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 $$ riangle A B C^{\sim} riangle P Q R $$ ? 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, $$ riangle A B C \sim riangle P Q R $$ 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, $$ riangle A B C \sim riangle P $$ 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, $$ riangle A B C \sim riangle P Q R $$ 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, $$ riangle A B C \sim riangle P Q R $$ 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, $$ riangle A B C \sim riangle P Q R $$ 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, $$ riangle A B C \sim riangle P Q R $$ 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, $$ riangle A B C \sim riangle P Q R $$ by $$ S A S \sim $$.

Question

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 $$ 	riangle A B C^{\sim} 	riangle P Q R $$ ? 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, $$ 	riangle A B C \sim 	riangle P Q R $$ 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, $$ 	riangle A B C \sim 	riangle P $$ 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, $$ 	riangle A B C \sim 	riangle P Q R $$ 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, $$ 	riangle A B C \sim 	riangle P Q R $$ 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, $$ 	riangle A B C \sim 	riangle P Q R $$ 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, $$ 	riangle A B C \sim 	riangle P Q R $$ 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, $$ 	riangle A B C \sim 	riangle P Q R $$ by $$ S A S \sim $$.

UpStudy AI · Accepted Answer

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.
$$ \triangle ABC \sim \triangle PQR $$ by SSS similarity.

Upstudy AI Solution

Answer

Solution

Step 1: Calculate the distances using the Distance Formula

For triangle :

For triangle :

Step 2: Compare the ratios of the sides

Step 3: Check for similarity using SSS similarity criterion

Conclusion

The Deep Dive

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