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Perimeter and Area: Chapter Test

20 multiple-choice questions and 10 core practice problems, drawn from across the chapter and mixed together.

Core practice

10 problems from across the chapter. 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 regions on a grid

A unit square is divided into seven rows and seven columns of congruent small squares. A rectangle contains ten columns and four rows of these same small squares.

A diagonal divides the rectangle into two triangles. Find the rectangle's area and the area of either triangle in square units. Justify both results from the unit square and the relationship between the triangles.

Every small tile is congruent to every other small tile.

Text description of this figure

A unit square has seven rows and seven columns of congruent tiles. Below, ten columns and four rows of the same tiles form a rectangle, divided by a diagonal. No area values are supplied.
1. Hint 1
  Start by assigning an area to one small square, using how many congruent pieces fill the unit square.
2. Hint 2
  Count the rectangle's tiles. Then compare the two triangles formed by its diagonal through a half turn about the diagonal's midpoint.
Answer

Rectangle:
$\frac{40}{49}\text{ square unit}.$
Either triangle:
$\frac{20}{49}\text{ square unit}.$

Full solution

The unit square contains 49 congruent small squares.
Congruence gives equal areas and addition gives total area one, so each tile has area $1/49$ square unit.

The rectangle contains 40 tiles.
$A_R=40\left(\frac1{49}\right)$ $A_R=\frac{40}{49}\text{ square unit}.$

A half turn about the diagonal's midpoint carries one triangular half to the other.
They are congruent and their interiors do not overlap, so their equal areas add to the rectangle area.

Each triangle therefore has area
$A_T=\frac12\left(\frac{40}{49}\right)$ $A_T=\frac{20}{49}\text{ square unit}.$

Answer

Rectangle:
$\frac{40}{49}\text{ square unit}.$
Either triangle:
$\frac{20}{49}\text{ square unit}.$

Key idea

A unit subdivision fixes each tile area, while congruent halves share the total equally.
Problem 2 Two angle records

Triangle ABC has sides AB of length 18 cm and AC of length 13 cm enclosing an obtuse angle at A. The perpendicular CH to line AB is 12 cm, with H on the extension beyond A.

Triangle DEF has DE of length 7 cm and DF of length 26 cm enclosing an angle at D equal to angle A. Find its perpendicular height from F to line DE and its area. Explain the height comparison without finding the included angle's numerical measure.

The matching arcs represent equal given angles; the panels have independent scales.

Text description of this figure

Separate schematic triangles ABC and DEF have equal obtuse included angles at A and D, marked by matching arcs. AB is 18 cm, AC is 13 cm, and exterior perpendicular CH is 12 cm. DE is 7 cm and DF is 26 cm. The second height is not constructed. The panels use independent drawing scales.
1. Hint 1
  The perpendicular feet for the obtuse angles lie on extensions of the base lines. Compare the resulting right triangles.
2. Hint 2
  Those right triangles have equal acute angles supplementary to the equal included angles. Their hypotenuses are AC and DF.
3. Hint 3
  Use the hypotenuse ratio for the heights, then pair the second height with DE.
Answer

Height from F:
$24\text{ cm}.$
Area of DEF:
$84\text{ cm}^2.$

Full solution

The right triangles formed by the perpendiculars have equal acute angles at A and D, each supplementary to the same obtuse angle.
They are similar by AA.
Their height ratio equals the ratio of hypotenuses DF and AC.

Writing h for the second height gives
$\frac{h}{12}=\frac{26}{13}$ $h=24\text{ cm}.$

Use this perpendicular height with the 7 cm side DE.
$A_{DEF}=\frac12(7)(24)$ $A_{DEF}=84\text{ cm}^2.$

The base AB was not needed to transfer heights, but it confirms the first area as 108 square cm.
The area ratio 84 to 108 is 7 to 9, agreeing with the ordered side-product ratio $7(26)$ to $18(13)$ .

Answer

Height from F:
$24\text{ cm}.$
Area of DEF:
$84\text{ cm}^2.$

Key idea

Equal included angles supply a height ratio through similar right triangles, even with exterior feet.
Problem 3 An opening in the boundary

A 34 cm by 23 cm rectangle has a rectangular notch of width x cm and depth 6 cm opening into the middle of its top edge. All boundary segments are horizontal or vertical. The retained area is 710 square cm.

Find x and the perimeter of the retained polygon. Check that the width and depth fit the described region.

Schematic; not drawn to scale.

Text description of this figure

Schematic 34 by 23 cm rectangle with a middle top notch of unknown width x and known depth 6 cm. The notch is open to the outside. No derived length is labeled. Not drawn to scale.
1. Hint 1
  Compare the complete rectangle area with the retained area to identify the absent region.
2. Hint 2
  The missing area is six times the notch width.
3. Hint 3
  Trace the boundary: the notch bottom replaces a missing top segment, while both depth edges are new.
Answer

$x=12\text{ cm}.$ $P=126\text{ cm}.$
The notch dimensions fit within the rectangle.

Full solution

The complete rectangle has area
$34(23)=782\text{ cm}^2.$
Its missing rectangular notch therefore has area 72 square cm.

The 6 cm depth and unknown width give
$6x=72$ $x=12.$
The width is positive and less than 34 cm, and the depth is less than 23 cm.
Thus it can open into the middle of the top edge as stated.

The original rectangle perimeter is 114 cm.
The missing top width is replaced by an equal notch bottom, and the two 6 cm depth edges are additional.
$P=114+2(6)$ $P=126\text{ cm}.$

Answer

$x=12\text{ cm}.$ $P=126\text{ cm}.$
The notch dimensions fit within the rectangle.

Key idea

An area deficit reveals a missing dimension, but perimeter still comes from the exposed boundary.
Problem 4 A removed corner region

Convex kite ABCD has equal-side pairs meeting at A and C. Its diagonals meet at O, and AO, OC, and BO have lengths 9 cm, 16 cm, and 12 cm, respectively.

The triangular region AOB is removed. Find the area and perimeter of the remaining region, including the newly exposed edges. Justify the complete kite area calculation from the given diagonal information.

Schematic; not drawn to scale.

Text description of this figure

Schematic convex kite ABCD. AC and BD are perpendicular at O. AO is 9 cm, OC is 16 cm, and BO is 12 cm. The region AOB is labeled removed. The other regions remain. Not drawn to scale.
1. Hint 1
  The kite properties tell you the angle between its diagonals and which diagonal is bisected.
2. Hint 2
  Determine the complete kite area, then remove the one specified triangle.
3. Hint 3
  The side AB disappears from the original boundary. AO and OB become exposed; the other three original sides remain.
Answer

Retained area:
$246\text{ cm}^2.$
Perimeter:
$76\text{ cm}.$

Full solution

AC perpendicularly bisects BD, so OD is 12 cm and the full diagonals are 25 cm and 24 cm.
Convexity lets AC divide the complete kite into triangles ABC and ADC with disjoint interiors.
They share base AC of length 25 cm, and their perpendicular heights BO and DO are each 12 cm.
Each triangle has area
$\frac12(25)(12)=150\text{ cm}^2.$
Their areas add to give
$A_K=150+150$ $A_K=300\text{ cm}^2.$

AOB is a right triangle with legs 9 cm and 12 cm.
Its area is 54 square cm.
Subtraction gives
$A=300-54$ $A=246\text{ cm}^2.$

Pythagoras in AOB gives AB and AD as 15 cm each.
Pythagoras in BOC gives BC and CD as 20 cm each.
The required perimeter contains AD, DC, CB, BO, and OA.

Add precisely these exposed edges.
$P=15+20+20+12+9$ $P=76\text{ cm}.$
The old edge AB is absent, and OC and OD remain internal.

Schematic; not drawn to scale.

Answer

Retained area:
$246\text{ cm}^2.$
Perimeter:
$76\text{ cm}.$

Key idea

Removing a triangular region changes both the area inventory and the list of exposed edges.
Problem 5 A ten-sided region

A convex regular decagon has side length 15 cm. Find its apothem and radius to the nearest tenth of a centimeter and its area to the nearest tenth of a square centimeter. Use degree mode and keep full calculator precision until each final result.

Explain why the center triangles can be added, and check that your unrounded distances fit one right half-triangle.

Schematic; not drawn to scale.

Text description of this figure

Schematic convex regular decagon with center O, segments to every vertex, and one perpendicular apothem a. Each side is 15 cm. No computed half-angle, radius, apothem length, or area is supplied. Not drawn to scale.
1. Hint 1
  Start with the center and the midpoint of one side. Identify the angle at the center in the resulting right triangle.
2. Hint 2
  The half-side is opposite the half-central angle. Tangent relates it to the apothem, and sine relates it to the radius.
3. Hint 3
  All ten center-triangle bases are already included in the perimeter. Check Pythagoras with the exact trigonometric expressions.
Answer

$a\approx23.1\text{ cm}.$ $r\approx24.3\text{ cm}.$ $A\approx1731.2\text{ cm}^2.$
The unrounded lengths satisfy $r^2=a^2+7.5^2$ .

Full solution

The central angle is 36 degrees, so the right half-triangle uses 18 degrees and half-side 7.5 cm.
Tangent gives
$a=\frac{7.5}{\tan18^\circ}.$
This is approximately 23.1 cm.

Sine uses the radius as hypotenuse.
$r=\frac{7.5}{\sin18^\circ}.$
This is approximately 24.3 cm.

Convexity keeps the center triangles inside the polygon.
Its rotational symmetry makes them congruent; their consecutive sectors fill one turn with disjoint interiors, so their areas add.
The full perimeter is 150 cm.

Use the exact apothem expression.
$A=75\left(\frac{7.5}{\tan18^\circ}\right)$ $A\approx1731.2\text{ cm}^2.$

For the unrounded check, the right-triangle ratios give
$\frac ar=\cos18^\circ.$ $\frac{7.5}{r}=\sin18^\circ.$
The Pythagorean identity makes the squared ratios add to one, hence
$r^2=a^2+7.5^2.$
Rounded lengths may leave a small numerical difference.

Answer

$a\approx23.1\text{ cm}.$ $r\approx24.3\text{ cm}.$ $A\approx1731.2\text{ cm}^2.$
The unrounded lengths satisfy $r^2=a^2+7.5^2$ .

Key idea

A regular polygon needs one compatible right half-triangle and one complete sum of center-triangle areas.
Problem 6 An inner partition

In nondegenerate triangle ABC, E lies inside AB and F lies inside AC, with EF parallel to BC. The ratio AE to AB is 3 to 8. G is the centroid of AEF and H is the midpoint of EF. Triangle EGH has area 9 square cm.

D is the midpoint of BC. Find the areas of ABC and ABD, justifying the geometric relationships you use.

The upper median partition belongs to AEF; D belongs to the whole triangle.

Text description of this figure

Schematic triangle ABC with E on AB and F on AC, and EF parallel to BC. D is the midpoint of BC. The three medians of AEF meet at G; H is the midpoint of EF on the median from A. Triangle EGH has given area 9 square cm. The diagram has no measuring scale.
1. Hint 1
  First place EGH among the regions formed by the three medians of AEF.
2. Hint 2
  Parallel EF and BC give a similarity between AEF and ABC. Keep the inner-to-whole length ratio in the same order when comparing areas.
3. Hint 3
  After recovering the whole area, consider what the midpoint D does to the two bases on BC.
Answer

$A_{ABC}=384\text{ cm}^2.$ $A_{ABD}=192\text{ cm}^2.$

Full solution

Within AEF, triangle EGH is one of the six equal-area median regions.
Therefore
$A_{AEF}=6(9)$ $A_{AEF}=54\text{ cm}^2.$

The triangles share angle A, and parallel EF and BC give matching corresponding angles.
Thus AEF is similar to ABC by AA.
The inner-to-whole length factor is 3/8.

The area ratio is its square.
$\frac{54}{A_{ABC}}=\frac9{64}$ $A_{ABC}=54\left(\frac{64}{9}\right)$ $A_{ABC}=384\text{ cm}^2.$

Since D is the midpoint of BC, BD and DC are equal bases with the common height from A to BC.
The median AD bisects the area.
$A_{ABD}=\frac12(384)$ $A_{ABD}=192\text{ cm}^2.$

Answer

$A_{ABC}=384\text{ cm}^2.$ $A_{ABD}=192\text{ cm}^2.$

Key idea

A centroid subdivision, a similarity, and a median each transfer a different area relationship.
Problem 7 A cut and a turn

Let ABC be any nondegenerate triangle. M is the midpoint of AB and N is the midpoint of AC. Cut along MN. Turn triangle AMN through a half turn about N, and call the image of M point X. Leave the other piece in place.

Describe the region filled by the two pieces after the turn. Justify its sides, its perpendicular height, and why its area equals the original triangle's area. If BC has length b and the height from A to line BC is h, derive the original triangle's area from this reconstruction.

Schematic; not drawn to scale.

Text description of this figure

Schematic nondegenerate triangle ABC with midpoint M on AB and midpoint N on AC. A single-tick pattern marks AM equal to MB; a double-tick pattern marks AN equal to NC. The cut MN is drawn. The turned location is not supplied.
1. Hint 1
  A half turn about the midpoint N interchanges A and C. Track where AM and MN go.
2. Hint 2
  The midpoint theorem makes MN parallel to BC and half its length. The original small triangle has half the corresponding height.
3. Hint 3
  Check the new outer boundary and the shared join before applying the parallelogram area rule.
Answer

The pieces fill parallelogram BCXM, with base b and perpendicular height h/2. The original triangle area is
$A=\frac12bh.$

Full solution

Since N is the midpoint of AC, the half turn carries A to C and M to X.
It preserves lengths and parallelism.
Hence CX is parallel to AM and equal to AM.
Because M is the midpoint of AB, AM and MB are equal and collinear; thus CX is parallel and equal to MB.

The midpoint theorem gives MN parallel to BC and half as long.
The turn sends MN onto NX in the opposite direction, so M, N, X lie in that order on one line.
Consequently MX has length b and is parallel to BC.
These opposite-side relationships identify parallelogram BCXM.

Triangle AMN is similar to ABC with length factor one half, from its two midpoint sides and common angle A.
Its corresponding height is h/2.
The line MN therefore lies halfway in perpendicular distance between A and BC, leaving perpendicular separation h/2 between MN and BC.
This argument uses length similarity, without assuming an area scaling rule.

The unturned quadrilateral MBCN lies inside BCXM, and diagonal CN splits off exactly triangle CNX on its other side.
The turned triangle is that missing triangle, so the two pieces fill BCXM without gaps or overlapping interiors.
Congruence preserves the moved piece area, and addition preserves the total.

Apply the parallelogram formula to base b and perpendicular height h/2.
$A=b\left(\frac h2\right)$ $A=\frac12bh.$

The red dashed triangle is the half-turn image; the shared join is CN.

Answer

The pieces fill parallelogram BCXM, with base b and perpendicular height h/2. The original triangle area is
$A=\frac12bh.$

Key idea

A reconstruction proves an area rule only when the moved piece fits exactly and the new height is justified.
Problem 8 Two related triangles

Two equilateral triangles have areas $121\sqrt3$ square cm and $400\sqrt3$ square cm, in first-to-second order.

Find the positive length scale factor taking the first to the second. Find the second triangle's perimeter and apothem exactly, and check the apothem against a right half of a center triangle.
1. Hint 1
  Equilateral triangles are similar, so their area ratio can determine a positive length ratio.
2. Hint 2
  Use an exact equilateral-triangle area to recover a side length; keep the side-to-apothem relationship separate from the full altitude.
3. Hint 3
  A right half of a center triangle uses half the side opposite 60 degrees. The apothem is the shorter leg.
Answer

$k=\frac{20}{11}.$ $P_2=120\text{ cm}.$ $a_2=\frac{20\sqrt3}{3}\text{ cm}.$

Full solution

Both triangles have three 60-degree angles, so they are similar by AA.
Their positive length factor is
$k=\sqrt{\frac{400\sqrt3}{121\sqrt3}}$ $k=\frac{20}{11}.$

For the first equilateral triangle, its side s satisfies
$\frac{s^2\sqrt3}{4}=121\sqrt3$ $s^2=484$ $s=22\text{ cm}.$
Its perimeter is 66 cm.

Scale that perimeter by the length factor.
$P_2=\frac{20}{11}(66)$ $P_2=120\text{ cm}.$
Each second-triangle side is therefore 40 cm.

The apothem is the perpendicular center-to-side distance.
The area rule gives
$a_2=\frac{2(400\sqrt3)}{120}$ $a_2=\frac{20\sqrt3}{3}\text{ cm}.$

For the geometric check, the second triangle's half-side is 20 cm, opposite a 60-degree half-central angle.
The shorter leg is $20/\sqrt3$ cm, which equals the apothem found above.
The full altitude starts at a vertex rather than the center and has a different length.

Schematic; not drawn to scale.

Answer

$k=\frac{20}{11}.$ $P_2=120\text{ cm}.$ $a_2=\frac{20\sqrt3}{3}\text{ cm}.$

Key idea

Establish similarity before reversing an area ratio, then check regular-polygon dimensions in the same right triangle.
Problem 9 Two joined panels

A flat rectangular panel is 0.46 m wide and 30 cm high. A second rectangular panel is w cm wide and 140 mm high. Their lower edges are aligned, their interiors are disjoint, and they join along the second panel's entire 140 mm side, with no other contact. The combined area is 0.1688 square meters.

Find the combined area in square centimeters, w in centimeters, and the outside boundary length of the joined panels in meters. Use 100 cm equal to 1 m and 10 mm equal to 1 cm.

Schematic; not drawn to scale.

Text description of this figure

Schematic joined rectangular panels with aligned bottom edges and a dashed internal vertical join. The first has width 0.46 m and height 30 cm. The second has unknown width w cm and height 140 mm. No computed lengths or areas are labeled. Not drawn to scale.
1. Hint 1
  Before combining the rectangles, put the lengths and the given area into consistent units.
2. Hint 2
  A square meter contains 100 centimeters along each of two directions. The smaller rectangle's area is the combined area minus the first rectangle's area.
3. Hint 3
  For boundary length, the join appears in each separate perimeter but is internal in the union. Remove both copies before converting to meters.
Answer

Combined area:
$1688\text{ cm}^2.$ $w=22\text{ cm}.$
Outside boundary:
$1.96\text{ m}.$

Full solution

The first panel is 46 cm wide, and the second panel is 14 cm high.
A square meter contains 10000 square centimeters, so the combined area is
$0.1688(10000)=1688\text{ cm}^2.$

The first rectangle area is 1380 square cm.
Disjoint interiors make the second rectangle account for the remaining area.
$14w=1688-1380$ $14w=308$ $w=22\text{ cm}.$

The first panel perimeter is 152 cm, and the second is 72 cm.
Their shared 14 cm segment is counted twice in that sum and contributes no exposed edge.
$P=152+72-2(14)$ $P=196\text{ cm}.$

A boundary is a length, so divide its centimeter measure by 100 only once.
$P=1.96\text{ m}.$
The squared factor belonged to the area conversion, not this boundary conversion.

Answer

Combined area:
$1688\text{ cm}^2.$ $w=22\text{ cm}.$
Outside boundary:
$1.96\text{ m}.$

Key idea

Convert square units twice, recover the missing piece, and remove both counted copies of an internal join.
Problem 10 Three rectangles

Rectangle R measures 11 cm by 28 cm. Rectangle S is similar to R and has area 693 square cm. Rectangle T measures 7 cm by 99 cm.

Find the dimensions and perimeter of S. Then determine whether S and T have equal areas, are similar, and are congruent. Justify each conclusion.

Independent schematics; every numerical comparison comes from the stated givens.

Text description of this figure

Three separate schematic rectangles, drawn at independent scales. R is 11 by 28 cm. S is similar to R and has given area 693 square cm; its side lengths are not labeled. T is 7 by 99 cm. No computed R area or S lengths are supplied.
1. Hint 1
  The given similarity links R's and S's area ratio to their corresponding length ratio.
2. Hint 2
  Find the positive square root of that ratio before resizing the two side lengths.
3. Hint 3
  For the final comparison, an equal product does not establish equal corresponding side ratios. Compare longer side to shorter side in each rectangle.
Answer

S: 16.5 cm by 42 cm; perimeter 117 cm. S and T have equal area, 693 square cm, but are neither similar nor congruent.

Full solution

The area of R is 308 square cm.
Because R and S are already known to be similar, their length factor satisfies
$k^2=\frac{693}{308}$ $k^2=\frac94$ $k=\frac32.$

Apply that positive factor to the corresponding lengths.
The shorter side of S is
$11\left(\frac32\right)=16.5\text{ cm}.$
The longer side is
$28\left(\frac32\right)=42\text{ cm}.$

Add the four boundary sides.
$P_S=2(16.5+42)$ $P_S=117\text{ cm}.$
The product of the two sides is 693 square cm, checking the given area.

Rectangle T also has area $7(99)$ , or 693 square cm.
Thus the areas of S and T agree.

Their longer-to-shorter ratios differ.
$\frac{42}{16.5}=\frac{28}{11}$ $\frac{28}{11}\ne\frac{99}{7}.$
Therefore S and T are not similar.
Since congruent rectangles would be similar, they are not congruent either.

Answer

S: 16.5 cm by 42 cm; perimeter 117 cm. S and T have equal area, 693 square cm, but are neither similar nor congruent.

Key idea

An established similarity permits inverse area scaling; equal area alone still leaves shape undetermined.

Multiple choice

Core practice