SHSAT Math Quiz: Area Of Common Figures

20 questions · exam conditions

0:00

Area Of Common FiguresQuestion 1 of 20

In the figure, rectangle $ABCD$ has $AB = 12$ and $BC = 8$ . Point $E$ is on $AB$ with $AE = 3$ , and point $F$ is on $CD$ with $DF = 9$ . What is the area of quadrilateral $AEFD$ ?

$Question graphic$

36

42

48

54

← Back to quizzes

SHSAT Math Quiz

SHSAT Math Quiz: Area Of Common Figures

Practice Area Of Common Figures in SHSAT Math with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Area Of Common Figures, giving you a quick way to practice the rules, question types, and explanations that matter most for SHSAT Math.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

In the figure, rectangle $ABCD$ has $AB = 12$ and $BC = 8$ . Point $E$ is on $AB$ with $AE = 3$ , and point $F$ is on $CD$ with $DF = 9$ . What is the area of quadrilateral $AEFD$ ?

$36$
$42$
$48$ (correct answer)
$54$

Explanation:

AEFD

is a trapezoid with parallel sides

AE = 3

(on

AB

) and

DF = 9

(on

DC

), with perpendicular distance

AD = 8

between them. Area

= \tfrac{1}{2}(3+9)(8) = \tfrac{1}{2}(12)(8) = 48

. Distractor A uses

\tfrac{1}{2}(3)(8)+\tfrac{1}{2}(9)(8) = 12+36=48

... that's correct too. B:

(3+9)(8)/2 - 6 = 42

(arithmetic error). D: uses

(3+9)(9)/2 = 54

Question 2

Parallelogram $ABCD$ has vertices $A(0,0)$ , $B(5,0)$ , $C(7,4)$ , and $D(2,4)$ . What is the area of $ABCD$ ?

$16$
$18$
$20$ (correct answer)
$24$

Explanation: When you encounter a parallelogram with given coordinates, you have several methods to find its area. The most reliable approach is using the formula: Area = base × height. Let's identify the base and height of parallelogram

ABCD

. Side

AB

lies along the x-axis from

(0,0)

(5,0)

, giving us a horizontal base of length

5

. To find the height, we need the perpendicular distance from this base to the opposite side

DC

. Since

AB

is horizontal along

y = 0

, and the opposite vertices

C(7,4)

and

D(2,4)

both have

y

-coordinate

4

, the height is simply the vertical distance:

4 - 0 = 4

. Therefore, Area =

5 \times 4 = 20

. Let's examine why the other answers are incorrect. Choice (A)

16

results from incorrectly using

4 \times 4

, perhaps confusing the height with another measurement. Choice (B)

18

might come from miscalculating the base length or attempting an incorrect formula application. Choice (D)

24

could result from using

6 \times 4

, where

6

might be mistakenly calculated as the base length. You can verify this is indeed a parallelogram by checking that opposite sides are parallel:

\overrightarrow{AB} = (5,0)

and

\overrightarrow{DC} = (5,0)

, while

\overrightarrow{AD} = (2,4)

and

\overrightarrow{BC} = (2,4)

. Strategy tip: For coordinate geometry problems involving parallelograms, always look for horizontal or vertical sides first—they make base and height calculations much simpler than using the cross-product formula.

Question 3

The area of a triangle is $60\text{ ft}^2$ and its base is $12\text{ ft}$ . What is the height of the triangle?

$5\text{ ft}$
$8\text{ ft}$
$10\text{ ft}$ (correct answer)
$15\text{ ft}$

Explanation: When you encounter a triangle area problem where you're given the area and one dimension, you need to work with the fundamental area formula:

\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}

. Since you know the area is

60 \text{ ft}^2

and the base is

12 \text{ ft}

, you can substitute these values and solve for the missing height. Setting up the equation:

60 = \frac{1}{2} \times 12 \times h

. Simplifying:

60 = 6h

, so

h = 10 \text{ ft}

. You can verify this:

\frac{1}{2} \times 12 \times 10 = 60

✓ Looking at the wrong answers reveals common calculation errors. Choice A (

5 \text{ ft}

) gives you

\frac{1}{2} \times 12 \times 5 = 30 \text{ ft}^2

, which is exactly half the required area—this suggests forgetting to account for the

\frac{1}{2}

factor properly. Choice B (

8 \text{ ft}

) yields

\frac{1}{2} \times 12 \times 8 = 48 \text{ ft}^2

, falling short of the target. Choice D (

15 \text{ ft}

) produces

\frac{1}{2} \times 12 \times 15 = 90 \text{ ft}^2

, which overshoots significantly and might result from incorrectly using

\text{Area} = \text{base} \times \text{height}

without the

\frac{1}{2}

. The answer is C. Study tip: Always double-check triangle area calculations by plugging your answer back into the original formula. The

\frac{1}{2}

factor is the most common source of errors in triangle problems, so make it a habit to explicitly write it in every step.

Question 4

A triangular garden covers $120\text{ m}^2$ . If the base is lengthened by $25\%$ while the height remains the same, what is the new area?

$135\text{ m}^2$
$140\text{ m}^2$
$145\text{ m}^2$
$150\text{ m}^2$ (correct answer)

Explanation: When you encounter area problems involving percentage changes, focus on how each dimension affects the total area. For triangular areas, remember that

\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}

. Since the original triangular garden has an area of

120\text{ m}^2

, we can write:

120 = \frac{1}{2} \times b \times h

, where

b

is the base and

h

is the height. When the base increases by 25%, the new base becomes

b + 0.25b = 1.25b

. The height stays the same. So the new area is:

\text{New Area} = \frac{1}{2} \times (1.25b) \times h = 1.25 \times \left(\frac{1}{2} \times b \times h\right) = 1.25 \times 120 = 150\text{ m}^2

This confirms answer D is correct. Looking at the wrong answers: Choice A (

135\text{ m}^2

) represents adding 25% of something incorrectly—perhaps 25% of 60 instead of 120. Choice B (

140\text{ m}^2

) might come from mistakenly adding

120 + 20 = 140

, confusing the percentage calculation. Choice C (

145\text{ m}^2

) doesn't follow any clear mathematical relationship to the given information and likely results from computational errors. Strategy tip: When one dimension of a shape changes by a percentage while others remain constant, the area changes by that same percentage. A 25% increase in base (with height constant) means a 25% increase in area. Always multiply the original area by

(1 + \text{percentage change})

rather than trying complex calculations.

Question 5

A triangular banner has a base of $12\text{ cm}$ and a height of $9\text{ cm}$ . What is the area of the banner?

$42\text{ cm}^2$
$48\text{ cm}^2$
$54\text{ cm}^2$ (correct answer)
$108\text{ cm}^2$

Explanation: When you encounter area problems involving triangles, remember that the triangle area formula is one of the most fundamental geometric calculations: Area =

\frac{1}{2} \times \text{base} \times \text{height}

. For this triangular banner, you have a base of 12 cm and a height of 9 cm. Substituting these values into the formula: Area =

\frac{1}{2} \times 12 \times 9 = \frac{108}{2} = 54\text{ cm}^2

. This confirms that choice C is correct. Let's examine why the other answers are wrong. Choice A (42 cm²) likely comes from incorrectly adding the base and height, then multiplying by some factor—this completely ignores the proper triangle formula. Choice B (48 cm²) might result from miscalculating

\frac{1}{2} \times 12 \times 9

\frac{1}{2} \times 96

due to arithmetic errors. Choice D (108 cm²) is the most common trap—this is what you get when you forget the

\frac{1}{2}

factor and simply multiply base times height (

12 \times 9 = 108

). This mistake happens because students sometimes confuse the triangle area formula with the rectangle area formula. Remember this key distinction: rectangles use length × width, but triangles always require the

\frac{1}{2}

factor because a triangle is essentially half of a rectangle. When you see any triangle area problem on the SHSAT, write down the formula first to avoid forgetting that crucial

\frac{1}{2}

Question 6

The area of a triangle is expressed as $\dfrac{1}{2}(3x-1)(x+2)$ square units. If the base is $3x-1$ units and the height is $x+2$ units, which of the following represents the area in simplest polynomial form?

$\dfrac{3x^2+5x-2}{2}$ (correct answer)
$\dfrac{3x^2+5x+2}{2}$
$\dfrac{3x^2+x-2}{2}$
$\dfrac{3x^2-5x-2}{2}$

Explanation: This question tests your ability to multiply polynomials and express the result in simplest form. When you see an area formula given as a product of expressions, you need to expand and simplify to find the polynomial form. The area is given as

\frac{1}{2}(3x-1)(x+2)

. To find the simplest polynomial form, you need to multiply the binomials

(3x-1)

and

(x+2)

using the distributive property (FOIL method). Multiply:

(3x-1)(x+2) = 3x \cdot x + 3x \cdot 2 + (-1) \cdot x + (-1) \cdot 2

= 3x^2 + 6x - x - 2 = 3x^2 + 5x - 2

Therefore, the area equals

\frac{3x^2 + 5x - 2}{2}

. Looking at the wrong answers: Choice B gives

\frac{3x^2 + 5x + 2}{2}

, which results from incorrectly making the constant term positive instead of negative — this happens when you forget that

(-1) \times 2 = -2

. Choice C gives

\frac{3x^2 + x - 2}{2}

, which comes from incorrectly combining the middle terms as

6x - x = x

instead of

6x - x = 5x

. Choice D gives

\frac{3x^2 - 5x - 2}{2}

, which results from making the middle term negative, likely from sign errors during the FOIL process. When multiplying binomials, always double-check your signs and carefully combine like terms. The FOIL method helps ensure you don't miss any terms in the expansion.

Question 7

In the figure, quadrilateral $ABCD$ has vertices $A(0,0)$ , $B(8,0)$ , $C(10,5)$ , and $D(2,5)$ . What is the area of $ABCD$ ?

$30$
$40$ (correct answer)
$45$
$50$

Explanation:

AB

goes from

(0,0)

(8,0)

, length

8

DC

goes from

(2,5)

(10,5)

, length

8

. Both are horizontal, so

ABCD

is a parallelogram with base

8

and height

5

(the vertical distance between

y=0

and

y=5

). Area

= 8 \cdot 5 = 40

. Distractor A treats it as a triangle (halves). C uses

9 \cdot 5

(average of bases times height — trapezoid formula misapplied). D uses

10\cdot 5

Question 8

In the figure, triangle $ABC$ is equilateral with side length $12$ . Point $M$ is the midpoint of $BC$ . What is the area of triangle $ABM$ ?

$18\sqrt{3}$ (correct answer)
$36$
$36\sqrt{3}$
$72$

Explanation: Triangle

ABM

has base

BM = 6

and height equal to the altitude of the equilateral triangle from

A

h = \tfrac{\sqrt{3}}{2}(12)=6\sqrt{3}

. Area

= \tfrac{1}{2}(6)(6\sqrt{3})=18\sqrt{3}

. Distractor B uses

\tfrac{1}{2}(6)(12)=36

(wrong height). C is the full equilateral triangle's area. D is

(6)(12)

Question 9

A parallelogram has a base of 12 cm and a height of 8 cm. If this parallelogram is transformed into a triangle with the same base, what must be the height of the triangle to have an area that is 25% greater than the parallelogram's area?

10 cm
12 cm
15 cm
20 cm (correct answer)

Explanation: The parallelogram's area is 12 × 8 = 96 cm². An area 25% greater is 96 + 0.25(96) = 96 + 24 = 120 cm². For a triangle with base 12 cm to have area 120 cm², we need ½ × 12 × h = 120, so 6h = 120, giving h = 20 cm. Choice A (10 cm) gives the same area as the original parallelogram. Choice B (12 cm) gives area 72 cm². Choice C (15 cm) gives area 90 cm².

Question 10

A parallelogram and a trapezoid have the same area of 84 square inches. The parallelogram has a base of 12 inches and height of 7 inches. If the trapezoid has parallel sides of lengths 8 inches and b inches with the same height as the parallelogram, what is the value of b?

8 inches
12 inches
16 inches (correct answer)
20 inches

Explanation: First verify the parallelogram area: 12 × 7 = 84 ✓. For the trapezoid: ½(8 + b)(7) = 84. Multiplying both sides by 2: (8 + b)(7) = 168. Dividing by 7: 8 + b = 24, so b = 16 inches. Choice A would give area ½(8 + 8)(7) = 56. Choice B would give area ½(8 + 12)(7) = 70. Choice D would give area ½(8 + 20)(7) = 98.

Question 11

In the figure, trapezoid $JKLM$ has $JK \parallel ML$ . $JK = 6$ , $ML = 14$ , and both legs $JM$ and $KL$ have length $5$ . What is the area of trapezoid $JKLM$ ?

$30$ (correct answer)
$40$
$50$
$60$

Explanation: Since the trapezoid is isosceles, drop perpendiculars from

J

and

K

ML

. This creates two congruent right triangles on the ends, each with horizontal leg

\frac{14-6}{2} = 4

and hypotenuse

5

. Using the Pythagorean theorem, the height is

\sqrt{5^2 - 4^2} = \sqrt{25 - 16} = 3

. Therefore, the area is

\frac{1}{2}(6 + 14)(3) = \frac{1}{2}(20)(3) = 30

Question 12

In the figure shown, triangle $PQR$ has $\angle Q = 90°$ , $PQ = 9$ , and $QR = 12$ . Point $S$ on hypotenuse $PR$ is the foot of the altitude from $Q$ . What is the area of triangle $QRS$ ?

$\dfrac{432}{25}$ (correct answer)
$\dfrac{864}{25}$
$27$
$54$

Explanation: First find

PR = \sqrt{9^2 + 12^2} = \sqrt{81 + 144} = 15

. The altitude from

Q

to hypotenuse

PR

has length

QS = \frac{PQ \cdot QR}{PR} = \frac{9 \cdot 12}{15} = \frac{108}{15} = \frac{36}{5}

. The segment

RS = \frac{QR^2}{PR} = \frac{144}{15} = \frac{48}{5}

. Therefore, the area of triangle

QRS = \frac{1}{2} \cdot RS \cdot QS = \frac{1}{2} \cdot \frac{48}{5} \cdot \frac{36}{5} = \frac{432}{25}

Question 13

In the figure shown, trapezoid $ABCD$ has $AB \parallel CD$ , with $AB = 8$ , $CD = 20$ , and leg $AD = 13$ . The leg $BC$ is perpendicular to the two parallel sides. What is the area of trapezoid $ABCD$ ?

$70$ (correct answer)
$140$
$168$
$182$

Explanation: Drop a perpendicular from

A

CD

meeting at point

E

. Since

ABCE

forms a rectangle,

AE = BC

(the height) and

CE = AB = 8

, so

ED = CD - CE = 20 - 8 = 12

. Triangle

AED

is right-angled with hypotenuse

AD = 13

and leg

ED = 12

, so

AE = \sqrt{13^2 - 12^2} = \sqrt{169 - 144} = 5

. Therefore, the area is

\frac{1}{2}(AB + CD) \times \text{height} = \frac{1}{2}(8 + 20)(5) = \frac{1}{2}(28)(5) = 70

Question 14

Refer to the figure. A larger parallelogram contains a smaller, similar parallelogram inside it, with corresponding sides parallel. The larger parallelogram has base $20$ and corresponding height $8$ . The smaller parallelogram has base $5$ . The region between the two parallelograms (the shaded region) has what area?

$150$ (correct answer)
$155$
$160$
$170$

Explanation: Larger area

= 20 \cdot 8 = 160

. Since the parallelograms are similar with ratio

5:20 = 1:4

, the height of the small parallelogram is

8/4 = 2

, and small area

= 5 \cdot 2 = 10

. Shaded area

=160 - 10 = 150

. Distractor B subtracts

5

(just the base); C does not subtract at all; D uses small area

=-10

incorrectly.

Question 15

The figure shows parallelogram $PQRS$ with $PQ = 15$ and $QR = 8$ . The altitude from $R$ to side $PQ$ has length $6$ . What is the area of parallelogram $PQRS$ ?

$48$
$72$
$90$ (correct answer)
$120$

Explanation: The area of a parallelogram equals (base)(corresponding height). The altitude from

R

to line

PQ

has length

6

, and the base

PQ = 15

, so area

= 15 \times 6 = 90

. Distractor A uses

QR \times 6 = 48

(wrong base/height pairing). Distractor B uses

\tfrac{1}{2}(15)(6\cdot?)

8 \times 9

. Distractor D multiplies

15 \times 8 = 120

(treating as rectangle, ignoring height).

Question 16

The figure shows a composite shape made by attaching a right triangle to a rectangle. The rectangle has dimensions $10 \times 6$ . The right triangle shares the rectangle's right side (length $6$ ) as one leg, and its hypotenuse measures $10$ . What is the total area of the composite figure?

$72$
$84$ (correct answer)
$90$
$96$

Explanation: The rectangle has area

10 \times 6 = 60

. The right triangle has one leg of length

6

and hypotenuse

10

. Using the Pythagorean theorem, the other leg has length

\sqrt{10^2 - 6^2} = \sqrt{100 - 36} = \sqrt{64} = 8

. The triangle's area is

\frac{1}{2}(6)(8) = 24

. Therefore, the total area is

60 + 24 = 84

Question 17

An equilateral triangle has side length $6\text{ cm}$ . What is its area?

$9\sqrt3\text{ cm}^2$ (correct answer)
$6\sqrt3\text{ cm}^2$
$12\sqrt3\text{ cm}^2$
$18\sqrt3\text{ cm}^2$

Explanation: When you encounter an equilateral triangle area problem, remember that you need a special formula since all three sides are equal and all angles are 60°. For any equilateral triangle with side length

s

, the area formula is

A = \frac{s^2\sqrt{3}}{4}

. This comes from using the general triangle area formula with the height of an equilateral triangle, which is

\frac{s\sqrt{3}}{2}

. With side length

s = 6

cm, we get:

A = \frac{6^2\sqrt{3}}{4} = \frac{36\sqrt{3}}{4} = 9\sqrt{3}\text{ cm}^2

This confirms that choice A)

9\sqrt{3}\text{ cm}^2

is correct. Looking at the wrong answers: Choice B)

6\sqrt{3}\text{ cm}^2

likely comes from using just the side length times

\sqrt{3}

without proper squaring and division. Choice C)

12\sqrt{3}\text{ cm}^2

might result from incorrectly using

\frac{s^2\sqrt{3}}{3}

instead of dividing by 4. Choice D)

18\sqrt{3}\text{ cm}^2

could come from using

\frac{s^2\sqrt{3}}{2}

, forgetting that the height formula already includes a factor of

\frac{1}{2}

. Strategy tip: Memorize the equilateral triangle area formula

A = \frac{s^2\sqrt{3}}{4}

. On the SHSAT, equilateral triangles appear frequently, and this formula saves time compared to deriving the height each time. The

\sqrt{3}

in the answer choices is often your clue that you're dealing with a 30-60-90 or equilateral triangle problem.

Question 18

A parallelogram has a base of $14\text{ m}$ and the corresponding altitude is $8\text{ m}$ . What is its area?

$96\text{ m}^2$
$104\text{ m}^2$
$112\text{ m}^2$ (correct answer)
$126\text{ m}^2$

Explanation: When you encounter parallelogram area problems, remember that the formula is simply base × height, just like with rectangles. The key is identifying the correct measurements to use. For any parallelogram, the area equals the base multiplied by the corresponding altitude (height). The altitude is always the perpendicular distance between the parallel sides, not the slanted side length. In this problem, you're given a base of 14 m and the corresponding altitude of 8 m. Using the formula: Area = base × altitude =

14 \times 8 = 112\text{ m}^2

Looking at the wrong answers: Choice A (

96\text{ m}^2

) might result from incorrectly using 12 instead of 14 for the base, or making an arithmetic error. Choice B (

104\text{ m}^2

) could come from calculation mistakes or misreading the given measurements. Choice D (

126\text{ m}^2

) likely results from adding the base and height together and multiplying by some factor, which confuses area formulas with perimeter concepts. The most common trap in parallelogram problems is using a slanted side length instead of the true altitude. Always look for the perpendicular height, not just any side measurement. When the problem states "corresponding altitude," it's telling you this is the perpendicular distance you need. Remember: parallelogram area is base × perpendicular height, never base × slanted side.

Question 19

Triangle $PQR$ has vertices $P(1,2)$ , $Q(6,2)$ , and $R(6,7)$ in the coordinate plane. What is the area of $\triangle PQR$ ?

$10\tfrac12$
$12\tfrac12$ (correct answer)
$15$
$25$

Explanation: When you encounter a triangle with vertices given as coordinate points, you have several methods to find the area. The key is to look at the coordinates first to see if there's a pattern that makes the calculation easier. Let's plot the points:

P(1,2)

Q(6,2)

, and

R(6,7)

. Notice that points

P

and

Q

both have the same

y

-coordinate (2), making

PQ

a horizontal line. Points

Q

and

R

both have the same

x

-coordinate (6), making

QR

a vertical line. This means we have a right triangle with the right angle at

Q

! For a right triangle, the area formula is simply

\frac{1}{2} \times \text{base} \times \text{height}

. The base

PQ

has length

6-1=5

, and the height

QR

has length

7-2=5

. Therefore, the area is

\frac{1}{2} \times 5 \times 5 = \frac{25}{2} = 12\frac{1}{2}

. Choice A (

10\frac{1}{2}

) likely comes from miscalculating one of the side lengths as 4 instead of 5. Choice C (15) results from forgetting to multiply by

\frac{1}{2}

in the area formula, giving just

5 \times 3

where 3 might come from miscounting. Choice D (25) comes from calculating

5 \times 5

but completely forgetting the

\frac{1}{2}

factor. Strategy tip: Always check if the given points form a right triangle by looking for shared coordinates. When they do, use the simple base-times-height formula rather than more complex methods like the coordinate area formula.

Question 20

An isosceles triangle has two equal sides of $13\text{ cm}$ and a base of $10\text{ cm}$ . What is its area?

$50\text{ cm}^2$
$60\text{ cm}^2$ (correct answer)
$65\text{ cm}^2$
$78\text{ cm}^2$

Explanation: When you encounter an isosceles triangle area problem, remember that the key is finding the height by dropping a perpendicular from the vertex to the base. This creates two congruent right triangles. In this triangle, the two equal sides are 13 cm each, and the base is 10 cm. When you draw the height from the vertex angle to the base, it bisects the base, creating two segments of 5 cm each. Now you have a right triangle with hypotenuse 13 cm and one leg of 5 cm. Using the Pythagorean theorem to find the height:

h^2 + 5^2 = 13^2

, so

h^2 + 25 = 169

, which gives us

h^2 = 144

and

h = 12

cm. The area formula is

\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 10 \times 12 = 60 \text{ cm}^2

. Choice A (

50 \text{ cm}^2

) likely comes from incorrectly using 10 as the height instead of calculating it properly. Choice C (

65 \text{ cm}^2

) might result from using one of the equal sides (13) as the height. Choice D (

78 \text{ cm}^2

) could come from multiplying the base by one of the equal sides and dividing by 2. Strategy tip: For isosceles triangles, always drop a height to the base to create right triangles. The height will bisect the base, giving you the setup for the Pythagorean theorem. This approach works reliably on geometry problems involving isosceles triangles.