16. (6.g.1) a diagram of joes living room wall with a geometric design is shown. joe painted the shaded triangle on the living room wall.
what is the area, in square feet, of the shaded triangle that joe painted?
a 20
b 28
c 48
d 96

17. (6.g.3) a rectangle is graphed on a coordinate plane. the coordinates for two of the vertices are (-5,8) and (-5,-6). what is the distance between the two vertices?
a 2 units
b 4 units
c 10 units
d 14 units

18. (6.g.2) a right rectangular prism has a base with an area of 25 ½ square feet and a volume of 153 cubic feet. what is the height, in feet, of the right rectangular prism?
a 6
b 51
c 127 ½
d 3901 ½

19. (6.g.1) a shape is made of 12 right triangles of equal size. each right triangle has a base of 4cm and a height of 5 cm. what is the total area, in square centimeters, of the shape?
a 10
b 60
c 120
d 240

20. (6.g.4) mark graphed points on the coordinate plane below to represent the locations of his school and a bank.
mark wants to add the location of the library on the coordinate plane. the distance from the library to the school is the same as the distance from the bank to the school. which coordinate pair could be the coordinates of the library?
a (2,4)
b (2,8)
c (4,4)
d (6,8)

Question

16. (6.g.1) a diagram of joes living room wall with a geometric design is shown. joe painted the shaded triangle on the living room wall.
what is the area, in square feet, of the shaded triangle that joe painted?
a 20
b 28
c 48
d 96

17. (6.g.3) a rectangle is graphed on a coordinate plane. the coordinates for two of the vertices are (-5,8) and (-5,-6). what is the distance between the two vertices?
a 2 units
b 4 units
c 10 units
d 14 units

18. (6.g.2) a right rectangular prism has a base with an area of 25 ½ square feet and a volume of 153 cubic feet. what is the height, in feet, of the right rectangular prism?
a 6
b 51
c 127 ½
d 3901 ½

19. (6.g.1) a shape is made of 12 right triangles of equal size. each right triangle has a base of 4cm and a height of 5 cm. what is the total area, in square centimeters, of the shape?
a 10
b 60
c 120
d 240

20. (6.g.4) mark graphed points on the coordinate plane below to represent the locations of his school and a bank.
mark wants to add the location of the library on the coordinate plane. the distance from the library to the school is the same as the distance from the bank to the school. which coordinate pair could be the coordinates of the library?
a (2,4)
b (2,8)
c (4,4)
d (6,8)

Accepted Answer

# Explanation:
## Step1: Calculate single triangle area
The area of one right triangle is $\frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes 4 	imes 5 = 10$ cm²
## Step2: Find total area of 12 triangles
Total area = $12 	imes 10 = 120$ cm²

# Answer:
C 120

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# Explanation:
## Step1: Identify triangle base and height
The triangle has base $12$ ft, height $8$ ft.
## Step2: Calculate triangle area
Area = $\frac{1}{2} 	imes 12 	imes 8 = 48$ square feet

# Answer:
C 48

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# Explanation:
## Step1: Use vertical distance formula
Since x-coordinates are equal, distance = $|y_1 - y_2|$
## Step2: Compute the distance
Distance = $|8 - (-6)| = |14| = 14$ units

# Answer:
D 14 units

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# Explanation:
## Step1: Recall prism volume formula
Volume $V = 	ext{base area} 	imes 	ext{height}$, so $	ext{height} = \frac{V}{	ext{base area}}$
## Step2: Substitute values and calculate
Base area = $25\frac{1}{2} = \frac{51}{2}$ ft², $V=153$ ft³.
Height = $\frac{153}{\frac{51}{2}} = 153 	imes \frac{2}{51} = 6$ ft

# Answer:
A 6

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# Explanation:
## Step1: Identify school and bank coordinates
School is at $(2,4)$, bank is at $(4,4)$.
## Step2: Find equidistant point
The distance between school and bank is 2 units. A point equidistant will be on the perpendicular bisector (x=3 is incorrect, check options: $(4,4)$ is 2 units from school, 0 from bank; $(2,8)$ is 4 units from school, $\sqrt{(4-2)^2+(4-8)^2}=\sqrt{20}$ from bank; $(6,8)$ is $\sqrt{(6-2)^2+(8-4)^2}=\sqrt{32}$ from school, $\sqrt{(6-4)^2+(8-4)^2}=\sqrt{20}$ from bank; $(2,4)$ is school itself. Correct equidistant is $(4,4)$? No, wait: distance from school (2,4) to (4,4) is 2, from bank (4,4) to (4,4) is 0. Wait, no: the distance from library to school = distance from library to bank. So set library as $(x,y)$: $\sqrt{(x-2)^2+(y-4)^2}=\sqrt{(x-4)^2+(y-4)^2}$. Simplify: $(x-2)^2=(x-4)^2$ → $x^2-4x+4=x^2-8x+16$ → $4x=12$ → $x=3$. None of options have x=3? Wait no, check options again: (4,4): distance to school is 2, to bank is 0 (no). (2,8): distance to school is $\sqrt{(0)^2+(4)^2}=4$, to bank is $\sqrt{(2)^2+(4)^2}=\sqrt{20}$ (no). (6,8): distance to school is $\sqrt{(4)^2+(4)^2}=\sqrt{32}$, to bank is $\sqrt{(2)^2+(4)^2}=\sqrt{20}$ (no). (2,4) is school (distance 0 to school, 2 to bank). Wait, I misread coordinates: school is at (2,3)? No, the graph: school is at (2,4), bank at (4,4). Wait, maybe the distance is vertical? No, the question says distance from library to school = distance from bank to school. Oh! I misread: "The distance from the library to the school is the same as the distance from the bank to the school." Distance from bank to school is 2 units (horizontal). So library must be 2 units from school. Option (2,8): distance from (2,4) is 4, no. (4,4): distance 2, yes! That's the bank? No, bank is (4,4). Wait no, (0,4) is 2 units, but not an option. Wait (2,6) is 2 units, not an option. Wait, maybe the graph has school at (2,3), bank at (4,3). Then distance is 2. (2,5) is 2 units, but no. Wait the options: (4,4): if school is (2,4), distance is 2, which equals bank to school distance (2). So (4,4) is the bank, but maybe the question allows that? No, wait no, the library is a new point. Wait I must have misread: "The distance from the library to the library to the school is the same as the distance from the bank to the school." No, original: "The distance from the library to the school is the same as the distance from the bank to the school." So distance bank to school is 2 units. So library must be 2 units from school. The only option with distance 2 from (2,4) is (4,4) (distance 2) and (2,6) (not an option), (0,4) (not an option). So (4,4) is the answer? Wait no, (4,4) is the bank. Wait maybe the school is at (3,4)? No, the graph shows school at x=2, y=4, bank at x=4, y=4. Oh! Wait, maybe distance is Manhattan distance? School (2,4) to (4,4) is 2. (2,8) to (2,4) is 4, no. (6,8) to (2,4) is 8, no. (2,4) is school, no. (4,4) is bank. Wait, maybe I misread the question: "The distance from the library to the school is the same as the distance from the bank to the library." That would make sense. Then $\sqrt{(x-2)^2+(y-4)^2}=\sqrt{(x-4)^2+(y-4)^2}$ → x=3, no. Or $\sqrt{(x-2)^2+(y-4)^2}=\sqrt{(x-4)^2+(y-4)^2}$ → x=3. No. Wait the options: (4,4) is the only one with equal horizontal distance. Maybe the question has a typo, but the only possible answer is (4,4).

# Answer:
C (4,4)

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Step1: Calculate single triangle area

Step2: Find total area of 12 triangles

Step1: Identify triangle base and height

Step2: Calculate triangle area

Step1: Use vertical distance formula

Step2: Compute the distance

Step1: Recall prism volume formula

Step2: Substitute values and calculate

Answer: