Same Base, Same Height, and Area Ratios

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Learning goals

Identify the base and height conditions that guarantee equal triangle areas
Calculate area ratios for triangles whose perpendicular heights are equal
Derive shared-angle area ratios from corresponding perpendicular height ratios
Justify area bisection by a median using equal bases and a common height
Find areas in the six equal regions formed by the three medians

Equal bases and equal heights

Equal base lengths and equal corresponding heights guarantee equal triangle areas. For instance, a base of 9 cm and a perpendicular height of 4 cm give 18 square cm, wherever the height meets the base’s supporting line. The area formula from Area from Rectangles to Triangles applies to acute, right, and obtuse triangles.

Write $A_{ABC}$ for the area of triangle $ABC$ ; the subscript names the region. All triangles here are nondegenerate, so their bases, heights, and areas are positive.

Why equal bases and heights give equal areas#

First, with base 12 and height 7, either triangle has area $\frac12(12)(7)$ , or 42 square units. Changing the apex sideways changes neither factor.

In general, two triangles with equal chosen base lengths $b$ and equal corresponding perpendicular heights $h$ both have area

A=\frac12 bh.

The numbers 12 and 7 had no special role. The formula depends only on these two measurements.

Apexes on a parallel have equal distances to the base line, including outside feet.

In this figure, the common base is $BC$ . Because $AD$ is parallel to $BC$ , the perpendicular distances from $A$ and $D$ to the line $BC$ are equal. The foot from $D$ lies outside segment $BC$ , but it still gives the height. Thus the two triangles have equal areas.

The bases may instead be separate segments with equal lengths; equality of their matching heights still suffices. Conversely, equal areas alone do not establish equal bases and equal heights. A base of 3 with height 12 and a base of 6 with height 6 both give 18 square units.

Area with a fixed base and height

Vertex A

A is above the interior of BC. BC = 8 units. The perpendicular separation is 5 units. Area ABC = 20 square units.

BC and the perpendicular separation h stay fixed as A moves along the upper parallel.

Worked example 1 Move an apex along a parallel

Problem. Segment $BC$ is 11 cm long. Points $A$ and $D$ lie on a line parallel to $BC$ , at perpendicular distance 6 cm from it. The perpendicular from $D$ meets the line $BC$ beyond $C$ . Find $A_{ABC}$ and $A_{DBC}$ .

Solution. Both triangles use base $BC$ and height 6 cm, including the triangle with the outside foot.

\begin{aligned} A_{ABC}&=\frac12(11)(6)\\ A_{ABC}&=33\text{ cm}^2. \end{aligned}

The second triangle has the same two factors, so

A_{DBC}=33\text{ cm}^2.

The parallel configuration is the one shown above; the new measurements do not change its reasoning.

Check your understanding

Which pair of conditions guarantees equal triangle areas?

A Equal base lengths and equal corresponding perpendicular heights B Equal base lengths alone C Equal corresponding heights alone D Equal perimeters

A divided base transfers its ratio to area

Triangles with equal heights have areas in the ratio of their bases. For example, bases 3 and 8 with a common height of 4 give areas 6 and 16. Both areas are the base multiplied by 2, so their ratio is $3:8$ .

Both bases lie on the same line, so the perpendicular height is shared.

Let $D$ lie strictly between $B$ and $C$ , and let $A$ lie off line $BC$ . The two triangles use bases $BD$ and $DC$ . Their common height is the perpendicular distance from $A$ to the same line $BC$ , whether its foot is inside or outside either base segment.

Why common-height areas follow the base ratio#

For a numerical run, bases 7 and 10 with common height 6 give areas 21 and 30. Dividing cancels the shared factor 3, leaving $7/10$ .

For arbitrary positive bases $b_1,b_2$ and common positive height $h$ ,

\frac{A_1}{A_2}=\frac{\frac12 b_1h}{\frac12 b_2h}.

Cancel the nonzero common factor $\frac12 h$ :

\frac{A_1}{A_2}=\frac{b_1}{b_2}.

Here $A_1$ and $A_2$ name the two areas in the same order as the bases. In the partition above, this gives

\frac{A_{ABD}}{A_{ADC}}=\frac{BD}{DC}.

No particular height was needed, only equality of the two heights. Likewise, for equal positive bases, the area ratio equals the corresponding height ratio, by canceling the common factor $\frac12 b$ instead.

Because $D$ is inside $BC$ , the two small triangles have disjoint interiors and fill $ABC$ . To split its area, share the total according to the base ratio.

Area ratios from a divided base

Length BD

BD = 3 units. DC = 5 units. The perpendicular separation is 5 units. Area ABD = 7.5 square units. Area ADC = 12.5 square units. Area ABC = 20 square units. The area ratio ABD to ADC is 3 to 5.

D stays between B and C. Compare the two base lengths with the areas of triangles ABD and ADC, which share the same perpendicular height.

Worked example 2 Use a ratio without finding the height

Problem. In triangle $ABC$ , $D$ lies on $BC$ , with $BD=5$ cm and $DC=9$ cm. The total area is 63 square cm. Find $A_{ABD}$ and $A_{ADC}$ .

Solution. The common-height ratio is $5:9$ . There are 14 equal ratio parts, so one part is

63\div14=4.5\text{ cm}^2.

The area beside the 5 cm base is

\begin{aligned}A_{ABD}&=5(4.5)\\A_{ABD}&=22.5\text{ cm}^2.\end{aligned}

The area beside the 9 cm base is

\begin{aligned}A_{ADC}&=9(4.5)\\A_{ADC}&=40.5\text{ cm}^2.\end{aligned}

The areas add to 63 square cm. The partition figure above shows which region goes with each base.

Check your understanding

D lies inside BC in triangle ABC. The ratio BD to DC is 4 to 11. If triangle ABD has area 18 square cm, what is the area of triangle ADC?

A 72 square cm B 6.55 square cm C 49.5 square cm D 31.5 square cm

A shared angle lets us compare heights

When two triangles have equal included angles, their areas are in the ratio of the products of the two sides enclosing those angles. The angle condition is essential: two specified side lengths alone leave the height free to change.

For a concrete configuration, take two rays from $O$ . On the first ray, let $OA=7$ and $OC=10$ . On the second ray, let $OB=6$ and $OD=9$ . Suppose the perpendicular heights from $B$ and $D$ to the first ray’s line are 4 and 6. Similar right triangles explain the matching ratios $4:6$ and $6:9$ . The areas are 14 and 30, so their ratio is $7:15$ , also the ratio of the products $7(6)$ and $10(9)$ .

Shared-angle area ratios through heights#

First use enclosing side pairs 4 and 10, then 9 and 15. If the first perpendicular height is 6, AA similarity of the right triangles gives a second height of 9: their height ratio is $2:3$ , the same as $10:15$ . The areas are 12 and 40.5, whose ratio is $8:27$ . The enclosing side products are 40 and 135, with that same ratio. These particular lengths are not needed in the following argument.

Arrange the two nondegenerate triangles by a rigid motion so their equal included angles occupy the same two rays from $O$ . Write $a,b,c,d$ for the enclosing side lengths $OA,OB,OC,OD$ , respectively. Let the perpendicular heights from $B$ and $D$ to the full line $OC$ be $h_B$ and $h_D$ .

Acute included angle. Let the feet be $H$ and $K$ . Right triangles $OBH$ and $ODK$ have equal acute angles at $O$ , so AA makes them similar. Their corresponding heights and hypotenuses give

\frac{h_B}{h_D}=\frac{b}{d}.

Acute case: the two perpendicular right triangles have equal angles at O.

Right included angle. The feet coincide with $O$ . There are no small right triangles to compare: the heights are the sides $OB$ and $OD$ themselves, so the same ratio follows directly.

Right case: the enclosing sides OB and OD are already the heights.

Obtuse included angle. The feet $H$ and $K$ lie on the opposite ray of the base line. Right triangles $OBH$ and $ODK$ have equal acute angles at $O$ , each supplementary to the given obtuse angle. AA again gives the same positive height ratio.

Obtuse case: extend the base line behind O to reach both feet.

These cases include a foot beyond the end of a chosen base segment: the height always reaches its supporting line. Using the area formula, then the height ratio, gives

\begin{aligned} \frac{A_{OAB}}{A_{OCD}}&=\frac{a h_B}{c h_D}\\ \frac{A_{OAB}}{A_{OCD}}&=\frac{ab}{cd}. \end{aligned}

The shared factor from the angle cancels through similar right triangles. We did not need its numerical measure.

Worked example 3 Match the included angle before multiplying

Problem. Two triangles have equal included angles of $110^\circ$ . The sides enclosing that angle in the first triangle are 9 cm and 14 cm; those in the second are 6 cm and 21 cm. Find the first-to-second area ratio.

Solution. The two sides in each product enclose the equal angle, so the height argument applies, including its obtuse case.

\frac{A_1}{A_2}=\frac{9(14)}{6(21)}.

Both products are 126, giving area ratio $1:1$ . Equal areas do not require the individual enclosing sides to match.

Check your understanding

Two triangles have equal included angles. Their enclosing sides are 4 cm and 15 cm in the first triangle, and 6 cm and 10 cm in the second. What is the first-to-second area ratio?

A 2 to 3 B 3 to 2 C 4 to 9 D 1 to 1

A median halves area

A median gives two equal areas because it cuts the opposite base into equal lengths. For a triangle of area 54 square units, the two regions made by any median therefore have area 27 each. This does not say the two triangles are congruent.

The midpoint provides equal bases; no right angle is needed.

Why a median bisects triangle area#

For example, if $BD$ and $DC$ are each 6 units, their common height of 5 gives area 15 square units on each side of $AD$ .

In any nondegenerate triangle $ABC$ , let $D$ be the midpoint of $BC$ . The bases $BD$ and $DC$ are equal. The two heights both measure the distance from $A$ to line $BC$ . Equal-base, equal-height areas give

A_{ABD}=A_{ADC}.

The triangles fill $ABC$ , so each is half its area. No perpendicular or angle-bisector condition on $AD$ was used.

Conversely, if $D$ is inside $BC$ and $AD$ gives equal subareas, their common-height ratio forces $BD=DC$ . In this specific configuration, equal-area bisection implies that $AD$ is a median.

Worked example 4 Bisect area in an obtuse triangle

Problem. Triangle $ABC$ is obtuse and has area 86 square cm. Segment $AD$ is a median. Find the two subareas and explain whether the obtuse angle changes the answer.

An obtuse angle places the perpendicular foot outside the base; the median still reaches its midpoint.

Solution. The midpoint supplies equal bases, and both heights are measured from $A$ to line $BC$ . Thus

A_{ABD}=43\text{ cm}^2.

Also,

A_{ADC}=43\text{ cm}^2.

The argument uses a distance to a line, so it remains valid for an obtuse triangle. A median need not be a perpendicular height.

Check your understanding

D lies strictly inside BC, and AD divides triangle ABC into two equal areas. What must follow?

A AD is perpendicular to BC B D is the midpoint of BC C AB and AC have equal lengths D AD bisects angle A

The centroid gives six equal areas

The three medians divide a triangle into six equal-area triangles. A total area of 72 square units, for example, gives 12 in each small region. The equality follows from the median and centroid facts already proved: the medians meet inside the triangle, and a centroid divides each median in the vertex-to-midpoint ratio $2:1$ .

Distinct tick patterns identify each side midpoint; the medians form six regions.

Why the six median regions have equal areas#

Let $D,E,F$ be the midpoints of $BC,CA,AB$ , and let $G$ be the centroid. First run the argument with a total area of 108. Median $AD$ leaves area 54 in triangle $ABD$ . Inside that triangle, use bases $GD$ and $AD$ , both on the median’s line, with common vertex $B$ . Since $GD$ is one third of $AD$ , triangle $BGD$ has area 18. Equal bases $BD$ and $DC$ , with common vertex $G$ , give area 18 in $GDC$ too. Repeating at the other side midpoints gives 18 in each of the six regions.

For the general argument, let the whole area be $T$ . The median theorem gives

A_{ABD}=\frac{T}{2}.

The earlier centroid theorem gives $GD/AD=1/3$ . Triangles $BGD$ and $BAD$ share the height from $B$ to line $AD$ , so

\begin{aligned} A_{BGD}&=\frac13 A_{BAD}\\ A_{BGD}&=\frac{T}{6}. \end{aligned}

Triangles $GBD$ and $GDC$ have equal bases on $BC$ and common vertex $G$ , so they have equal areas. Repeating the same argument with median $BE$ gives $T/6$ for $CGE$ and $EGA$ ; with median $CF$ it gives $T/6$ for $AGF$ and $FGB$ . These six nonoverlapping triangles fill $ABC$ because $G$ is inside and each side has been divided at its midpoint.

Worked example 5 Reassemble regions around the centroid

Problem. In triangle $ABC$ , all three medians meet at $G$ , with $D,E,F$ the midpoints of $BC,CA,AB$ . Triangle $GBD$ has area 7.5 square cm. Find the whole area, the area of $BGC$ , and the area of quadrilateral $AEGF$ .

Solution. Each small region has the given area, so the whole area is

\begin{aligned}A_{ABC}&=6(7.5)\\A_{ABC}&=45\text{ cm}^2.\end{aligned}

Triangle $BGC$ combines two small regions:

\begin{aligned}A_{BGC}&=2(7.5)\\A_{BGC}&=15\text{ cm}^2.\end{aligned}

Quadrilateral $AEGF$ combines $AEG$ and $AGF$ , also two small regions:

A_{AEGF}=15\text{ cm}^2.

Trace these regions in the centroid diagram above. Equal area here does not mean equal shape.

Check your understanding

The three medians of a triangle form six small triangles. If the whole triangle has area 102 square cm, what is the area of one small triangle?

A 34 square cm B 51 square cm C 17 square cm D 68 square cm

Same Base, Same Height, and Area Ratios

Learning goals

Equal bases and equal heights

A divided base transfers its ratio to area

A shared angle lets us compare heights

A median halves area

The centroid gives six equal areas

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Core practice

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