QuestionIf $\triangle \mathrm{AFG} \sim \triangle \mathrm{DRH}$ , find the missing value in $\frac{D R}{A F}=\frac{D H}{\square}$ . Choices: HD, AH, AG.

Studdy Solution

STEP 1

Assumptions1. $\triangle AFG \sim \triangle DRH$ means that triangle AFG is similar to triangle DRH. . In similar triangles, the ratios of the corresponding sides are equal.
3. We are given that $\frac{DR}{AF} = \frac{DH}{\square}$ , and we are asked to find the value of $\square$ for each of the options HD, AH, AG.

STEP 2

First, let's consider the option HD. We substitute HD for $\square$ in the given ratio.
$\frac{DR}{AF} = \frac{DH}{HD}$

STEP 3

In this case, the ratio $\frac{DH}{HD}$ simplifies to1, because any number divided by itself equals1.
$\frac{DR}{AF} =1$

STEP 4

This implies that DR = AF, because any number multiplied by1 equals the number itself.

STEP 5

Next, let's consider the option AH. We substitute AH for $\square$ in the given ratio.
$\frac{DR}{AF} = \frac{DH}{AH}$

STEP 6

In this case, we can't simplify the ratio $\frac{DH}{AH}$ , because DH and AH are sides of different triangles. So, we can't find a specific numerical value for $\square$ .

STEP 7

Finally, let's consider the option AG. We substitute AG for $\square$ in the given ratio.
$\frac{DR}{AF} = \frac{DH}{AG}$

STEP 8

In this case, we can't simplify the ratio $\frac{DH}{AG}$ , because DH and AG are sides of different triangles. So, we can't find a specific numerical value for $\square$ .
In conclusion, the only option for which we can find a specific value for $\square$ is HD, for which $\square$ equals AF.

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QuestionIf $\triangle \mathrm{AFG} \sim \triangle \mathrm{DRH}$ , find the missing value in $\frac{D R}{A F}=\frac{D H}{\square}$ . Choices: HD, AH, AG.

Studdy Solution

STEP 1

STEP 2

First, let's consider the option HD. We substitute HD for $\square$ in the given ratio.
$\frac{DR}{AF} = \frac{DH}{HD}$

STEP 3

In this case, the ratio $\frac{DH}{HD}$ simplifies to1, because any number divided by itself equals1.
$\frac{DR}{AF} =1$

STEP 4

This implies that DR = AF, because any number multiplied by1 equals the number itself.

STEP 5

Next, let's consider the option AH. We substitute AH for $\square$ in the given ratio.
$\frac{DR}{AF} = \frac{DH}{AH}$

STEP 6

In this case, we can't simplify the ratio $\frac{DH}{AH}$ , because DH and AH are sides of different triangles. So, we can't find a specific numerical value for $\square$ .

STEP 7

Finally, let's consider the option AG. We substitute AG for $\square$ in the given ratio.
$\frac{DR}{AF} = \frac{DH}{AG}$

STEP 8

In this case, we can't simplify the ratio $\frac{DH}{AG}$ , because DH and AG are sides of different triangles. So, we can't find a specific numerical value for $\square$ .
In conclusion, the only option for which we can find a specific value for $\square$ is HD, for which $\square$ equals AF.

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QuestionIf △AFG∼△DRH\triangle \mathrm{AFG} \sim \triangle \mathrm{DRH}△AFG∼△DRH, find the missing value in DRAF=DH□\frac{D R}{A F}=\frac{D H}{\square}AFDR​=□DH​. Choices: HD, AH, AG.

Studdy Solution

STEP 1

STEP 2

First, let's consider the option HD. We substitute HD for □\square□ in the given ratio.DRAF=DHHD\frac{DR}{AF} = \frac{DH}{HD}AFDR​=HDDH​

STEP 3

In this case, the ratio DHHD\frac{DH}{HD}HDDH​ simplifies to1, because any number divided by itself equals1.DRAF=1\frac{DR}{AF} =1AFDR​=1

STEP 4

This implies that DR = AF, because any number multiplied by1 equals the number itself.

STEP 5

Next, let's consider the option AH. We substitute AH for □\square□ in the given ratio.DRAF=DHAH\frac{DR}{AF} = \frac{DH}{AH}AFDR​=AHDH​

STEP 6

In this case, we can't simplify the ratio DHAH\frac{DH}{AH}AHDH​, because DH and AH are sides of different triangles. So, we can't find a specific numerical value for □\square□.

STEP 7

Finally, let's consider the option AG. We substitute AG for □\square□ in the given ratio.DRAF=DHAG\frac{DR}{AF} = \frac{DH}{AG}AFDR​=AGDH​

STEP 8

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QuestionIf $\triangle \mathrm{AFG} \sim \triangle \mathrm{DRH}$ , find the missing value in $\frac{D R}{A F}=\frac{D H}{\square}$ . Choices: HD, AH, AG.

First, let's consider the option HD. We substitute HD for $\square$ in the given ratio.
$\frac{DR}{AF} = \frac{DH}{HD}$

In this case, the ratio $\frac{DH}{HD}$ simplifies to1, because any number divided by itself equals1.
$\frac{DR}{AF} =1$

Next, let's consider the option AH. We substitute AH for $\square$ in the given ratio.
$\frac{DR}{AF} = \frac{DH}{AH}$

In this case, we can't simplify the ratio $\frac{DH}{AH}$ , because DH and AH are sides of different triangles. So, we can't find a specific numerical value for $\square$ .

Finally, let's consider the option AG. We substitute AG for $\square$ in the given ratio.
$\frac{DR}{AF} = \frac{DH}{AG}$