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Area from Rectangles to Triangles: 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 Small-square units

A unit square is divided into four rows and four columns of congruent squares. A rectangle is made from five columns and three rows of these small squares.

Find its area in square units.

Compare the tile count with the unit square before assigning a tile area.

Text description of this figure

A rectangle has five columns and three rows of small squares. Beside it, a unit square has four columns and four rows of congruent small squares. Every small square in both grids is the same size.
1. Hint 1
  Find the area of one small square from the unit square.
2. Hint 2
  Count the small squares in the rectangle, then combine their areas.
Answer

$\tfrac{15}{16}$ square unit.

Full solution

The unit square contains 16 congruent small squares.
By congruence and addition, each small square has area $\tfrac1{16}$ square unit.

The rectangle contains 15 of them.
$A=15\cdot\frac1{16}$ $A=\frac{15}{16}$
square unit.

Answer

$\tfrac{15}{16}$ square unit.

Key idea

The unit square fixes the area of a tile before the tiles are counted.
Problem 2 A perpendicular measurement

A triangle has a 16 cm base and a corresponding perpendicular height of 9 cm.

Find its area.

The square mark identifies the perpendicular height.

Text description of this figure

Schematic triangle, not drawn to scale. Base 16 cm, perpendicular height 9 cm.
1. Hint 1
  Relate the triangle to two congruent copies in a parallelogram.
2. Hint 2
  Take half the product of the matching base and height.
Answer

$72\text{ cm}^2$ .

Full solution

A parallelogram with the same base and height is filled by two congruent copies of the triangle.
Its area is $16\cdot9$ square centimeters.

One triangle has half that area.
$A=\frac12\cdot16\cdot9$ $A=72\text{ cm}^2.$

Answer

$72\text{ cm}^2$ .

Key idea

A triangle has half the area of its paired parallelogram.
Problem 3 Two kinds of measure

A rectangle has sides $\tfrac{17}{4}$ cm and $\tfrac74$ cm. Find its area and perimeter.

The same rectangle supplies two different measurements.

Text description of this figure

Schematic rectangle, not drawn to scale, with sides 17/4 cm and 7/4 cm.
1. Hint 1
  The two requests measure the interior and the boundary separately.
2. Hint 2
  Multiply the side lengths for area and add two of each side for perimeter.
Answer

$\begin{aligned}A&=\frac{119}{16}\text{ cm}^2,\\P&=12\text{ cm}.\end{aligned}$

Full solution

For area, multiply the side lengths.
$A=\frac{17}{4}\cdot\frac74$ $A=\frac{119}{16}\text{ cm}^2.$

For perimeter, add the four sides.
$P=2\left(\frac{17}{4}+\frac74\right)$ $P=12\text{ cm}.$

Answer

$\begin{aligned}A&=\frac{119}{16}\text{ cm}^2,\\P&=12\text{ cm}.\end{aligned}$

Key idea

Perimeter uses length units while enclosed area uses square units.
Problem 4 Parallelogram dimensions

A parallelogram has area $152\text{ cm}^2$ , perpendicular height 8 cm, and adjacent side 10 cm.

Find its chosen base length and perimeter.

The height is an auxiliary segment, not an extra boundary side.

Text description of this figure

Schematic parallelogram, not drawn to scale, with base b, perpendicular height 8 cm, and adjacent side 10 cm.
1. Hint 1
  The area and perpendicular height determine the matching base.
2. Hint 2
  Divide the area by the height, then use the two pairs of opposite sides for perimeter.
Answer

$\begin{aligned}b&=19\text{ cm},\\P&=58\text{ cm}.\end{aligned}$

Full solution

The unknown base b matches the 8 cm perpendicular height.
$152=8b$ $b=19\text{ cm}.$

The adjacent side is 10 cm, and opposite sides of a parallelogram are equal.
$P=2(19+10)$ $P=58\text{ cm}.$

Answer

$\begin{aligned}b&=19\text{ cm},\\P&=58\text{ cm}.\end{aligned}$

Key idea

Area and perpendicular height determine the base; the boundary side lengths determine perimeter.
Problem 5 A height beyond the side

Triangle ABC has side AB of length 10 cm. The perpendicular from C to line AB meets it at H beyond B. The lengths BH and CH are 5 cm and 12 cm, respectively.

Find the area of triangle ABC. Explain whether the segment BH is included in the chosen base.

H lies beyond B; BH is 5 cm.

Text description of this figure

Schematic triangle ABC with base AB 10 cm and perpendicular CH 12 cm. The foot H is beyond B on the extension of AB. The drawing is not to scale.
1. Hint 1
  Keep the chosen side AB separate from its supporting line.
2. Hint 2
  The perpendicular CH is the height, even though its foot is outside the triangle.
Answer

$A=60\text{ cm}^2.$
Segment BH is not included in base AB.

Full solution

The chosen base AB has length 10 cm.
Extending its line locates the foot H but does not change that side length.

The matching height CH is 12 cm.
$A=\frac12\cdot10\cdot12$ $A=60\text{ cm}^2.$

Answer

$A=60\text{ cm}^2.$
Segment BH is not included in base AB.

Key idea

Extend a side line to locate the height without extending the measured base.
Problem 6 Choose the hypotenuse

A right triangle has legs 7 cm and 24 cm, and hypotenuse 25 cm.

Find its area, then find its altitude to the hypotenuse.

The hypotenuse is 25 cm; the shown legs are perpendicular.

Text description of this figure

Schematic triangle, not drawn to scale. Base 24 cm, perpendicular height 7 cm.
1. Hint 1
  One pair of perpendicular sides is already known.
2. Hint 2
  After finding the area with the legs, use the hypotenuse as a new base.
Answer

$\begin{aligned}A&=84\text{ cm}^2,\\h&=\frac{168}{25}\text{ cm}.\end{aligned}$
Here h is the altitude to the hypotenuse.

Full solution

The legs give the area directly.
$A=\frac12\cdot7\cdot24$ $A=84\text{ cm}^2.$

Let h be the altitude to the 25 cm hypotenuse.
The area stays the same when the base changes.
$84=\frac12\cdot25h$ $h=\frac{168}{25}\text{ cm}.$

The new height is perpendicular to the hypotenuse.

Answer

$\begin{aligned}A&=84\text{ cm}^2,\\h&=\frac{168}{25}\text{ cm}.\end{aligned}$
Here h is the altitude to the hypotenuse.

Key idea

A new chosen base must use a new matching height.
Problem 7 Exact irrational lengths

A rectangle has positive side lengths $\sqrt5$ cm and $\sqrt{20}$ cm. Find its exact area. State whether finite rational-square tiling proves the rectangle formula for these side lengths in this lesson.

The exact radical labels are given measurements.

Text description of this figure

Schematic rectangle, not drawn to scale, with sides √5 cm and √20 cm.
1. Hint 1
  Distinguish the proved rational case from the accepted extension.
2. Hint 2
  Multiply the positive lengths and simplify the product of the square roots.
Answer

$A=10\text{ cm}^2.$
The rectangle formula for these irrational lengths is accepted here; finite rational-square tiling does not prove this case.

Full solution

Both side lengths are irrational.
Our finite rational tiling proof therefore does not establish the formula for this rectangle; we use the explicitly accepted extension.

Multiply the side lengths.
$A=\sqrt5\cdot\sqrt{20}$ $A=\sqrt{100}$ $A=10\text{ cm}^2.$

Answer

$A=10\text{ cm}^2.$
The rectangle formula for these irrational lengths is accepted here; finite rational-square tiling does not prove this case.

Key idea

State the accepted foundation when a proof does not cover the lengths being used.
Problem 8 A base expression

A triangle has area $69\text{ cm}^2$ , perpendicular height 6 cm, and matching base $(2x+1)$ cm.

Find $x$ and the base length.

The schematic shows the matching base and height.

Text description of this figure

Schematic triangle, not drawn to scale. Base (2x+1) cm, perpendicular height 6 cm.
1. Hint 1
  The base expression and height must reproduce the stated area.
2. Hint 2
  Substitute into the triangle formula, then solve the resulting linear equation.
Answer

$\begin{aligned}x&=11,\\b&=23\text{ cm}.\end{aligned}$

Full solution

Substitute the base expression and height.
$69=\frac12(2x+1)6$ $69=3(2x+1)$

Solve the equation.
$23=2x+1$ $x=11.$

Substitute into the base expression.
$b=2(11)+1$ $b=23\text{ cm}.$

Answer

$\begin{aligned}x&=11,\\b&=23\text{ cm}.\end{aligned}$

Key idea

Recover the length from the area equation before interpreting its variable.
Problem 9 Two side choices

In a triangle, side AB is 18 cm long and its perpendicular height from C is 10 cm. Side AC is 15 cm long.

Find the triangle area and the perpendicular height from B to line AC.

The displayed height starts at C; the requested height starts at B.

Text description of this figure

Schematic triangle, not drawn to scale. Base AB: 18 cm, perpendicular height 10 cm, side AC: 15 cm.
1. Hint 1
  An area calculation remains valid when a different side is chosen as base.
2. Hint 2
  Find twice the area first, then divide it by the new base length.
Answer

$\begin{aligned}A&=90\text{ cm}^2,\\h&=12\text{ cm}.\end{aligned}$
Here h is the height from B to line AC.

Full solution

Use base AB and its given height.
$A=\frac12\cdot18\cdot10$ $A=90\text{ cm}^2.$

For base AC, let $h$ be the perpendicular height from B.
$90=\frac12\cdot15h$ $h=12\text{ cm}.$

The derived height to AC is shown only in the solution.

Answer

$\begin{aligned}A&=90\text{ cm}^2,\\h&=12\text{ cm}.\end{aligned}$
Here h is the height from B to line AC.

Key idea

Each side has its own height, while the region has one area.
Problem 10 Reconstruct the region

Triangle ABC has horizontal base AB of length 12 units. The perpendicular CH has length 9 units, with H on AB and AH of length 3 units.

Draw a congruent copy by a half turn about the midpoint of BC, naming the image of A point D. Describe a cut and translation that turns the paired region into a rectangle. Find the original triangle area and justify why this reconstruction gives it.

H lies on AB, with AH equal to 3 units.

Text description of this figure

Schematic triangle ABC with base AB 12 units and perpendicular CH 9 units. The foot H is on side AB. The drawing is not to scale.
1. Hint 1
  Track the images of both endpoints of each side under the half turn.
2. Hint 2
  The paired region is a parallelogram. Its upper left corner is directly above H, which lies inside AB.
3. Hint 3
  A perpendicular cut removes a corner triangle that fits on the other end after a translation by the base length.
Answer

The copies form parallelogram ABDC. Cut along CH and translate the corner triangle ACH by 12 units toward B to complete a 12 by 9 rectangle. The original triangle area is $54$ square units.

Full solution

The half turn interchanges B and C and sends A to D.
It sends AB to parallel side DC and AC to parallel side DB, so the two congruent triangles fill parallelogram ABDC without interior overlap.

Since AH is 3 units and AB is 12 units, the cut CH lies within the parallelogram.
Translate triangle ACH 12 units along AB.
Parallel sides make its horizontal width and perpendicular height match the corner missing on the right, so the pieces fill a rectangle with sides 12 and 9.

Congruence preserves each moved piece area, and addition preserves their total.
The paired region has area
$12\cdot9=108$
square units.

The original triangle and its half-turn copy are congruent, so their areas are equal.
The original therefore has area
$108\div2=54$
square units.

The solution constructs the mate, then slides a corner to complete a rectangle.

Answer

The copies form parallelogram ABDC. Cut along CH and translate the corner triangle ACH by 12 units toward B to complete a 12 by 9 rectangle. The original triangle area is $54$ square units.

Key idea

A reconstruction needs congruent pieces, no gaps or overlaps, and the correct total number of copies.

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