Pregunta
Given points \( A(-2,0), B(-5,3, C(-9,-1), P(7,6), Q(4,0) \), and \( R(-4,4) \), which of the following proves that \( \triangle A B C^{\sim} \triangle P Q R \) ? By the Distance Formula, \( \mathrm{AB}=50, \mathrm{BC}=32 \), and \( \mathrm{CA}=18 \). \( \mathrm{Also}, \mathrm{PQ}=125, \mathrm{QR}=90 \), and \( \mathrm{RP}=45 \). Therefore, ABPQ \( =B C Q R=C A R P=25 \), and therefore, \( \triangle A B C \sim \triangle P Q R \) by \( S S S \sim \). \( \operatorname{By} \) the Distance Formula, \( A B=50, B C=32 \), and \( C A=18 . A 1 s 0, P Q=125, Q R=90 \), and \( R P=45 \). Therefore, \( A B P Q=B C Q R=C A R P=25 \), and therefore, \( \triangle A B C \sim \triangle P \) QR by SSS ~. By the Distance Formula, \( A B=32-v, B C=42-v \), and \( C A=52-v . A l s o, P Q=35-V Q R=45-v \), and \( R P=55- \) V.Therefore, \( A B P Q=B C Q R=C A R P=2 V 5 V=10 v 5 \), and therefore, \( \triangle A B C \sim \triangle P Q R \) by \( S S S \sim \). \( \operatorname{By} \) the Distance For mula, \( \mathrm{AB}=32, \mathrm{BC}=42 \), and \( \mathrm{CA}=52 \). Also, \( \mathrm{PQ}=35 \mathrm{QR}=45 \), and \( \mathrm{RP}=55 \). Therefore, \( \mathrm{ABPQ}=\mathrm{BCQR}=\mathrm{CARP}=25=10 \) 5 , and therefore, \( \triangle A B C \sim \triangle P Q R \) by SSS \( \sim \). By the Distance Formula, \( A B=32-v, B C=42-v \), and \( C A=52-v . A l s o, P Q=35-v, Q R=45-v \), and \( R P=55- \) V.Therefore, \( A B P Q=B C Q R=C A R P=2 \mathrm{~V} 5 \mathrm{~V}=10 \mathrm{~V} 5 \), and therefore, \( \triangle A B C \sim \triangle P Q R \) by SAS \( \sim \).By the Distance Fo rmula, \( \mathrm{AB}=32, \mathrm{BC}=42 \), and \( \mathrm{CA}=52 \). Also, \( \mathrm{PQ}=35, \mathrm{QR}=45 \), and \( \mathrm{RP}=55 \). Therefore, \( \mathrm{ABPQ}=\mathrm{BCQR}=\mathrm{CARP}=25=1 \) 05 , and therefore, \( \triangle A B C \sim \triangle P Q R \) by \( S A S \sim \). \( B y \) the Distance Formula, \( A B=18, B C=32 \), and \( C A=50 \). Also, \( P Q=45, Q R=90 \), and \( R P=125 \). Therefore, \( A B P Q \) \( =B C Q R=C A R P=25 \), andtherefore, \( \triangle A B C \sim \triangle P Q R \) by \( S A S \sim \).
Ask by Peterson Greene. in the United States
Jan 23,2025
Solución de tutoría real
Respuesta verificada por el tutor
Responder
\( \triangle ABC \sim \triangle PQR \) by SSS similarity.
Solución
To determine whether \( \triangle ABC \sim \triangle PQR \), we need to analyze the given distances and the similarity criteria.
### Step 1: Calculate the distances using the Distance Formula
The Distance Formula between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by:
\[
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\]
#### For triangle \( ABC \):
1. **Distance \( AB \)**:
\[
AB = \sqrt{((-5) - (-2))^2 + (3 - 0)^2} = \sqrt{(-3)^2 + 3^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}
\]
2. **Distance \( BC \)**:
\[
BC = \sqrt{((-9) - (-5))^2 + ((-1) - 3)^2} = \sqrt{(-4)^2 + (-4)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}
\]
3. **Distance \( CA \)**:
\[
CA = \sqrt{((-2) - (-9))^2 + (0 - (-1))^2} = \sqrt{(7)^2 + (1)^2} = \sqrt{49 + 1} = \sqrt{50} = 5\sqrt{2}
\]
#### For triangle \( PQR \):
1. **Distance \( PQ \)**:
\[
PQ = \sqrt{(4 - 7)^2 + (0 - 6)^2} = \sqrt{(-3)^2 + (-6)^2} = \sqrt{9 + 36} = \sqrt{45} = 3\sqrt{5}
\]
2. **Distance \( QR \)**:
\[
QR = \sqrt{((-4) - 4)^2 + (4 - 0)^2} = \sqrt{(-8)^2 + (4)^2} = \sqrt{64 + 16} = \sqrt{80} = 4\sqrt{5}
\]
3. **Distance \( RP \)**:
\[
RP = \sqrt{(7 - (-4))^2 + (6 - 4)^2} = \sqrt{(11)^2 + (2)^2} = \sqrt{121 + 4} = \sqrt{125} = 5\sqrt{5}
\]
### Step 2: Compare the ratios of the sides
Now we have the distances:
- For \( \triangle ABC \):
- \( AB = 3\sqrt{2} \)
- \( BC = 4\sqrt{2} \)
- \( CA = 5\sqrt{2} \)
- For \( \triangle PQR \):
- \( PQ = 3\sqrt{5} \)
- \( QR = 4\sqrt{5} \)
- \( RP = 5\sqrt{5} \)
### Step 3: Check for similarity using SSS similarity criterion
To check if the triangles are similar by the SSS criterion, we need to compare the ratios of corresponding sides:
\[
\frac{AB}{PQ} = \frac{3\sqrt{2}}{3\sqrt{5}} = \frac{\sqrt{2}}{\sqrt{5}}
\]
\[
\frac{BC}{QR} = \frac{4\sqrt{2}}{4\sqrt{5}} = \frac{\sqrt{2}}{\sqrt{5}}
\]
\[
\frac{CA}{RP} = \frac{5\sqrt{2}}{5\sqrt{5}} = \frac{\sqrt{2}}{\sqrt{5}}
\]
Since all three ratios are equal, we conclude that:
\[
\triangle ABC \sim \triangle PQR \text{ by SSS similarity.}
\]
### Conclusion
The correct proof that \( \triangle ABC \sim \triangle PQR \) is based on the SSS similarity criterion, as all corresponding side ratios are equal.
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Simplifique esta solución The Deep Dive
Did you know that the similarity of triangles has been a key principle in geometry for thousands of years? Ancient Greeks, especially Euclid, laid the groundwork for our understanding of similar shapes. He showed that if two triangles have corresponding angles equal or the sides in proportion, they are similar! Fast forward to today, and we continue to build on these ideas, making connections between geometry and the world around us. In real-life situations, triangle similarity is commonly applied in navigation and architecture. For instance, architects use the principles of similar triangles when creating scaled models of buildings. By ensuring that the proportions are accurate, they can predict how the building will look and function while not needing to construct a full-size version first. It’s like planning a huge party in your backyard – you might sketch it out to make sure everything fits, avoiding any last-minute surprises!
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