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Same Base, Same Height, and Area Ratios: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 Two equal-base triangles

Triangles PQR and XYZ have equal chosen base lengths QR and YZ. Their corresponding perpendicular heights are both 6.5 cm. The area of PQR is 37.5 square cm. Find the area of XYZ and explain which measurements guarantee it.
1. Hint 1
  Identify the two factors in the triangle area formula.
2. Hint 2
  The given bases match, and so do their matching perpendicular heights.
Answer

The area of XYZ is 37.5 square cm; equal chosen bases and equal corresponding heights guarantee equality.

Full solution

The area of a triangle is half its chosen base times its corresponding perpendicular height.
Both factors match for the two triangles, so their areas are equal.

Therefore triangle XYZ has area 37.5 square cm.
The common height need not lie inside either triangle.

Answer

The area of XYZ is 37.5 square cm; equal chosen bases and equal corresponding heights guarantee equality.

Key idea

Equal bases and equal matching heights are sufficient for equal triangle areas.
Problem 2 A base partition

In nondegenerate triangle ABC, D lies strictly inside BC. The ratio BD to DC is 9 to 13. The area of ABC is 121 square cm.

Find the areas of ABD and ADC.

Two subtriangles have their bases on line BC.

Text description of this figure

A lies off line BC and D lies strictly between B and C. Segment AD divides the triangle. The schematic has no numerical scale.
1. Hint 1
  Compare the two heights when BD and DC are chosen as bases.
2. Hint 2
  The common-height area ratio has 22 parts in all. Find the area of one part.
Answer

$\begin{aligned}A_{ABD}&=49.5\text{ cm}^2\\A_{ADC}&=71.5\text{ cm}^2.\end{aligned}$

Full solution

The two heights are the distance from A to the same line BC.
Their ordered area ratio is therefore 9 to 13.

One part has area $121\div22=5.5$ square cm.
The first region contains nine parts:
$\begin{aligned}A_{ABD}&=9(5.5)\\A_{ABD}&=49.5\text{ cm}^2.\end{aligned}$

The second region contains thirteen parts:
$\begin{aligned}A_{ADC}&=13(5.5)\\A_{ADC}&=71.5\text{ cm}^2.\end{aligned}$

The two areas sum to 121 square cm, as required.

Answer

$\begin{aligned}A_{ABD}&=49.5\text{ cm}^2\\A_{ADC}&=71.5\text{ cm}^2.\end{aligned}$

Key idea

A common height transfers a divided-base ratio directly to the subareas.
Problem 3 Recover a height

Two nondegenerate triangles have equal chosen bases. Their areas are 21 and 35 square cm, in first-to-second order. The corresponding height of the second triangle is 15 cm.

Find the corresponding height of the first triangle.
1. Hint 1
  With equal bases, compare the heights through the area ratio.
2. Hint 2
  The first height is three fifths of the second height.
Answer

The first height is 9 cm.

Full solution

Equal bases let the common positive factor one half times the base cancel.
The height ratio is the area ratio, 21 to 35, or 3 to 5.

Multiply the second height by three fifths:
$\begin{aligned}h_1&=\frac35(15)\\h_1&=9\text{ cm}.\end{aligned}$

As a check, the common base obtained from the second triangle is $2(35)/15=14/3$ cm.
Half of this base times 9 gives 21 square cm.

Answer

The first height is 9 cm.

Key idea

For equal bases, the ordered area ratio equals the ordered height ratio.
Problem 4 An obtuse median partition

Triangle ABC is obtuse and has area 94 square cm. D is the midpoint of BC. Find the areas of ABD and ADC. Explain why AD does not have to be perpendicular to BC.

An outside foot does not change the median area argument.

Text description of this figure

ABC is obtuse at C. D is the midpoint of BC, shown by equal ticks. AD is a median. The perpendicular from A reaches H beyond C on the base line.
1. Hint 1
  Use the equality supplied by the midpoint.
2. Hint 2
  The bases BD and DC lie on the same line, so their heights both come from A.
Answer

$\begin{aligned}A_{ABD}&=47\text{ cm}^2\\A_{ADC}&=47\text{ cm}^2.\end{aligned}$
The median need not be perpendicular.

Full solution

D is the midpoint, so BD and DC have equal lengths.
Both subtriangles have the same perpendicular height from A to line BC.

Each subtriangle therefore occupies half the total area:
$94\div2=47\text{ cm}^2.$

AD provides the cut, not necessarily the height.
The actual height can reach a supporting-line extension, so the obtuse angle does not invalidate the comparison.

Answer

$\begin{aligned}A_{ABD}&=47\text{ cm}^2\\A_{ADC}&=47\text{ cm}^2.\end{aligned}$
The median need not be perpendicular.

Key idea

A median halves area through equal bases, without an altitude condition.
Problem 5 A side-pair counterexample

Give two nondegenerate triangles, each with two sides of lengths 10 cm and 13 cm, whose areas are unequal. Specify enough information to establish each area, and compute both.
1. Hint 1
  The two given sides need not enclose the same angle in both triangles.
2. Hint 2
  One construction can make the two given sides perpendicular. For another, a 13 cm side can be the hypotenuse of a right triangle with legs 5 cm and 12 cm.
Answer

One valid pair has areas 65 square cm and 60 square cm. The first has perpendicular 10 cm and 13 cm sides; the second has a 10 cm base and a 13 cm side whose perpendicular height is 12 cm.

Full solution

For the first triangle, make the 10 cm and 13 cm sides perpendicular.
Then
$\begin{aligned}A_1&=\frac12(10)(13)\\A_1&=65\text{ cm}^2.\end{aligned}$

For the second triangle, start with a 10 cm base.
Put the perpendicular foot at its midpoint, 5 cm from the left endpoint, and put the apex 12 cm above that foot.

The side from the left endpoint to the apex has length 13 cm because
$5^2+12^2=13^2.$
Its area is
$\begin{aligned}A_2&=\frac12(10)(12)\\A_2&=60\text{ cm}^2.\end{aligned}$

The side pairs agree, but the included angles and the matching heights differ.
Other valid constructions with unequal computed areas are also acceptable.

A side pair can occur with different heights.

Answer

One valid pair has areas 65 square cm and 60 square cm. The first has perpendicular 10 cm and 13 cm sides; the second has a 10 cm base and a 13 cm side whose perpendicular height is 12 cm.

Key idea

Two side lengths alone do not determine a triangle area.
Problem 6 The two directions of bisection

Let ABC be any nondegenerate triangle, and let D lie strictly inside BC.

Forward direction: prove that if D is the midpoint of BC, then AD divides ABC into two equal areas.

Converse direction: prove that if ABD and ADC have equal areas, then D is the midpoint of BC.

Two subtriangles have their bases on line BC.

Text description of this figure

A lies off line BC and D lies strictly between B and C. Segment AD divides the triangle. The schematic has no numerical scale.
1. Hint 1
  Both directions use bases BD and DC on the same supporting line.
2. Hint 2
  Write the area ratio as BD divided by DC. For the forward direction use equal bases; for the converse use equal areas.
Answer

Both directions hold under the stated hypotheses, so AD bisects the area exactly when D is the midpoint of BC.

Full solution

Forward direction: if D is the midpoint, then BD and DC are equal.
Both heights measure the distance from A to line BC.
The equal-base, equal-height rule gives equal subareas.

Converse direction: the common-height ratio gives
$\frac{A_{ABD}}{A_{ADC}}=\frac{BD}{DC}.$
Equal positive areas make the left side 1, so BD and DC are equal.
Since D lies inside BC, D is its midpoint.

The condition that D lies inside the side ensures that the two regions actually partition ABC.
This converse is about this specified vertex-to-side cut, rather than arbitrary equal-area figures.

Answer

Both directions hold under the stated hypotheses, so AD bisects the area exactly when D is the midpoint of BC.

Key idea

A converse needs its own argument and the same stated configuration.
Problem 7 A shared obtuse angle

Two triangles have equal obtuse included angles. The sides enclosing that angle in the first triangle are 8 cm and 15 cm, and its area is 30 square cm.

The sides enclosing the equal angle in the second triangle are 12 cm and x cm, and its area is 63 square cm. Find x and explain why an obtuse angle still permits the ratio.

Match the angle between the two specified sides.

Text description of this figure

Two schematic triangles with their included angles marked equal. The first is OAB and the second is PCD. Numerical side lengths are supplied in the text, not by measuring the drawing.
1. Hint 1
  Compare the products of the sides enclosing the equal angles.
2. Hint 2
  The second-to-first area ratio is 63 to 30. Match that to 12x divided by the first side product.
Answer

The missing side is 21 cm. The ratio remains valid because the perpendicular feet may lie on the base-line extensions.

Full solution

For equal included angles, the perpendicular heights have the same ratio as one pair of enclosing sides.
When the angles are obtuse, the right triangles used for this comparison have their feet on the opposite base-line rays.

Compare the side products in the same second-to-first order as the areas:
$\frac{12x}{8(15)}=\frac{63}{30}.$

Multiply by 120 and solve:
$\begin{aligned}12x&=252\\x&=21.\end{aligned}$

The positive result checks the ratio: $12(21)/120=21/10$ , the same as $63/30$ .

Answer

The missing side is 21 cm. The ratio remains valid because the perpendicular feet may lie on the base-line extensions.

Key idea

Shared-angle side products come from proportional perpendicular heights, including obtuse cases.
Problem 8 Regroup the median regions

In ABC, all three medians meet at G, and D, E and F are the midpoints of BC, CA and AB. Triangle GDC has area 9.75 square cm.

Find the area of ABC.

Find the area of quadrilateral AEGF.

Find the area inside ABC but outside triangle BGC.

All six median regions remain visible.

Text description of this figure

D, E and F are the midpoints of BC, CA and AB. Distinct one-, two- and three-tick patterns identify these separate pairs. The medians meet at G.
1. Hint 1
  Count the six equal regions and identify which ones belong to each requested shape.
2. Hint 2
  AEGF contains two small regions. BGC also contains two, leaving four outside it.
Answer

$\begin{aligned}A_{ABC}&=58.5\text{ cm}^2\\A_{AEGF}&=19.5\text{ cm}^2\\A_{\text{outside }BGC}&=39\text{ cm}^2.\end{aligned}$

Full solution

Whole triangle: there are six equal small regions.
$\begin{aligned}A_{ABC}&=6(9.75)\\A_{ABC}&=58.5\text{ cm}^2.\end{aligned}$

Quadrilateral AEGF: its two regions are AEG and AGF.
$\begin{aligned}A_{AEGF}&=2(9.75)\\A_{AEGF}&=19.5\text{ cm}^2.\end{aligned}$

Outside BGC: that triangle uses GBD and GDC, leaving four of the six regions outside it.
$\begin{aligned}A_{\text{outside }BGC}&=4(9.75)\\A_{\text{outside }BGC}&=39\text{ cm}^2.\end{aligned}$

Answer

$\begin{aligned}A_{ABC}&=58.5\text{ cm}^2\\A_{AEGF}&=19.5\text{ cm}^2\\A_{\text{outside }BGC}&=39\text{ cm}^2.\end{aligned}$

Key idea

Count the small median regions belonging to a requested larger region.
Problem 9 Equal area with changing side lengths

Two nondegenerate triangles have equal included angles and equal areas. The sides enclosing that angle in the first triangle are $x+1$ cm and 12 cm. Those in the second are 18 cm and $x-1$ cm.

Find x and both pairs of enclosing side lengths. Check positivity.

Match the angle between the two specified sides.

Text description of this figure

Two schematic triangles with their included angles marked equal. The first is OAB and the second is PCD. Numerical side lengths are supplied in the text, not by measuring the drawing.
1. Hint 1
  Equal included angles let equal areas impose equality of the enclosing side products.
2. Hint 2
  Set 12 times x + 1 equal to 18 times x - 1, placing parentheses around each full side expression.
Answer

$x=5$ . First side pair: 6 cm and 12 cm. Second side pair: 18 cm and 4 cm. All lengths are positive.

Full solution

The area ratio is 1, so the products of the enclosing sides are equal:
$12(x+1)=18(x-1).$

Expand and solve:
$\begin{aligned}12x+12&=18x-18\\30&=6x\\x&=5.\end{aligned}$

The first side pair is 6 cm and 12 cm.
The second is 18 cm and 4 cm.
Every length is positive, and both side products are 72 square cm.
With the given common nonzero included angle, both triangles exist and their areas match.

Answer

$x=5$ . First side pair: 6 cm and 12 cm. Second side pair: 18 cm and 4 cm. All lengths are positive.

Key idea

Equal included angles turn an equal-area condition into equality of side products.
Problem 10 Transfer a partition twice

In nondegenerate triangle ABC, D lies strictly inside BC, and the ratio BD to DC is 7 to 8. The area of ABC is 210 square cm.

E lies strictly inside AD, with AE to ED equal to 3 to 4. Segments BE and CE are drawn.

Find the areas of BDE and CDE separately, then find the area of BCE. Justify each ratio used.

A second partition changes which common height is useful.

Text description of this figure

D is inside BC and E is inside AD. Segments AD, BE and CE show the two linked partitions. No computed area labels are shown.
1. Hint 1
  First split ABC using the bases BD and DC.
2. Hint 2
  Next compare triangles BDE and BDA using bases ED and AD on the same line. Do the same with vertex C.
3. Hint 3
  ED is four of the seven parts of AD. Add the two lower subareas only after finding them.
Answer

$\begin{aligned}A_{BDE}&=56\text{ cm}^2\\A_{CDE}&=64\text{ cm}^2\\A_{BCE}&=120\text{ cm}^2.\end{aligned}$

Full solution

Triangles ABD and ADC have the common height from A to line BC.
Their area ratio is 7 to 8.
Fifteen parts total 210 square cm, so one part is 14 square cm.

The first split gives
$\begin{aligned}A_{ABD}&=7(14)\\A_{ABD}&=98\text{ cm}^2\\A_{ADC}&=8(14)\\A_{ADC}&=112\text{ cm}^2.\end{aligned}$

Since AE to ED is 3 to 4, the ratio ED to AD is 4 to 7.
Triangles BDE and BDA share vertex B and use bases ED and AD on the same line.
Hence
$\begin{aligned}A_{BDE}&=\frac47(98)\\A_{BDE}&=56\text{ cm}^2.\end{aligned}$

Triangles CDE and CDA instead share the height from C to line AD.
The same base ratio gives
$\begin{aligned}A_{CDE}&=\frac47(112)\\A_{CDE}&=64\text{ cm}^2.\end{aligned}$

Because D lies on BC and E is inside AD, BDE and CDE fill BCE without overlapping interiors.
Add their areas:
$\begin{aligned}A_{BCE}&=56+64\\A_{BCE}&=120\text{ cm}^2.\end{aligned}$

Answer

$\begin{aligned}A_{BDE}&=56\text{ cm}^2\\A_{CDE}&=64\text{ cm}^2\\A_{BCE}&=120\text{ cm}^2.\end{aligned}$

Key idea

Choose the relevant supporting line anew each time a base ratio transfers to an area ratio.

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