Question

name: seina daltro date: per: 5 geometry a unit 4 – relationships in triangles calculators may be used on this portion of the test. 1. if (overline{jk}congoverline{kl}), find (jm). 14x + 15 19x - 5 2. find (qr). 4x - 7 2x + 9 3. find (mangle abd). (3x + 10)° (6x + 4)° 4. find (mn). 10x + 6 18x + 22 5. find (mangle1). 52° 89°

Explanation:

1. For the first problem (if $\overline{JK}\cong\overline{KL}$, find $JM$):

• Explanation:
• Step 1: Set up the equation based on congruence

Since $\overline{JK}\cong\overline{KL}$, we have $14x + 15=19x - 5$.

• Step 2: Solve for $x$

Subtract $14x$ from both sides: $15 = 5x-5$.
Add 5 to both sides: $20 = 5x$.
Divide both sides by 5: $x = 4$.

• Step 3: Find $JM$

Substitute $x = 4$ into the expression for $JM$ (which is $14x + 15$).
$JM=14\times4 + 15=56 + 15=71$.

• Answer: $JM = 71$

1. For the second problem (find $QR$):

• Explanation:
• Step 1: Set up the equation based on the right - triangle congruence (HL or other relevant congruence in right - triangles)

Assuming some congruence property (e.g., if $\triangle PQS\cong\triangle RQS$), we have $4x-7=2x + 9$.

• Step 2: Solve for $x$

Subtract $2x$ from both sides: $2x-7=9$.
Add 7 to both sides: $2x=16$.
Divide both sides by 2: $x = 8$.

• Step 3: Find $QR$

Substitute $x = 8$ into the expression for $QR$ (which is $2x + 9$).
$QR=2\times8 + 9=16 + 9=25$.

• Answer: $QR = 25$

1. For the third problem (find $m\angle ABD$):

• Explanation:
• Step 1: Set up the equation based on angle - bisector property

If $BD$ is an angle - bisector, then $3x + 10=6x+4$.

• Step 2: Solve for $x$

Subtract $3x$ from both sides: $10 = 3x+4$.
Subtract 4 from both sides: $6 = 3x$.
Divide both sides by 3: $x = 2$.

• Step 3: Find $m\angle ABD$

Substitute $x = 2$ into the expression for $m\angle ABD$ (which is $3x + 10$).
$m\angle ABD=3\times2+10=6 + 10=16^{\circ}$.

• Answer: $m\angle ABD = 16^{\circ}$

1. For the fourth problem (find $MN$):

• Explanation:
• Step 1: Set up the equation based on mid - segment or similar triangle property

Assuming some triangle similarity or mid - segment property, if $MN$ is related to the given side lengths, and we have $10x+6=\frac{1}{2}(18x + 22)$.
Multiply both sides by 2 to get $20x+12=18x + 22$.

• Step 2: Solve for $x$

Subtract $18x$ from both sides: $2x+12=22$.
Subtract 12 from both sides: $2x = 10$.
Divide both sides by 2: $x = 5$.

• Step 3: Find $MN$

Substitute $x = 5$ into the expression for $MN$ (which is $10x + 6$).
$MN=10\times5+6=50 + 6=56$.

• Answer: $MN = 56$

1. For the fifth problem (find $m\angle1$):

• Explanation:
• Step 1: Use the angle - sum property of a triangle

First, find the third angle in the larger triangle. Let's assume we know some other angles in the triangle.
The sum of angles in a triangle is $180^{\circ}$.
Let the third angle in the larger triangle be $\theta$.
We know two angles: $52^{\circ}$ and $89^{\circ}$.
So, $\theta=180-(52 + 89)=39^{\circ}$.
Since $\angle1$ is related to this angle (e.g., vertical angles, corresponding angles etc., assuming some geometric relationship), if $\angle1$ and $\theta$ are equal (depending on the figure), then $m\angle1 = 39^{\circ}$.

• Answer: $m\angle1 = 39^{\circ}$

Answer:

1. For the first problem (if $\overline{JK}\cong\overline{KL}$, find $JM$):

• Explanation:
• Step 1: Set up the equation based on congruence

Since $\overline{JK}\cong\overline{KL}$, we have $14x + 15=19x - 5$.

• Step 2: Solve for $x$

Subtract $14x$ from both sides: $15 = 5x-5$.
Add 5 to both sides: $20 = 5x$.
Divide both sides by 5: $x = 4$.

• Step 3: Find $JM$

Substitute $x = 4$ into the expression for $JM$ (which is $14x + 15$).
$JM=14\times4 + 15=56 + 15=71$.

• Answer: $JM = 71$

1. For the second problem (find $QR$):

• Explanation:
• Step 1: Set up the equation based on the right - triangle congruence (HL or other relevant congruence in right - triangles)

Assuming some congruence property (e.g., if $\triangle PQS\cong\triangle RQS$), we have $4x-7=2x + 9$.

• Step 2: Solve for $x$

Subtract $2x$ from both sides: $2x-7=9$.
Add 7 to both sides: $2x=16$.
Divide both sides by 2: $x = 8$.

• Step 3: Find $QR$

Substitute $x = 8$ into the expression for $QR$ (which is $2x + 9$).
$QR=2\times8 + 9=16 + 9=25$.

• Answer: $QR = 25$

1. For the third problem (find $m\angle ABD$):

• Explanation:
• Step 1: Set up the equation based on angle - bisector property

If $BD$ is an angle - bisector, then $3x + 10=6x+4$.

• Step 2: Solve for $x$

Subtract $3x$ from both sides: $10 = 3x+4$.
Subtract 4 from both sides: $6 = 3x$.
Divide both sides by 3: $x = 2$.

• Step 3: Find $m\angle ABD$

Substitute $x = 2$ into the expression for $m\angle ABD$ (which is $3x + 10$).
$m\angle ABD=3\times2+10=6 + 10=16^{\circ}$.

• Answer: $m\angle ABD = 16^{\circ}$

1. For the fourth problem (find $MN$):

• Explanation:
• Step 1: Set up the equation based on mid - segment or similar triangle property

Assuming some triangle similarity or mid - segment property, if $MN$ is related to the given side lengths, and we have $10x+6=\frac{1}{2}(18x + 22)$.
Multiply both sides by 2 to get $20x+12=18x + 22$.

• Step 2: Solve for $x$

Subtract $18x$ from both sides: $2x+12=22$.
Subtract 12 from both sides: $2x = 10$.
Divide both sides by 2: $x = 5$.

• Step 3: Find $MN$

Substitute $x = 5$ into the expression for $MN$ (which is $10x + 6$).
$MN=10\times5+6=50 + 6=56$.

• Answer: $MN = 56$

1. For the fifth problem (find $m\angle1$):

• Explanation:
• Step 1: Use the angle - sum property of a triangle

First, find the third angle in the larger triangle. Let's assume we know some other angles in the triangle.
The sum of angles in a triangle is $180^{\circ}$.
Let the third angle in the larger triangle be $\theta$.
We know two angles: $52^{\circ}$ and $89^{\circ}$.
So, $\theta=180-(52 + 89)=39^{\circ}$.
Since $\angle1$ is related to this angle (e.g., vertical angles, corresponding angles etc., assuming some geometric relationship), if $\angle1$ and $\theta$ are equal (depending on the figure), then $m\angle1 = 39^{\circ}$.

• Answer: $m\angle1 = 39^{\circ}$