frac{6^2\sqrt{3}}{4} = 9\sqrt{3}\] \[\text{Segment area} = 6\pi - 9\sqrt{3}\]Each leaf consists of 2 such segments, and there are 3 leaves:
\(\text{Total} = 6(6\pi - 9\sqrt{3}) = \boxed{36\pi - 54\sqrt{3}}\)
Exercises
Exercise 15-1. If two triangles have the same base, prove that the ratio of their areas equals the ratio of their altitudes.
Problems to Solve for Chapter 15
Problem 252. Sides $AB$, $BC$, $CD$, and $DA$ of convex quadrilateral $ABCD$ have lengths 3, 4, 12, and 13, respectively; and $\angle CBA$ is a right angle. What is the area of the quadrilateral? (AHSME 1980)
Problem 253. Find the total area of the figure with right angles and segment measures as shown (a staircase-shaped polygon with measurement 17 and 25). (MAΘ 1990)
Problem 254. Find the ratio of the area of an equilateral triangle inscribed in a circle to the area of a square circumscribed about the same circle. (MAΘ 1987)
Problem 255. If a square is inscribed in a semicircle of radius $r$ and the square has area 8, find the area of a square inscribed in a circle of radius $r$. (MAΘ 1987)
Problem 256. Points $D$, $E$, and $F$ are midpoints of the sides of equilateral triangle $ABC$. The shaded central triangle is formed by connecting the midpoints of the sides of $\triangle DEF$. What fraction of the total area of $ABC$ is shaded? (MATHCOUNTS 1992)
Problem 257. A cow is tied to the corner of a 20 ft by 15 ft shed with a 30 ft rope. Find her total grazing area. (MAΘ 1992)
Problem 258. Find the ratio of the area of $\triangle ACE$ to the area of rectangle $ABCD$. (MATHCOUNTS 1986)
Problem 259. Find the area of the largest triangle that can be inscribed in a semicircle of radius $r$. (Vertices are on the semicircle or its diameter.) (AHSME 1950)
Problem 260. Given hexagon $ABCDEF$ with sides of length 6, six congruent $30°$-$60°$-$90°$ triangles are drawn. Find the ratio of the area of the smaller hexagon formed to the area of the original hexagon. (MATHCOUNTS 1988)
Problem 261. In the figure, $\triangle ABC$ and $\triangle ADE$ are both equilateral with side length 4. Segment $AD$ is perpendicular to $BC$. Find the area of the region common to both triangles. (MAΘ 1992)
Problem 262. A rhombus is formed by two radii and two chords of a circle whose radius is 16 feet. What is the area of the rhombus in square feet? (AHSME 1956)
Problem 263. A square has sides of length 9 cm. The radius of an interior circle is 2 cm. What is the area of the shaded region? (MATHCOUNTS 1992)
Problem 264. In the diagram, $ABCD$ and $DEFG$ are squares of area 16. If $H$ is the midpoint of $BC$ and $EF$, find the total area of $ABHFGD$. (MAΘ 1987)
Problem 265. Let $M$, $N$, $P$ be the midpoints of sides $BC$, $CA$, $AB$ of $\triangle ABC$ respectively. Prove that segments $MN$, $NP$, $PM$ divide $\triangle ABC$ into four triangles of equal area. (M&IQ 1992)
Problem 266. In rectangle $ABCD$, interior point $E$ is chosen at random. Prove that the sum of the areas of triangles $AEB$ and $EDC$ is the same regardless of where $E$ is chosen. (MAΘ 1990)
Problem 267. Find the number of square units in the area of the inscribed pentagon with right angle and dimensions as shown. (MATHCOUNTS 1988)
Problem 268. Let $N$ be an arbitrary point on the median $CM$ of $\triangle ABC$. Prove that $[AMN] = [NMB]$ and $[ANC] = [BNC]$. (M&IQ 1992)
Problem 269. A 3-meter square and a 4-meter square overlap. $D$ is the center of the 3-meter square. Find the area of the shaded region $DGFE$. (MAΘ 1987)
Problem 270. Square $ABCD$ is inscribed in a circle. Point $X$ lies on minor arc $AB$ such that $[XCD] = 993$ and $[XAB] = 1$. Find $[XAD] + [XBC]$. (Mandelbrot #3)
Problem 271. The convex pentagon $ABCDE$ has $\angle A = \angle B = 120°$, $EA = AB = BC = 2$ and $CD = DE = 4$. What is the area of $ABCDE$? (AHSME 1993)
Problem 272. A triangle is inscribed in a circle. The vertices divide the circle into three arcs of lengths 3, 4, and 5. What is the area of the triangle? (AHSME 1989)
Problem 273. Point $A$ is the center of a 100 cm × 100 cm square. Find $x$ such that the shaded region has an area that is one-fifth of the area of the square. (MATHCOUNTS 1992)
Problem 274. Let $M$ be any point on diagonal $AC$ of rectangle $ABCD$. Show that $[ADM] = [AMB]$. (M&IQ 1992)
Problem 275. $ABCD$ is a square with $AB = 5$ and $AE = AF = CG = CH$. The shaded region is $\frac{5}{9}$ the area of $ABCD$. Find $AF$. (MATHCOUNTS 1992)
Problem 276. The medians to the legs of an isosceles triangle are perpendicular to each other. If the base of the triangle is 4, find its area. (MAΘ 1990)
Problem 277. A square has side 6 and curved paths are arcs of circles centered at vertices $A$ and $B$. Find the area of the shaded section. (Mandelbrot #3)
The Big Picture. We’ve seen how to calculate areas of figures made of straight lines and circular arcs. One of the great accomplishments of calculus is enabling us to find the areas of far more complex figures. The key idea — breaking a region into tiny rectangles — is the foundation of the integral. Calculus builds complex ideas from these simple building blocks.