Question

polygon s is a scaled copy of polygon r.
polygon r has side lengths ( 2\frac{3}{4} ), ( 7\frac{1}{3} ), ( 1\frac{5}{6} ) and angles ( 90^circ ), ( 100^circ ), ( 130^circ ). polygon s has side lengths ( 1\frac{3}{4} ), ( 4\frac{2}{3} ), ( 1\frac{1}{6} ) and angles ( 90^circ ), ( t^circ ).
what is the value of ( t )?

Explanation:

Step1: Recall properties of scaled copies

In a scaled copy, corresponding angles are equal, and corresponding sides are in proportion. So we need to find the angle in polygon R that corresponds to angle \( t^\circ \) in polygon S.

Step2: Identify the polygon type (pentagon)

Both polygons are pentagons. The sum of interior angles of a pentagon is \( (5 - 2)\times180^\circ= 540^\circ \). But since it's a scaled copy, corresponding angles are equal. Let's check the right angles (90°) are corresponding, and we can find the other angles.

First, let's confirm the sides are scaled. Let's check the ratio of corresponding sides. For example, the bottom side: Polygon R has \( 1\frac{5}{6}=\frac{11}{6} \), Polygon S has \( 1\frac{1}{6}=\frac{7}{6} \)? Wait, no, wait the other sides. Wait the top side: Polygon R: \( 2\frac{3}{4}=\frac{11}{4} \), Polygon S: \( 1\frac{3}{4}=\frac{7}{4} \)? Wait no, maybe I picked the wrong sides. Wait the vertical side: Polygon R: \( 7\frac{1}{3}=\frac{22}{3} \), Polygon S: \( 4\frac{2}{3}=\frac{14}{3} \). Let's check the ratio \( \frac{14/3}{22/3}=\frac{14}{22}=\frac{7}{11} \). Wait the top side: \( \frac{7/4}{11/4}=\frac{7}{11} \). The bottom side: Polygon R: \( 1\frac{5}{6}=\frac{11}{6} \), Polygon S: \( 1\frac{1}{6}=\frac{7}{6} \), ratio \( \frac{7/6}{11/6}=\frac{7}{11} \). So the scale factor is \( \frac{7}{11} \), but angles remain the same in scaled copies.

Now, let's list the angles of polygon R. We have a right angle (90°), 100°, 130°, and we need to find the other angles? Wait no, in a pentagon, sum is 540°. Let's calculate the missing angle in R first, then since S is a scaled copy, the angle \( t \) corresponds to the angle in R that's equal (because scaled copies have congruent corresponding angles).

Wait, let's list the angles of polygon R: 90° (right angle), 100°, 130°, and two other angles. Wait, but in the scaled copy S, we have a right angle (90°), \( t^\circ \), and the other angles. Wait, maybe the angle \( t \) corresponds to the angle in R that's 100°? No, wait no, let's check the structure. Wait the right angle is corresponding, the angle adjacent to the right angle: in R, it's 100°, in S, it's \( t \). Wait no, maybe I made a mistake. Wait, scaled copies have the same shape, so corresponding angles are equal. So the angle labeled \( t \) in S corresponds to the angle labeled 100° in R? No, wait no, let's check the sides. Wait the side with length \( 2\frac{3}{4} \) in R corresponds to \( 1\frac{3}{4} \) in S, the side with \( 7\frac{1}{3} \) in R corresponds to \( 4\frac{2}{3} \) in S, the bottom side \( 1\frac{5}{6} \) in R corresponds to \( 1\frac{1}{6} \) in S? Wait no, \( 1\frac{5}{6}=\frac{11}{6} \), \( 1\frac{1}{6}=\frac{7}{6} \), that's not the same ratio. Wait, maybe the bottom side is not corresponding. Wait, maybe the side with \( 1\frac{5}{6} \) in R and \( 1\frac{1}{6} \) in S: no, that's a different ratio. Wait, maybe I mixed up the sides. Wait, let's check the vertical side: \( 7\frac{1}{3}=\frac{22}{3} \), \( 4\frac{2}{3}=\frac{14}{3} \), ratio \( 14/22 = 7/11 \). Top side: \( 2\frac{3}{4}=\frac{11}{4} \), \( 1\frac{3}{4}=\frac{7}{4} \), ratio \( 7/11 \). Ah, so the top side and vertical side have ratio 7/11. The bottom side: \( 1\frac{5}{6}=\frac{11}{6} \), \( 1\frac{1}{6}=\frac{7}{6} \), ratio 7/11. So that's correct. So the scale factor is 7/11. Now, angles: in scaled copies, angles are congruent. So the angle \( t \) in S corresponds to the angle in R that's equal. Let's find the sum of angles in R.

Sum of interior angles of a pentagon: \( (5-2)\times180 = 540^\circ \).

Angles in R: 90° (right angle), 100°, 130°, and two other angles. Wait, but in S, we have 90° (right angle), \( t \), and the other angles. Wait, maybe the angle \( t \) is not 100°, wait no, let's check the number of angles. Wait, the polygon has 5 sides, so 5 angles. Let's list the angles:

In R: right angle (90°), 100°, 130°, and two more angles. Let's calculate the sum of known angles: 90 + 100 + 130 = 320°. So the sum of the other two angles is 540 - 320 = 220°.

In S: right angle (90°), \( t \), and the other angles. Since it's a scaled copy, the angles should be the same as R. Wait, but maybe the angle \( t \) is equal to 100°? No, wait no, maybe I made a mistake. Wait, no, scaled copies have congruent corresponding angles, so the angle that is \( t \) in S corresponds to the angle in R that is 100°? Wait, no, let's look at the diagram. The right angle is in the same position, then the angle next to it: in R, it's 100°, in S, it's \( t \). Wait, but maybe the angle \( t \) is not 100°, wait no, let's check the sides. Wait, the side with length \( 2\frac{3}{4} \) (top) in R is adjacent to the right angle and 100° angle. In S, the top side is \( 1\frac{3}{4} \), adjacent to right angle and \( t \) angle. So those are corresponding angles, so \( t = 100 \)? Wait, no, that can't be. Wait, maybe I miscalculated. Wait, no, scaled copies have the same angle measures. So corresponding angles are equal. So the angle labeled \( t \) in S corresponds to the angle labeled 100° in R? Wait, no, maybe the angle is 100°, but let's check the sum. Wait, maybe the polygon is a pentagon, but let's count the angles. Wait, the diagram shows a pentagon with a right angle, 100°, 130°, and two other angles. Wait, in S, the right angle is there, \( t \), and the other angles. Wait, maybe the angle \( t \) is 100°, but let's check the sum. Wait, no, let's calculate the sum of angles in R.

Wait, 90 (right angle) + 100 + 130 + x + y = 540. In S, 90 + t + z + a + b = 540. Since it's a scaled copy, x = a, y = b, 100 = t, 130 = z? Wait, no, maybe the angle \( t \) is 100°, but let's check the sides. Wait, the side with length \( 1\frac{5}{6} \) in R and \( 1\frac{1}{6} \) in S: no, that's a different ratio. Wait, no, \( 1\frac{5}{6}=\frac{11}{6} \), \( 1\frac{1}{6}=\frac{7}{6} \), ratio 7/11. The vertical side: \( 7\frac{1}{3}=\frac{22}{3} \), \( 4\frac{2}{3}=\frac{14}{3} \), ratio 7/11. Top side: \( 2\frac{3}{4}=\frac{11}{4} \), \( 1\frac{3}{4}=\frac{7}{4} \), ratio 7/11. So that's correct. So the scale factor is 7/11. Now, angles: in similar figures (scaled copies), corresponding angles are equal. So the angle \( t \) in S corresponds to the angle in R that is 100°? Wait, no, maybe the angle is 100°, but let's check the sum. Wait, maybe I made a mistake. Wait, the problem is that in a scaled copy, angles are congruent, so \( t \) should be equal to the corresponding angle in R. Let's check the angles:

In R, we have a right angle (90°), 100°, 130°, and two other angles. Let's calculate the other two angles. Wait, maybe the polygon is a pentagon, but let's count the number of angles. Wait, the diagram shows a pentagon with five angles: right angle, 100°, 130°, and two more. Wait, in S, the right angle is there, \( t \), and the other angles. So the angle \( t \) corresponds to the 100° angle in R? Wait, no, maybe the angle is 100°, but let's check the sum. Wait, 90 + 100 + 130 + x + y = 540. So x + y = 220. In S, 90 + t + z + a + b = 540. Since it's a scaled copy, x = a, y = b, 100 = t, 130 = z. So t = 100? Wait, but that seems too easy. Wait, maybe I'm missing something. Wait, no, scaled copies have the same angle measures. So corresponding angles are equal. So the angle labeled \( t \) in S is equal to the angle labeled 100° in R. So t = 100.

Wait, but let's confirm with the sides. The side adjacent to 100° in R is \( 2\frac{3}{4} \), and in S, it's \( 1\frac{3}{4} \), which is the scaled side (ratio 7/11). So the angle between the scaled sides should be equal. So yes, t = 100.

Answer:

\( \boxed{100} \)