TACHS Quiz: Measurement Problem Solving

17 questions · exam conditions

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Measurement Problem SolvingQuestion 1 of 17

Use the figure to answer the question. A parallelogram has a base of $16$ cm and a corresponding height of $7$ cm. A triangle has the same area as this parallelogram and a base of $14$ cm. What is the height of the triangle?

$Question graphic$

8

14

16

28

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TACHS Quiz

TACHS Quiz: Measurement Problem Solving

Practice Measurement Problem Solving in TACHS 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 Measurement Problem Solving, giving you a quick way to practice the rules, question types, and explanations that matter most for TACHS.

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

$8$ cm
$14$ cm
$16$ cm (correct answer)
$28$ cm

Explanation: Parallelogram area

=16\times7=112

. Triangle:

112=\tfrac12(14)h\Rightarrow h=16

. (A) forgets factor of

2

in reverse. (B) equals the base. (D) doubles unnecessarily.

Question 2

Use the figure to solve. A rectangular swimming pool is $25$ meters long and $10$ meters wide. It is surrounded by a tiled deck of uniform width. The total area of the pool and deck together is $594$ square meters. What is the width of the deck?

$2$ meters
$3$ meters
$4$ meters (correct answer)
$6$ meters

Explanation: Let

w

= deck width.

(25+2w)(10+2w)=594

. Expand:

250+70w+4w^2=594

, so

4w^2+70w-344=0

2w^2+35w-172=0

. Using quadratic:

w=\frac{-35+\sqrt{1225+1376}}{4}=\frac{-35+\sqrt{2601}}{4}=\frac{-35+51}{4}=4

. (A), (B), (D) don't satisfy equation.

Question 3

Refer to the figure. A rectangular poster is $2$ feet $6$ inches wide and $3$ feet $4$ inches tall. A decorative ribbon will be placed along the entire border of the poster. How many inches of ribbon are needed?

$70$ inches
$120$ inches
$140$ inches (correct answer)
$160$ inches

Explanation: Width

=30

in, height

=40

in. Perimeter

=2(30+40)=140

in. (A) adds

30+40

only. (B) mixes units wrong. (D) computes

2(30)+2(50)

Question 4

Refer to the figure. The hexagonal figure is made by attaching a square to an equilateral triangle along one side. If each side of the square is $8$ cm, what is the perimeter of the combined figure?

$32$ cm
$40$ cm (correct answer)
$48$ cm
$56$ cm

Explanation: The shared side is internal. Perimeter uses 3 sides of square

+

2 sides of triangle

=3(8)+2(8)=40

cm. (A) only the square. (C) doesn't remove the shared sides. (D) adds both shapes' full perimeters minus one side.

Question 5

Refer to the figure. A square and an equilateral triangle have the same perimeter. The side of the square is $9$ inches. How long is each side of the triangle?

$9$ inches
$12$ inches (correct answer)
$13.5$ inches
$18$ inches

Explanation: Square perimeter

=36

in. Triangle side

=36/3=12

in. (A) equates sides not perimeters. (C) divides by

2.67

or confuses with half. (D) divides by

2

Question 6

Refer to the figure. Each square in the grid has side length $1\text{ m}$ . What is the area of the shaded garden?

$20\text{ m}^2$
$22\text{ m}^2$ (correct answer)
$24\text{ m}^2$
$26\text{ m}^2$

Explanation: Counting the shaded unit squares gives 22, so the area is

22\text{ m}^2

.
Distractors: A. Counts one fewer column (20 squares). C. Adds an extra 2 squares that are not shaded (24 squares). D. Miscounts by adding 4 nonexistent squares (26 squares).

Question 7

Refer to the figure. A rectangular living-room floor measures $15$ feet by $12$ feet. Carpet is sold in square yards. How many square yards of carpet are needed to exactly cover the floor?

$9$ square yards
$20$ square yards (correct answer)
$60$ square yards
$180$ square yards

Explanation: Area

=15\times12=180

sq ft.

1

sq yd

=9

sq ft, so

180/9=20

sq yd. (A) divides perimeter by something. (C) divides by

3

not

9

. (D) leaves in square feet.

Question 8

The L-shaped figure shown is made up of two rectangles. Use the figure to find the perimeter of the entire L-shape.

$46$ cm
$52$ cm (correct answer)
$58$ cm
$64$ cm

Explanation: The L-shape has outer dimensions

16\times10

with a

6\times4

rectangle removed from the top-right corner. Sides:

16+10+10+4+6+6=52

cm. (A) forgets one segment. (C) adds the removed rectangle's perimeter twice. (D) uses

2(16+10)+2(6+... )

over-counting.

Question 9

Refer to the figure. Parallelogram $EFGH$ has base $\overline{EH}=12\text{ cm}$ , side $\overline{FG}=8\text{ cm}$ , and altitude $h=6\text{ cm}$ . What is the area of $EFGH$ ?

$72\text{ cm}^2$ (correct answer)
$96\text{ cm}^2$
$60\text{ cm}^2$
$48\text{ cm}^2$

Explanation: Area of a parallelogram is

\text{base}\times\text{height}=12\times6=72\text{ cm}^2

. Distractors: B. Multiplies base by the slanted side (12×8=96). C. Averages side and height before multiplying. D. Multiplies side (8) by height (6).

Question 10

Refer to the figure. Rhombus $ABCD$ has diagonals $\overline{AC}=12\text{ cm}$ and $\overline{BD}=10\text{ cm}$ . What is the area of the rhombus?

$60\text{ cm}^2$ (correct answer)
$120\text{ cm}^2$
$48\text{ cm}^2$
$24\text{ cm}^2$

Explanation: Area of a rhombus

=\tfrac12(d_1\times d_2)=\tfrac12(12\times10)=60\text{ cm}^2

. Distractors: B. Multiplies the diagonals without halving. C. Halves only one diagonal before multiplying (\tfrac12

\times

\times

8). D. Halves both diagonals separately then multiplies.

Question 11

Refer to the figure. Rectangle $ABCD$ has a perimeter of $34\text{ m}$ . Two opposite sides are each $11\text{ m}$ long, and the other two opposite sides are each $x\text{ m}$ long. What is the value of $x$ ?

$6\text{ m}$ (correct answer)
$5\text{ m}$
$12\text{ m}$
$13\text{ m}$

Explanation: The perimeter of a rectangle is

2(\text{length}+\text{width})

. Here,

2(11+x)=34

, so

11+x=17

and

x=6\text{ m}

.
Distractors: B.

5\text{ m}

comes from subtracting incorrectly:

34-11=23\rightarrow x=5\text{ m}

. C.

12\text{ m}

mistakes perimeter for the missing side by doubling the given length. D.

13\text{ m}

results from dividing the perimeter by 2 and then adding the length instead of subtracting.

Question 12

Refer to the figure. Right triangle $XYZ$ has legs $XY=9\text{ cm}$ and $YZ=12\text{ cm}$ . What is the perimeter of the triangle?

$30\text{ cm}$
$34\text{ cm}$
$36\text{ cm}$ (correct answer)
$38\text{ cm}$

Explanation: Hypotenuse

XZ=\sqrt{9^2+12^2}=\sqrt{81+144}=\sqrt{225}=15\text{ cm}

. Perimeter

=9+12+15=36\text{ cm}

. Distractors: A. Adds only the legs and one miscalculated hypotenuse (9+12+9=30). B. Uses

\sqrt{225}\approx15

correctly but then rounds down one leg. D. Adds an over-rounded hypotenuse of 17 cm.

Question 13

Refer to the figure. A flower bed is composed of a rectangle and a semicircle. The rectangle measures $12\text{ m}$ by $8\text{ m}$ , and the semicircle is attached along the $12\text{ m}$ side, using it as the diameter. What is the total area of the flower bed to the nearest whole square meter?

$140\text{ m}^2$
$147\text{ m}^2$
$150\text{ m}^2$
$153\text{ m}^2$ (correct answer)

Explanation: Rectangle area

=12\times8=96\text{ m}^2

. Semicircle radius

r=6\text{ m}

, so area

=\tfrac12\pi r^2=\tfrac12\pi(36)=18\pi\approx56.5\text{ m}^2

. Total

\approx96+56.5=152.5\text{ m}^2\approx153\text{ m}^2

.
Distractors: A. Rounds the semicircle area too low. B. Calculates using

\pi\approx3

instead of

3.14

. C. Uses

\pi=3.14

but mistakenly uses radius 5.

Question 14

Refer to the figure. A square lawn $ABCD$ has side length $18\text{ m}$ . A uniform walkway surrounds the lawn on all four sides, forming a larger square $EFGH$ whose side length is $22\text{ m}$ . What is the area of just the walkway?

$160\text{ m}^2$ (correct answer)
$176\text{ m}^2$
$200\text{ m}^2$
$196\text{ m}^2$

Explanation: Area walkway

=(22)^2-(18)^2=484-324=160\text{ m}^2

. Distractors: B. Uses outer area minus one incorrect inner dimension 17 m (484-289=195) then rounds. C. Adds the two areas instead of subtracting. D. Squares an average side length (20 m) then subtracts inner area incorrectly.

Question 15

Refer to the figure. A right triangle has legs of $9$ cm and $12$ cm. A square is drawn with one side equal to the hypotenuse of this triangle. What is the perimeter of the square?

$42$ cm
$60$ cm (correct answer)
$84$ cm
$225$ cm

Explanation: Hypotenuse

=\sqrt{81+144}=\sqrt{225}=15

. Perimeter of square

=4\times15=60

. (A) uses

9+12+21

. (C) uses

4\times21

(sum of legs). (D) is

15^2

, the area not perimeter.

Question 16

Use the figure shown. A trapezoid has parallel sides of $14$ inches and $22$ inches, and a height of $9$ inches. A rectangle has the same area as this trapezoid and a width of $6$ inches. What is the length of the rectangle?

$24$ inches
$27$ inches (correct answer)
$30$ inches
$33$ inches

Explanation: Trapezoid area

=\tfrac12(14+22)(9)=\tfrac12(36)(9)=162

. Rectangle length

=162/6=27

. (A) uses only

14

as average. (C) uses the average

18\times... /6

incorrectly. (D) forgets the

\tfrac12

36\times9/6...

Question 17

Refer to the figure. A rectangular garden measures $24$ feet by $18$ feet. A uniform path of width $3$ feet is built around the outside of the garden. What is the area of the path alone, in square feet?

$252$ square feet
$288$ square feet (correct answer)
$324$ square feet
$180$ square feet

Explanation: Outer rectangle:

(24+6)(18+6)=30\times24=720

. Inner garden:

24\times18=432

. Path area

=720-432=288

. (A) results from forgetting to double the width on each dimension (

27\times26-432=270

... close miscount). (C) comes from using width

3

only once on each side incorrectly. (D) is the perimeter times width (

84\times3/...

) style error without corners.