ISEE Middle Level Mathematics Achievement Quiz: Coordinate Shape Analysis

What this quiz covers

This quiz focuses on Coordinate Shape Analysis, giving you a quick way to practice the rules, question types, and explanations that matter most for ISEE Middle Level Mathematics Achievement.

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.

Question 1

1. 4 units
2. 5 units
3. (\sqrt{41}) units (correct answer)
4. (\sqrt{52}) units

Explanation: The length of the diagonal AC can be found using the distance formula, (\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}). For points A(2, 1) and C(6, 6), the length is (\sqrt{(6-2)^2 + (6-1)^2} = \sqrt{4^2 + 5^2} = \sqrt{16 + 25} = \sqrt{41}).

Question 2

1. (1, 5)
2. (2, 2)
3. (7, 2) (correct answer)
4. (4, -2)

Explanation: Let the coordinates of B be ((x, y)). The midpoint M is the average of the coordinates of A and B. So, (3 = \frac{-1+x}{2}) and (4 = \frac{6+y}{2}). Solving for x: (6 = -1+x), so (x=7). Solving for y: (8 = 6+y), so (y=2). The coordinates of point B are (7, 2).

Question 3

1. 0 (correct answer)
2. 1
3. 2
4. The sum is undefined.

Explanation: The diagonals connect opposite vertices. The first diagonal connects (1,1) and (5,5). Its slope is (\frac{5-1}{5-1} = \frac{4}{4} = 1). The second diagonal connects (1,5) and (5,1). Its slope is (\frac{1-5}{5-1} = \frac{-4}{4} = -1). The sum of the slopes is (1 + (-1) = 0). This is a property of any rhombus, including a square, whose diagonals are not horizontal and vertical.

Question 4

1. (-0.5, 7.5) (correct answer)
2. (5.5, -2.5)
3. (-1.5, 6.5)
4. (2.5, 2.5)

Explanation: First, find the center of the original square. The center is the midpoint of a diagonal. Using the diagonal from (1,1) to (4,4), the midpoint is ((\frac{1+4}{2}, \frac{1+4}{2}) = (2.5, 2.5)). The translation moves every point ((x, y)) to ((x-3, y+5)). Applying this translation to the center's coordinates: ((2.5 - 3, 2.5 + 5) = (-0.5, 7.5)).

Question 5

1. Equilateral and right
2. Isosceles and right
3. Scalene and right
4. Isosceles and not right (correct answer)

Explanation: When you encounter a triangle classification problem with given vertices, you need to determine two things: the side lengths (to classify by sides) and whether any angles are 90° (to determine if it's a right triangle). First, calculate the distances between each pair of vertices using the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$:

• Side AB: $\sqrt{(5-(-1))^2 + (-1-(-1))^2} = \sqrt{36 + 0} = 6$
• Side BC: $\sqrt{(2-5)^2 + (3-(-1))^2} = \sqrt{9 + 16} = 5$
• Side AC: $\sqrt{(2-(-1))^2 + (3-(-1))^2} = \sqrt{9 + 16} = 5$

Question 6

1. a = 8, b = 3
2. a = 12, b = 2
3. a = 6, b = 4 (correct answer)
4. a = 5, b = 5

Explanation: The side lengths of the rectangle are 'a' and 'b'. The area is (ab = 24). The perimeter is (2(a+b) = 20), which simplifies to (a+b = 10). We need to find two numbers that multiply to 24 and add up to 10. Let's test the options. A) 83=24, 8+3=11 (Incorrect). B) 122=24, 12+2=14 (Incorrect). C) 64=24, 6+4=10 (Correct). D) 55=25 (Incorrect).

Question 7

1. (-2, 6)
2. (-1, 3) (correct answer)
3. (-4, 5)
4. (8, 10)

Explanation: The center of the circle is the midpoint of its diameter. The midpoint formula is ((\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})). Using the endpoints (-5, 8) and (3, -2), the center is ((\frac{-5+3}{2}, \frac{8+(-2)}{2}) = (\frac{-2}{2}, \frac{6}{2}) = (-1, 3)).

Question 8

1. 22 square units
2. 48 square units
3. 32 square units
4. 24 square units (correct answer)

Explanation: When you encounter a trapezoid with given coordinates, you need to identify the parallel sides and calculate the area using the formula: Area = $\frac{1}{2}(b_1 + b_2) \times h$, where $b_1$ and $b_2$ are the lengths of the parallel bases and $h$ is the height. First, plot or visualize the points: A(1, 2), B(9, 2), C(7, 6), and D(3, 6). Notice that points A and B both have y-coordinate 2, making AB a horizontal line. Similarly, points C and D both have y-coordinate 6, making CD horizontal. These are your parallel bases. Calculate the base lengths:

• Base AB: distance from (1, 2) to (9, 2) = $9 - 1 = 8$ units
• Base CD: distance from (3, 6) to (7, 6) = $7 - 3 = 4$ units

Question 9

1. (-2, 7) (correct answer)
2. (2, -7)
3. (-2, -7)
4. (7, 2)

Explanation: A reflection across the y-axis transforms a point ((x, y)) to ((-x, y)). The x-coordinate changes sign, and the y-coordinate remains the same. Applying this rule to vertex R(2, 7), the new coordinates will be (-2, 7).

Question 10

1. They form a right triangle.
2. They form an isosceles triangle.
3. They are collinear. (correct answer)
4. They form an equilateral triangle.

Explanation: To determine the relationship between the points, we can check the slopes between them. The slope between A and B is (\frac{8-2}{3-1} = \frac{6}{2} = 3). The slope between B and C is (\frac{14-8}{5-3} = \frac{6}{2} = 3). Since the slope between A and B is the same as the slope between B and C, the three points lie on the same straight line. Therefore, they are collinear and do not form a triangle.

Question 11

1. Parallelogram
2. Rectangle
3. Rhombus
4. Square (correct answer)

Explanation: Let's analyze the properties. Diagonals that bisect each other define a parallelogram. If the diagonals are also congruent, the parallelogram is a rectangle. If the diagonals are also perpendicular, the parallelogram is a rhombus. A shape that is both a rectangle and a rhombus must be a square. Therefore, a quadrilateral with diagonals that are congruent, perpendicular, and bisect each other is a square.

Question 12

1. (2, 1)
2. (3, 1) (correct answer)
3. (4, 1)
4. (13, 1)

Explanation: In a parallelogram ABCD, the vector from A to B is equal to the vector from D to C. The vector AB is ((6-1), (5-5)) = ((5, 0)). Let the coordinates of D be ((x, y)). The vector DC is ((8-x), (1-y)). Setting the vectors equal: (8-x = 5) and (1-y = 0). This gives (x = 3) and (y = 1). So, vertex D is at (3, 1). Alternatively, the midpoint of the diagonal AC must be the same as the midpoint of the diagonal BD. Midpoint of AC is ((\frac{1+8}{2}, \frac{5+1}{2}) = (4.5, 3)). Midpoint of BD is ((\frac{6+x}{2}, \frac{5+y}{2})). Setting them equal: (\frac{6+x}{2} = 4.5 \Rightarrow 6+x=9 \Rightarrow x=3), and (\frac{5+y}{2} = 3 \Rightarrow 5+y=6 \Rightarrow y=1).

Question 13

1. 3
2. 5
3. 8 (correct answer)
4. 9

Explanation: Equidistant means the distance from M to P is equal to the distance from M to Q. We can compare the squares of the distances to avoid square roots. Distance MP squared is ((5-2)^2 + (1-5)^2 = 3^2 + (-4)^2 = 9 + 16 = 25). Distance MQ squared is ((5-x)^2 + (1-5)^2 = (5-x)^2 + (-4)^2 = (5-x)^2 + 16). Set the squared distances equal: (25 = (5-x)^2 + 16). Subtract 16 from both sides: (9 = (5-x)^2). Take the square root of both sides: (\pm 3 = 5-x). This gives two possibilities: (3 = 5-x) or (-3 = 5-x). The first equation gives (x=2), which is excluded by the problem. The second equation gives (x=8).

Question 14

1. 16 square units (correct answer)
2. 20 square units
3. 32 square units
4. 36 square units

Explanation: First, find the square of the length of the diagonal connecting the opposite vertices. Using the distance formula, (d^2 = (4-0)^2 + (1-5)^2 = 4^2 + (-4)^2 = 16 + 16 = 32). The area of a square can be calculated as (\frac{d^2}{2}), where (d) is the length of the diagonal. So, the area is (\frac{32}{2} = 16) square units.

Question 15

1. 14 units
2. 24 units (correct answer)
3. 28 units
4. 48 units

Explanation: The triangle has a horizontal side of length 8 (from (0,0) to (8,0)) and a vertical side of length 6 (from (0,0) to (0,6)). These are the legs of the right triangle. The third side is the hypotenuse. Its length can be found using the Pythagorean theorem: (c = \sqrt{a^2+b^2} = \sqrt{8^2+6^2} = \sqrt{64+36} = \sqrt{100} = 10). The perimeter is the sum of the side lengths: (8 + 6 + 10 = 24) units.

Question 16

1. 10 square units
2. 12 square units (correct answer)
3. 16 square units
4. 24 square units

Explanation: The base of the triangle can be the horizontal segment connecting (2, 2) and (8, 2). The length of this base is (8 - 2 = 6) units. The height of the triangle is the perpendicular distance from the third vertex (4, 6) to the line containing the base (the line y=2). The height is (6 - 2 = 4) units. The area of a triangle is (\frac{1}{2} \times \text{base} \times \text{height}). So, the area is (\frac{1}{2} \times 6 \times 4 = 12) square units.