SWBAT apply & identify the equilateral and isosceles triangle ... - Free Printable
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Step-by-step solution for: SWBAT apply & identify the equilateral and isosceles triangle ...
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Show Answer Key & Explanations
Step-by-step solution for: SWBAT apply & identify the equilateral and isosceles triangle ...
To solve the problems in the image, we need to use geometric properties such as angle relationships, triangle angle sum properties, and parallel line theorems. Let's go through each problem step by step.
---
The diagram shows a triangle with angles labeled as \(6x + 3\), \(8x - 21\), and \(7x - 10\). The sum of the angles in a triangle is always \(180^\circ\).
\[
(6x + 3) + (8x - 21) + (7x - 10) = 180
\]
Combine like terms:
\[
6x + 8x + 7x + 3 - 21 - 10 = 180
\]
\[
21x - 28 = 180
\]
Add 28 to both sides:
\[
21x = 208
\]
Divide by 21:
\[
x = \frac{208}{21}
\]
So, the value of \(x\) is:
\[
\boxed{\frac{208}{21}}
\]
---
The diagram shows a triangle with angles labeled as \(2x + 10\), \(3x - 15\), and \(x - 7\). Again, the sum of the angles in a triangle is \(180^\circ\).
\[
(2x + 10) + (3x - 15) + (x - 7) = 180
\]
Combine like terms:
\[
2x + 3x + x + 10 - 15 - 7 = 180
\]
\[
6x - 12 = 180
\]
Add 12 to both sides:
\[
6x = 192
\]
Divide by 6:
\[
x = 32
\]
So, the value of \(x\) is:
\[
\boxed{32}
\]
---
The diagram shows a right triangle with one angle labeled \(x\) and another angle labeled \(y\). One of the angles is given as \(50^\circ\). In a right triangle, the sum of the two non-right angles is \(90^\circ\).
Since one angle is \(50^\circ\), the other non-right angle \(x\) must be:
\[
x = 90^\circ - 50^\circ = 40^\circ
\]
The right angle is \(90^\circ\), so \(y = 90^\circ\).
Thus, the values are:
\[
x = 40, \quad y = 90
\]
---
The diagram shows a triangle with angles labeled \(2x\), \(5x - 30\), and an unknown angle. The sum of the angles in a triangle is \(180^\circ\).
Let the third angle be \(z\). Then:
\[
2x + (5x - 30) + z = 180
\]
Combine like terms:
\[
7x - 30 + z = 180
\]
Solve for \(z\):
\[
z = 210 - 7x
\]
Without additional information, we cannot determine \(x\) or \(z\) uniquely. However, if we assume the triangle is isosceles or equilateral, we can solve further. For now, the relationship is:
\[
z = 210 - 7x
\]
---
The diagram shows two parallel lines cut by a transversal. One angle is given as \(45^\circ\), and the angles \(x\) and \(y\) are alternate interior angles. Alternate interior angles are equal when the lines are parallel.
Thus:
\[
x = 45^\circ
\]
The angle \(y\) is a corresponding angle to \(x\), so:
\[
y = 45^\circ
\]
So, the values are:
\[
x = 45, \quad y = 45
\]
---
The diagram shows a right triangle with one angle labeled \(3x + 1\) and another angle labeled \(4x - 20\). The sum of the two non-right angles in a right triangle is \(90^\circ\).
\[
(3x + 1) + (4x - 20) = 90
\]
Combine like terms:
\[
7x - 19 = 90
\]
Add 19 to both sides:
\[
7x = 109
\]
Divide by 7:
\[
x = \frac{109}{7}
\]
So, the value of \(x\) is:
\[
\boxed{\frac{109}{7}}
\]
---
The diagram shows a triangle with angles labeled \(x\), \(y\), and \(28^\circ\). The sum of the angles in a triangle is \(180^\circ\).
\[
x + y + 28 = 180
\]
Solve for \(x + y\):
\[
x + y = 152
\]
Without additional information, we cannot determine \(x\) and \(y\) uniquely. However, the relationship is:
\[
x + y = 152
\]
---
The diagram shows a triangle with angles labeled \(3x - 6\), \(x + 10\), and an unknown angle. The sum of the angles in a triangle is \(180^\circ\).
Let the third angle be \(z\). Then:
\[
(3x - 6) + (x + 10) + z = 180
\]
Combine like terms:
\[
4x + 4 + z = 180
\]
Solve for \(z\):
\[
z = 176 - 4x
\]
Without additional information, we cannot determine \(x\) or \(z\) uniquely. However, the relationship is:
\[
z = 176 - 4x
\]
---
The diagram shows a triangle with angles labeled \(54^\circ\), \(x\), and another angle. The sum of the angles in a triangle is \(180^\circ\).
Let the third angle be \(y\). Then:
\[
54 + x + y = 180
\]
Solve for \(x + y\):
\[
x + y = 126
\]
Without additional information, we cannot determine \(x\) and \(y\) uniquely. However, the relationship is:
\[
x + y = 126
\]
---
The diagram shows a triangle with angles labeled \(5x\), \(2x\), and another angle. The sum of the angles in a triangle is \(180^\circ\).
Let the third angle be \(z\). Then:
\[
5x + 2x + z = 180
\]
Combine like terms:
\[
7x + z = 180
\]
Solve for \(z\):
\[
z = 180 - 7x
\]
Without additional information, we cannot determine \(x\) or \(z\) uniquely. However, the relationship is:
\[
z = 180 - 7x
\]
---
The diagram shows a triangle with angles labeled \(3x + 10\), \(5x - 10\), and a right angle (\(90^\circ\)). The sum of the angles in a triangle is \(180^\circ\).
\[
(3x + 10) + (5x - 10) + 90 = 180
\]
Combine like terms:
\[
8x + 90 = 180
\]
Subtract 90 from both sides:
\[
8x = 90
\]
Divide by 8:
\[
x = \frac{90}{8} = \frac{45}{4}
\]
So, the value of \(x\) is:
\[
\boxed{\frac{45}{4}}
\]
---
The diagram shows a right triangle with one angle labeled \(x\) and a right angle (\(90^\circ\)). The sum of the two non-right angles in a right triangle is \(90^\circ\).
Thus:
\[
x + 90 = 180
\]
Solve for \(x\):
\[
x = 90
\]
So, the value of \(x\) is:
\[
\boxed{90}
\]
---
The diagram shows a triangle with angles labeled \(4x + 2\), \(6x - 30\), and \(60^\circ\). The sum of the angles in a triangle is \(180^\circ\).
\[
(4x + 2) + (6x - 30) + 60 = 180
\]
Combine like terms:
\[
10x + 32 = 180
\]
Subtract 32 from both sides:
\[
10x = 148
\]
Divide by 10:
\[
x = \frac{148}{10} = 14.8
\]
So, the value of \(x\) is:
\[
\boxed{14.8}
\]
---
The diagram shows a quadrilateral with one angle labeled \(100^\circ\) and opposite angles labeled \(x\). The sum of the interior angles of a quadrilateral is \(360^\circ\).
\[
100 + x + x + \text{(other two angles)} = 360
\]
Since the quadrilateral is not specified, we assume it is a parallelogram (opposite angles are equal). Thus:
\[
100 + x + x + 100 + x + x = 360
\]
Combine like terms:
\[
4x + 200 = 360
\]
Subtract 200 from both sides:
\[
4x = 160
\]
Divide by 4:
\[
x = 40
\]
So, the value of \(x\) is:
\[
\boxed{40}
\]
---
The diagram shows a triangle with angles labeled \(50^\circ\), \(x\), and another angle. The sum of the angles in a triangle is \(180^\circ\).
Let the third angle be \(y\). Then:
\[
50 + x + y = 180
\]
Solve for \(x + y\):
\[
x + y = 130
\]
Without additional information, we cannot determine \(x\) and \(y\) uniquely. However, the relationship is:
\[
x + y = 130
\]
---
The diagram shows a triangle with angles labeled \(5x\), \(3x + 20\), and another angle. The sum of the angles in a triangle is \(180^\circ\).
Let the third angle be \(z\). Then:
\[
5x + (3x + 20) + z = 180
\]
Combine like terms:
\[
8x + 20 + z = 180
\]
Solve for \(z\):
\[
z = 160 - 8x
\]
Without additional information, we cannot determine \(x\) or \(z\) uniquely. However, the relationship is:
\[
z = 160 - 8x
\]
---
\[
\boxed{\frac{208}{21}, 32, 40, 90, 210 - 7x, 45, 45, \frac{109}{7}, 152, 176 - 4x, 126, 180 - 7x, \frac{45}{4}, 90, 14.8, 40, 130, 160 - 8x}
\]
---
Problem 13:
The diagram shows a triangle with angles labeled as \(6x + 3\), \(8x - 21\), and \(7x - 10\). The sum of the angles in a triangle is always \(180^\circ\).
\[
(6x + 3) + (8x - 21) + (7x - 10) = 180
\]
Combine like terms:
\[
6x + 8x + 7x + 3 - 21 - 10 = 180
\]
\[
21x - 28 = 180
\]
Add 28 to both sides:
\[
21x = 208
\]
Divide by 21:
\[
x = \frac{208}{21}
\]
So, the value of \(x\) is:
\[
\boxed{\frac{208}{21}}
\]
---
Problem 14:
The diagram shows a triangle with angles labeled as \(2x + 10\), \(3x - 15\), and \(x - 7\). Again, the sum of the angles in a triangle is \(180^\circ\).
\[
(2x + 10) + (3x - 15) + (x - 7) = 180
\]
Combine like terms:
\[
2x + 3x + x + 10 - 15 - 7 = 180
\]
\[
6x - 12 = 180
\]
Add 12 to both sides:
\[
6x = 192
\]
Divide by 6:
\[
x = 32
\]
So, the value of \(x\) is:
\[
\boxed{32}
\]
---
Problem 15:
The diagram shows a right triangle with one angle labeled \(x\) and another angle labeled \(y\). One of the angles is given as \(50^\circ\). In a right triangle, the sum of the two non-right angles is \(90^\circ\).
Since one angle is \(50^\circ\), the other non-right angle \(x\) must be:
\[
x = 90^\circ - 50^\circ = 40^\circ
\]
The right angle is \(90^\circ\), so \(y = 90^\circ\).
Thus, the values are:
\[
x = 40, \quad y = 90
\]
---
Problem 16:
The diagram shows a triangle with angles labeled \(2x\), \(5x - 30\), and an unknown angle. The sum of the angles in a triangle is \(180^\circ\).
Let the third angle be \(z\). Then:
\[
2x + (5x - 30) + z = 180
\]
Combine like terms:
\[
7x - 30 + z = 180
\]
Solve for \(z\):
\[
z = 210 - 7x
\]
Without additional information, we cannot determine \(x\) or \(z\) uniquely. However, if we assume the triangle is isosceles or equilateral, we can solve further. For now, the relationship is:
\[
z = 210 - 7x
\]
---
Problem 17:
The diagram shows two parallel lines cut by a transversal. One angle is given as \(45^\circ\), and the angles \(x\) and \(y\) are alternate interior angles. Alternate interior angles are equal when the lines are parallel.
Thus:
\[
x = 45^\circ
\]
The angle \(y\) is a corresponding angle to \(x\), so:
\[
y = 45^\circ
\]
So, the values are:
\[
x = 45, \quad y = 45
\]
---
Problem 18:
The diagram shows a right triangle with one angle labeled \(3x + 1\) and another angle labeled \(4x - 20\). The sum of the two non-right angles in a right triangle is \(90^\circ\).
\[
(3x + 1) + (4x - 20) = 90
\]
Combine like terms:
\[
7x - 19 = 90
\]
Add 19 to both sides:
\[
7x = 109
\]
Divide by 7:
\[
x = \frac{109}{7}
\]
So, the value of \(x\) is:
\[
\boxed{\frac{109}{7}}
\]
---
Problem 19:
The diagram shows a triangle with angles labeled \(x\), \(y\), and \(28^\circ\). The sum of the angles in a triangle is \(180^\circ\).
\[
x + y + 28 = 180
\]
Solve for \(x + y\):
\[
x + y = 152
\]
Without additional information, we cannot determine \(x\) and \(y\) uniquely. However, the relationship is:
\[
x + y = 152
\]
---
Problem 20:
The diagram shows a triangle with angles labeled \(3x - 6\), \(x + 10\), and an unknown angle. The sum of the angles in a triangle is \(180^\circ\).
Let the third angle be \(z\). Then:
\[
(3x - 6) + (x + 10) + z = 180
\]
Combine like terms:
\[
4x + 4 + z = 180
\]
Solve for \(z\):
\[
z = 176 - 4x
\]
Without additional information, we cannot determine \(x\) or \(z\) uniquely. However, the relationship is:
\[
z = 176 - 4x
\]
---
Problem 21:
The diagram shows a triangle with angles labeled \(54^\circ\), \(x\), and another angle. The sum of the angles in a triangle is \(180^\circ\).
Let the third angle be \(y\). Then:
\[
54 + x + y = 180
\]
Solve for \(x + y\):
\[
x + y = 126
\]
Without additional information, we cannot determine \(x\) and \(y\) uniquely. However, the relationship is:
\[
x + y = 126
\]
---
Problem 22:
The diagram shows a triangle with angles labeled \(5x\), \(2x\), and another angle. The sum of the angles in a triangle is \(180^\circ\).
Let the third angle be \(z\). Then:
\[
5x + 2x + z = 180
\]
Combine like terms:
\[
7x + z = 180
\]
Solve for \(z\):
\[
z = 180 - 7x
\]
Without additional information, we cannot determine \(x\) or \(z\) uniquely. However, the relationship is:
\[
z = 180 - 7x
\]
---
Problem 23:
The diagram shows a triangle with angles labeled \(3x + 10\), \(5x - 10\), and a right angle (\(90^\circ\)). The sum of the angles in a triangle is \(180^\circ\).
\[
(3x + 10) + (5x - 10) + 90 = 180
\]
Combine like terms:
\[
8x + 90 = 180
\]
Subtract 90 from both sides:
\[
8x = 90
\]
Divide by 8:
\[
x = \frac{90}{8} = \frac{45}{4}
\]
So, the value of \(x\) is:
\[
\boxed{\frac{45}{4}}
\]
---
Problem 24:
The diagram shows a right triangle with one angle labeled \(x\) and a right angle (\(90^\circ\)). The sum of the two non-right angles in a right triangle is \(90^\circ\).
Thus:
\[
x + 90 = 180
\]
Solve for \(x\):
\[
x = 90
\]
So, the value of \(x\) is:
\[
\boxed{90}
\]
---
Problem 25:
The diagram shows a triangle with angles labeled \(4x + 2\), \(6x - 30\), and \(60^\circ\). The sum of the angles in a triangle is \(180^\circ\).
\[
(4x + 2) + (6x - 30) + 60 = 180
\]
Combine like terms:
\[
10x + 32 = 180
\]
Subtract 32 from both sides:
\[
10x = 148
\]
Divide by 10:
\[
x = \frac{148}{10} = 14.8
\]
So, the value of \(x\) is:
\[
\boxed{14.8}
\]
---
Problem 26:
The diagram shows a quadrilateral with one angle labeled \(100^\circ\) and opposite angles labeled \(x\). The sum of the interior angles of a quadrilateral is \(360^\circ\).
\[
100 + x + x + \text{(other two angles)} = 360
\]
Since the quadrilateral is not specified, we assume it is a parallelogram (opposite angles are equal). Thus:
\[
100 + x + x + 100 + x + x = 360
\]
Combine like terms:
\[
4x + 200 = 360
\]
Subtract 200 from both sides:
\[
4x = 160
\]
Divide by 4:
\[
x = 40
\]
So, the value of \(x\) is:
\[
\boxed{40}
\]
---
Problem 27:
The diagram shows a triangle with angles labeled \(50^\circ\), \(x\), and another angle. The sum of the angles in a triangle is \(180^\circ\).
Let the third angle be \(y\). Then:
\[
50 + x + y = 180
\]
Solve for \(x + y\):
\[
x + y = 130
\]
Without additional information, we cannot determine \(x\) and \(y\) uniquely. However, the relationship is:
\[
x + y = 130
\]
---
Problem 28:
The diagram shows a triangle with angles labeled \(5x\), \(3x + 20\), and another angle. The sum of the angles in a triangle is \(180^\circ\).
Let the third angle be \(z\). Then:
\[
5x + (3x + 20) + z = 180
\]
Combine like terms:
\[
8x + 20 + z = 180
\]
Solve for \(z\):
\[
z = 160 - 8x
\]
Without additional information, we cannot determine \(x\) or \(z\) uniquely. However, the relationship is:
\[
z = 160 - 8x
\]
---
Final Answers:
\[
\boxed{\frac{208}{21}, 32, 40, 90, 210 - 7x, 45, 45, \frac{109}{7}, 152, 176 - 4x, 126, 180 - 7x, \frac{45}{4}, 90, 14.8, 40, 130, 160 - 8x}
\]
Parent Tip: Review the logic above to help your child master the concept of isosceles triangle worksheet.
