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11. The \( x \)-axis is a line of symmetry for \( \triangle R S T \). The coordinates of \( R \) and \( S \) are \( (6,0) \) and \( (-2,-2) \), respectively. (a) What are the coordinates of \( T \) ? (b) What type of triangle is \( \triangle R S T \) classified by sides? (c) If the measure of \( \angle R \) is approximately \( 28^{\circ} \), then what is the approximate measure of \( \angle S \) ?

Ask by Garrett Lawson. in the United States
Jan 31,2025

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Answer

(a) \( T = (-2,\ 2) \) (b) Isosceles triangle (c) \( \angle S \approx 76^\circ \)

Solution

Let's solve each part step by step. ### (a) Coordinates of \( T \) Given that the \( x \)-axis is a line of symmetry for \( \triangle RST \), the reflection of any point across the \( x \)-axis will also be a point of the triangle. - Point \( R \) is \( (6, 0) \), which lies on the \( x \)-axis. Therefore, its reflection is itself. - Point \( S \) is \( (-2, -2) \). Reflecting \( S \) across the \( x \)-axis changes the sign of the \( y \)-coordinate, resulting in \( T = (-2, 2) \). **Answer:** \( T \) has coordinates \( (-2,\ 2) \). ### (b) Type of Triangle by Sides Let's calculate the lengths of the sides: 1. **Distance \( RS \):** \[ RS = \sqrt{(6 - (-2))^2 + (0 - (-2))^2} = \sqrt{8^2 + 2^2} = \sqrt{64 + 4} = \sqrt{68} = 2\sqrt{17} \] 2. **Distance \( RT \):** \[ RT = \sqrt{(6 - (-2))^2 + (0 - 2)^2} = \sqrt{8^2 + (-2)^2} = \sqrt{64 + 4} = \sqrt{68} = 2\sqrt{17} \] 3. **Distance \( ST \):** \[ ST = \sqrt{(-2 - (-2))^2 + (-2 - 2)^2} = \sqrt{0^2 + (-4)^2} = \sqrt{16} = 4 \] Since \( RS = RT \) and both are longer than \( ST \), the triangle has **two equal sides**. **Answer:** \( \triangle RST \) is an **isosceles** triangle. ### (c) Measure of \( \angle S \) In an isosceles triangle, the angles opposite the equal sides are also equal. Since \( RS = RT \), angles \( S \) and \( T \) are equal. Given: \[ \angle R = 28^\circ \] Sum of angles in a triangle: \[ \angle R + \angle S + \angle T = 180^\circ \] \[ 28^\circ + 2\angle S = 180^\circ \] \[ 2\angle S = 152^\circ \] \[ \angle S = 76^\circ \] **Answer:** (a) \( T = (-2,\ 2) \); (b) An isosceles triangle; (c) Angle \( S \) is approximately 76°.

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To find the coordinates of point \( T \) in \( \triangle RST \) where the \( x \)-axis is a line of symmetry, we know that \( T \) must have the same \( x \)-coordinate as its corresponding point reflected over the \( x \)-axis. Given that the \( y \)-coordinate of \( S \) is \(-2\), the reflected point \( T \) will have coordinates \((x_T, 2)\). To keep the triangle symmetric, point \( T \) will have the same \( x \)-coordinate as \( R \) (since it's above the \( x \)-axis) but also has to maintain the triangle's shape. With \( R = (6, 0) \), the best guess is \( T = (6, 2) \). For part (b), we can compare the lengths of the sides. The distance between \( R(6, 0) \) and \( S(-2, -2) \) can be calculated using the distance formula and will clearly show that \( RS \) is longer than both \( RT \) and \( ST \) which could lead to \( \triangle RST \) being classified as an isosceles triangle if \( RT \) and \( ST \) are equal. Now for part (c), if \( \angle R \) measures approximately \( 28^{\circ} \), we can use the fact that the angles in a triangle sum up to \( 180^{\circ} \). This means that \( \angle S \) would be \( 180^{\circ} - (28^{\circ} + 90^{\circ}) \) if we consider \( \triangle RST \) having a right angle. Calculating that gives \( \angle S \) a measure of approximately \( 62^{\circ} \). Fun Fact: Did you know that isosceles triangles have both angles opposite the equal sides that are also equal? This makes them super symmetrical and pretty cool to look at! Keep in mind that visualization helps. If you sketch out the triangle and mark the points, it’s like giving life to your geometric creation, helping you catch mistakes before they snowball into bigger problem areas!

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