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Perimeter and Area: 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 The shaded cells

Find the area of the shaded region in the figure in square cm.

A grid of equal squares, with a scale key of one square, 1 cm by 1 cm.

Text description of this figure

A rectangular grid of equal squares, four columns across and three rows down. Every square in the top two rows is shaded, eight squares in all. In the bottom row, the two middle squares are shaded completely. The first and last squares of the bottom row each have a diagonal drawn from the bottom-left corner to the top-right corner, and only the triangle below that diagonal is shaded; the other half is unshaded. Beside the grid is a separate unshaded square, the scale key, with one horizontal side labeled 1 cm and one vertical side labeled 1 cm.
1. Hint 1
  Area counts whole unit squares, and parts of unit squares, that the region covers.
2. Hint 2
  Pair the two shaded half squares to make one whole square.
Answer

$11$ square cm.

Full solution

The upper two rows contain $8$ fully shaded squares.
The bottom row contains $2$ more fully shaded squares and $2$ shaded half squares.

Count their combined area in square cm.
$A=8+2+\frac12+\frac12$ $A=11$
The two halves make one whole square, so the shaded area is $11$ square cm.

Answer

$11$ square cm.

Key idea

Whole and partial unit squares both contribute to a region's area.
Problem 1: Mark complete
Problem 2 The sign outline

Find the perimeter of the sign shown in the figure.

A flat sign with five straight sides.

Text description of this figure

A five-sided sign shaped like a house, drawn to scale. The bottom side is horizontal and labeled 6 cm. A vertical side rises from each end of the bottom, and each is labeled 4 cm. From the tops of those two sides, two slanted roof sides rise to meet at a point centered above the bottom, and each roof side is labeled 5 cm. There are no lines inside the shape.
1. Hint 1
  Trace the complete outer boundary once.
2. Hint 2
  Add the five labeled lengths, including both slanted sides.
Answer

$24$ cm.

Full solution

Every labeled side lies on the outside boundary, so add all five lengths.
$P=6+4+5+5+4$ $P=24$
The sign's perimeter is $24$ cm.
The lengths $4$ cm and $5$ cm each appear twice in the sum because the sign has two upright sides and two roof sides.

Answer

$24$ cm.

Key idea

A polygon's perimeter is the sum of all its boundary side lengths.
Problem 2: Mark complete
Problem 3 The chosen side

In the figure, side $AB$ is chosen as the base of the parallelogram. Which of the segments drawn in the figure gives the height for that base?

Parallelogram $ABCD$ with segments $CE$ and $AC$ .

Text description of this figure

A parallelogram ABCD. A is at the lower left and B is directly above A, so side AB is vertical. C is to the right of B and lower than B, and D is directly below C, so side DC is vertical too. Single chevrons mark sides AB and DC as parallel, and double chevrons mark sides BC and AD as parallel. Point E lies on side AB, level with C. A dashed segment runs from C across to E, with a small square showing a right angle where it meets AB. A second dashed segment runs from A to C. No lengths, angle measures, grid or scale are shown.
1. Hint 1
  The height for a base is the distance straight across from that base to the side opposite it, measured at a right angle.
2. Hint 2
  Look for the segment meeting $AB$ at a right angle.
Answer

$\overline{CE}$ or $\overline{EC}$ .

Full solution

The chosen base $AB$ is vertical, so a perpendicular height to it runs horizontally.
Segment $CE$ goes from the opposite side to $AB$ and meets $AB$ at the marked right angle.

Segment $CD$ runs along the opposite side, and segment $AC$ is slanted across the parallelogram.
Neither gives the perpendicular distance to base $AB$ .
The required height is $\overline{CE}$ .

Answer

$\overline{CE}$ or $\overline{EC}$ .

Key idea

For a parallelogram, the height for a base runs from the opposite side to the line of that base and meets it at a right angle.
Problem 3: Mark complete
Problem 4 The hanging sign

The figure shows a flat sign. Find its area and its perimeter.

The stem is centered.

Text description of this figure

The outline of a flat T-shaped sign, drawn to scale and not shaded. A horizontal bar sits on top: its top edge is labeled 10 cm and its right end is labeled 2 cm. A narrower stem hangs below the bar: the right side of the stem, below the bar, is labeled 5 cm, and the bottom of the stem is labeled 4 cm. Small squares mark right angles at the two top corners of the bar, the two outside lower corners of the bar, and the two bottom corners of the stem. No line is drawn where the stem joins the bar. Below the figure it is stated that the stem is centered.
1. Hint 1
  The sign can be split into its horizontal top bar and its stem.
2. Hint 2
  The two overhangs together make up the part of the bar's bottom not covered by the stem.
3. Hint 3
  Add the two rectangular areas, then trace just the outer boundary for perimeter.
Answer

Area: $40$ square cm. Perimeter: $34$ cm.

Full solution

The top bar is $10$ cm by $2$ cm, and the stem below it is $4$ cm by $5$ cm.
Their interiors do not overlap.
$A=10\times2+4\times5$ $A=40$
The area is $40$ square cm.

The two overhangs are the part of the bar's bottom that the stem does not cover, so together they total $10-4=6$ cm, wherever the stem sits.

Walking around the boundary includes the top, two bar ends, two overhangs, both sides of the stem, and its bottom.
$P=10+2(2)+6+2(5)+4$ $P=34$
The perimeter is $34$ cm; the shared edge between the stem and bar is inside the sign.

Answer

Area: $40$ square cm. Perimeter: $34$ cm.

Key idea

Add non-overlapping pieces for area and trace the outside edges for perimeter.
Problem 4: Mark complete
Problem 5 The trapezoid panel

A trapezoidal panel has an area of $84$ square cm and a perpendicular height of $7$ cm. One parallel side is twice as long as the other. Find the lengths of both parallel sides.
1. Hint 1
  The area and height determine the average length of the parallel sides.
2. Hint 2
  Divide the area by the height, then double to find the sum of the two parallel lengths.
3. Hint 3
  If the shorter side is $s$ , the longer side is $2s$ .
Answer

$8$ cm and $16$ cm.

Full solution

Area equals the average parallel-side length times the height, so the average length is
$84\div7=12$
cm.
Thus the two parallel lengths have a sum of $24$ cm.

Let $s$ be the shorter length in cm.
The longer length is $2s$ , so
$s+2s=24$ $3s=24$ $s=8$
The longer side is $16$ cm.

Check the area.
$\frac{8+16}{2}\times7=84$
square cm, and $16$ cm is twice $8$ cm.

Answer

$8$ cm and $16$ cm.

Key idea

A trapezoid's area and height give the sum of its parallel-side lengths, which can then be shared according to their relationship.
Problem 5: Mark complete
Problem 6 The two fabric pieces

The figure shows triangular fabric pieces $P$ and $Q$ . Tia says the two pieces have equal areas because each is marked with a $6$ cm base and a $5$ cm length. Explain Tia's mistake, and find how much more area $Q$ has than $P$ .

Fabric pieces $P$ and $Q$ , drawn to the same scale.

Text description of this figure

Two separate triangles drawn to the same scale, labeled P on the left and Q on the right. Triangle P has a horizontal base labeled 6 cm. Its top vertex lies beyond the right end of the base, and the slanted side from the right end of the base up to the top vertex is labeled 5 cm. The base line is extended to the right as a dashed line, and a dashed vertical segment labeled 4 cm drops from the top vertex to that extension, meeting it at a marked right angle outside the triangle. The length of the extension is not labeled. Triangle Q has a horizontal base labeled 6 cm and its top vertex directly above the middle of the base; a dashed vertical segment inside it, labeled 5 cm, drops from the top vertex to the base and meets it at a marked right angle.
1. Hint 1
  Choose the distance perpendicular to each base, whether it lies inside or outside the piece.
2. Hint 2
  Find each triangle's area using its base and matching perpendicular height, then subtract.
Answer

Tia used the $5$ cm slanted side of $P$ as its height, but the height of $P$ is $4$ cm. $Q$ has $3$ square cm more area than $P$ .

Full solution

For $P$ , the $4$ cm segment outside the triangle is perpendicular to the base's line.
Its $5$ cm slanted side is not that height, and using it as one is Tia's mistake, because it would give $P$ the same area as $Q$ .
$A_P=\frac12\times6\times4$ $A_P=12$
Piece $P$ covers $12$ square cm.

For $Q$ , the inside dashed height is $5$ cm.
$A_Q=\frac12\times6\times5$ $A_Q=15$
Piece $Q$ covers $15$ square cm.

The difference is
$15-12=3$
square cm.
Both pieces have base $6$ cm, so the $1$ cm height difference also gives half of $6\times1$ , confirming the result.

Answer

Tia used the $5$ cm slanted side of $P$ as its height, but the height of $P$ is $4$ cm. $Q$ has $3$ square cm more area than $P$ .

Key idea

A triangle's perpendicular height is measured to its base's line even when the height lies outside the triangle.
Problem 6: Mark complete
Problem 7 Paint for the cards

A supply of paint covers $1{,}500$ square cm. Each of six parallelogram cards has a base of $18$ cm, a perpendicular height of $0.12$ m, and a slanted side of $15$ cm. After the front face of each card is painted, how many square cm could the remaining paint cover? Use $100$ cm for $1$ m.
1. Hint 1
  The paint used depends on the area of each card.
2. Hint 2
  Convert the perpendicular height to cm before multiplying it by the base.
3. Hint 3
  Multiply one card's area by six, then subtract from the original coverage.
Answer

$204$ square cm.

Full solution

The perpendicular height is $0.12\times100=12$ cm.
The $15$ cm slanted side is not the height.
One card's area in square cm is
$A=18\times12$ $A=216$

Six cards use paint covering
$6\times216=1{,}296$
square cm.
The remaining coverage is
$1{,}500-1{,}296=204$
square cm.

Checking, the used coverage and remaining coverage add to $1{,}500$ square cm.

Answer

$204$ square cm.

Key idea

Put the base and perpendicular height in the same unit before finding how much surface can be covered.
Problem 7: Mark complete
Problem 8 The paper pieces

The figure shows the same paper pieces before and after a triangular piece is slid to a new position, with no gaps or overlaps. Leo says the perimeter must stay the same. Is Leo correct? Justify your answer with both perimeters, and find the area of the final parallelogram.

The same paper pieces, before and after the triangle is slid.

Text description of this figure

Two panels drawn to the same scale, one above the other. The top panel, Before, shows a horizontal rectangle with its bottom labeled 8 cm and its left side labeled 4 cm. A dot marks an unlabeled point on the top edge, near the right end. A dashed cut labeled 5 cm runs from that dot down to the bottom-right corner, cutting a lightly shaded triangle off the right end. A curved arrow leads from that triangle down to the bottom panel. The bottom panel, After, shows the same shaded triangle moved to the left end of the remaining piece, fitting against its vertical left edge. The join between them is a dashed vertical line labeled 4 cm, with a small square marking a right angle at its bottom. Together the pieces form a parallelogram whose top edge is shifted to the left of its bottom edge. Its bottom is labeled 8 cm and its slanted right side is labeled 5 cm.
1. Hint 1
  Sliding a piece preserves how much paper there is, but changes which edges lie on the outside.
2. Hint 2
  Find the area of the original rectangle and trace each outline separately.
3. Hint 3
  Opposite sides of a parallelogram are equal in length; trace each outline once, counting only its outside edges.
Answer

No. Area: $32$ square cm. Rectangle perimeter: $24$ cm. Parallelogram perimeter: $26$ cm.

Full solution

The original rectangle is $8$ cm by $4$ cm.
Sliding all its pieces into the parallelogram without gaps or overlaps preserves its area.
$A=8\times4$ $A=32$
The final base remains $8$ cm and its perpendicular height remains $4$ cm, so the rearrangement gives base times height for its area.

The original rectangle's perimeter is
$P_R=2(8+4)$ $P_R=24$
cm.
The final parallelogram has an $8$ cm base and $5$ cm slanted sides.
$P_P=2(8+5)$ $P_P=26$
cm.

The paper still covers $32$ square cm, but the new outer edges total $2$ cm more.
Leo's perimeter claim is false.

Answer

No. Area: $32$ square cm. Rectangle perimeter: $24$ cm. Parallelogram perimeter: $26$ cm.

Key idea

Rearranging all pieces without gaps or overlaps preserves area while the outer boundary may change.
Problem 8: Mark complete
Problem 9 Three paper triangles

Three identical right triangles are turned or flipped and joined without gaps or overlaps to make the trapezoid in the figure. Use the trapezoid's area to find the area of one triangle, and explain how this arrangement gives half the base times the height for that triangle.

Three identical right triangles joined into a trapezoid.

Text description of this figure

A trapezoid drawn to scale. Its bottom and top are horizontal and its left side is vertical. The top is labeled 4 cm and the left side is labeled 6 cm. The right end of the top sits directly above the middle of the bottom, and the right side slants from there down to the bottom-right corner. A segment runs from the bottom-left corner to the top-right corner, and a vertical segment runs from the top-right corner straight down to the middle of the bottom. Together these split the trapezoid into three right triangles, each with a different light fill: a colored tint, a very pale tint, and diagonal hatching. Small squares mark right angles at the top-left corner and on both sides of the foot of the vertical segment. Below the bottom, an inner bracket marks the left half of the bottom as 4 cm, and an outer bracket marks the whole bottom as 8 cm.
1. Hint 1
  The three pieces are identical, so they have equal areas.
2. Hint 2
  Find the trapezoid's area using its two parallel sides and its height.
3. Hint 3
  Share that total equally among the three pieces and compare with one triangle's base and height.
Answer

One triangle has area $12$ square cm.

Full solution

The trapezoid has parallel sides $4$ cm and $8$ cm and a perpendicular height of $6$ cm.
Its area is
$A=\frac{4+8}{2}\times6$ $A=36$
square cm.

Three identical pieces fill the shape, so one piece has area
$36\div3=12$
square cm.

Each right triangle has a base of $4$ cm and a perpendicular height of $6$ cm.
The top of the trapezoid is one triangle base, $4$ cm, and the bottom is two triangle bases, $8$ cm, so the average width is $6$ cm, which is one and a half bases.
With $b$ and $h$ for the triangle's base and height, the trapezoid's area is $\tfrac32\times b\times h$ , and one third of that is $\tfrac12\times b\times h$ .
$\frac12\times4\times6=12$
This agrees with the area obtained by sharing the trapezoid.

Answer

One triangle has area $12$ square cm.

Key idea

Joining identical copies lets you recover one shape's area by dividing the total area equally.
Problem 9: Mark complete
Problem 10 The two paper parts

The figure shows one trapezoid cut halfway up its height. The upper piece is turned and placed beside the lower piece without gaps or overlaps. Use this arrangement to find the original trapezoid's area, and explain why the arrangement gives the average of its parallel-side lengths times its height.

One trapezoid, before and after its upper piece is turned.

Text description of this figure

Two panels drawn to the same scale, one above the other. The top panel, Before, shows a trapezoid with a horizontal bottom labeled 10 cm and a shorter horizontal top labeled 4 cm, centered over the bottom. A dashed horizontal cut runs across the trapezoid halfway between the top and the bottom; its length is not labeled. The piece above the cut is lightly shaded and the piece below is plain. Outside the left side, a dashed vertical height rises from the left end of the bottom to the level of the top, meeting the bottom at a marked right angle. A small tick splits it into two halves, each labeled 3 cm, and a bracket beside the whole height is labeled 6 cm. The bottom panel, After, shows the lower piece in the same place, and the shaded piece turned through a half turn about the right end of the cut, shown by a curved arrow around that point. The shaded piece now sits to the right of the lower piece, and the slanted edge where they meet is dashed. Together they form a parallelogram. Below it, one bracket marks the original bottom as 10 cm and a separate bracket marks the turned top, now on the bottom, as 4 cm. A dashed vertical height inside the parallelogram is labeled 3 cm, with a small square marking the right angle at the bottom.
1. Hint 1
  No paper has been added or removed by the cut and turn.
2. Hint 2
  In the new parallelogram, the bottom edge combines the original bottom and top lengths.
3. Hint 3
  The new height is half the original height, so compare the two ways of grouping the multiplication.
Answer

$42$ square cm.

Full solution

The half turn is about a point halfway up, so the top edge, $3$ cm above that point, lands $3$ cm below it, on the base line, and the two halves of the slanted side fit together.
After the turn, the parallelogram's base combines the original $10$ cm and $4$ cm parallel sides.
Its base is
$10+4=14$
cm.
Its perpendicular height is half of $6$ cm, or $3$ cm.

All the original paper fills this parallelogram, so the original area in square cm is
$A=14\times3$ $A=42$

Adding the original parallel sides and multiplying by half the height gives the same result as halving their sum first and multiplying by the full height.
$\frac{10+4}{2}\times6=42$
Thus the rearrangement gives the average parallel-side length times the original height.

Answer

$42$ square cm.

Key idea

A trapezoid's area is the average of its parallel-side lengths times their perpendicular distance.
Problem 10: Mark complete

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