Question

8 in rectangle fghi, diagonals intersect at e. ( ie = 3x + 4 ), ( eg = 5x - 6 ). what is the length of...?
a. 5 units
b. 10 units
c. 19 units
d. 38 units

9 this diagram shows parallelogram ( mnpq ). ( angle n = (7x - 1)^circ ), ( angle m = (3x + 1)^circ ). what is ( mangle npq )?
a. ( 18^circ )
b. ( 35^circ )
c. ( 55^circ )
d. ( 125^circ )

1 this shows parallelogram ( mnpr ). ( angle m = (x + 34)^circ ), ( angle r = (3x - 2)^circ ). what is ( mangle npr )?
a. ( 18^circ )
b. ( 38^circ )
c. ( 52^circ )
d. ( 128^circ )

1 this shows a transformation function. ( j(x, y) = (x + 2, y - 6) ). how does this function transform a point?
a. the function translates a point 2 units down and 6 units left.
b. the function translates a point 2 units to the left and 6 units up.
c. the function translates a point 2 units up and 6 units to the right.
d. the function translates a point 2 units to the right and 6 units down.

Explanation:

Problem 8 (Rectangle Diagonals)

Step1: Recall rectangle diagonal property

In a rectangle, diagonals are equal and bisect each other. So \( IE = EG \).
\( 3x + 4 = 5x - 6 \)

Step2: Solve for \( x \)

Subtract \( 3x \) from both sides: \( 4 = 2x - 6 \)
Add 6 to both sides: \( 10 = 2x \)
Divide by 2: \( x = 5 \)

Step3: Find \( IE \) and \( EG \)

\( IE = 3(5) + 4 = 19 \), \( EG = 5(5) - 6 = 19 \)

Step4: Find \( FH \) (diagonal)

Diagonal \( FH = IE + EG = 19 + 19 = 38 \)? Wait, no—wait, the question is about \( \overline{FH} \)? Wait, the problem says "length of \( \overline{FH} \)"? Wait, no, the original problem: "What is the length of \( \overline{FH} \)?" Wait, in a rectangle, diagonals are equal, and \( E \) is the midpoint, so \( FH = 2 \times IE \). Wait, \( IE = 19 \), so \( FH = 2 \times 19 = 38 \)? Wait, no, wait: \( IE \) is half the diagonal? Wait, no—wait, in a rectangle, diagonals bisect each other, so \( IE = EG = \frac{1}{2}FH \). Wait, no: \( IE \) and \( EG \) are segments of the diagonal \( IG \)? Wait, the diagram: rectangle \( FGHI \), diagonals \( FH \) and \( IG \) intersect at \( E \). So diagonals \( FH \) and \( IG \) bisect each other, so \( IE = EG \), and \( FE = EH \), and \( FH = IG \). Wait, the problem says "length of \( \overline{FH} \)"? Wait, no, the options are 5,10,19,38. Wait, maybe I misread: the problem says "length of \( \overline{FH} \)"? Wait, no, let's recheck:

Wait, the problem: "In rectangle \( FGHI \), diagonals \( \overline{FH} \) and \( \overline{IG} \) intersect at \( E \). \( IE = 3x + 4 \), \( EG = 5x - 6 \). What is the length of \( \overline{FH} \)?" Wait, no—wait, \( IG \) is the diagonal, so \( IE = EG \), so \( 3x + 4 = 5x - 6 \), \( x = 5 \), so \( IE = 19 \), \( EG = 19 \), so \( IG = 38 \). But \( FH = IG \) (diagonals of rectangle are equal), so \( FH = 38 \)? But option D is 38. Wait, but maybe the question is about \( \overline{FH} \), but let's confirm.

Wait, no—wait, maybe the diagonal is \( FH \), and \( E \) is the midpoint, so \( FE = EH \), but the given segments are \( IE \) and \( EG \) (from diagonal \( IG \)). Wait, maybe the problem has a typo, but assuming that \( FH \) is equal to \( IG \), then \( IG = IE + EG = 19 + 19 = 38 \), so \( FH = 38 \). So answer D.

Step1: Recall parallelogram angle property

In a parallelogram, consecutive angles are supplementary, and opposite angles are equal. Also, adjacent angles (like \( \angle M \) and \( \angle N \)) are supplementary? Wait, no: in parallelogram \( MNPQ \), \( \angle M \) and \( \angle N \) are adjacent (consecutive) angles, so they are supplementary. Wait, \( \angle M = (3x + 1)^\circ \), \( \angle N = (7x - 1)^\circ \), so \( (3x + 1) + (7x - 1) = 180 \)

Step2: Solve for \( x \)

\( 3x + 1 + 7x - 1 = 180 \)
\( 10x = 180 \)
\( x = 18 \)

Step3: Find \( \angle M \)

\( \angle M = 3(18) + 1 = 55^\circ \)

Step4: Find \( \angle NPQ \)

In parallelogram, \( \angle NPQ \) is equal to \( \angle M \) (opposite angles)? Wait, no: \( \angle M \) and \( \angle P \) are opposite? Wait, \( MNPQ \): vertices in order, so \( M \)-\( N \)-\( P \)-\( Q \)-\( M \). So \( \angle M \) and \( \angle P \) are opposite, \( \angle N \) and \( \angle Q \) are opposite. Wait, \( \angle NPQ \) is at \( P \), between \( N \)-\( P \)-\( Q \). Wait, \( \angle M \) is at \( M \), between \( N \)-\( M \)-\( Q \). So \( \angle M \) and \( \angle NPQ \) are equal? Wait, no—wait, \( \angle M \) and \( \angle N \) are supplementary, \( \angle N \) and \( \angle P \) (i.e., \( \angle NPQ \)) are supplementary? Wait, no, let's clarify the sides: \( MN \parallel PQ \), \( MQ \parallel NP \). So \( \angle M \) and \( \angle N \) are consecutive angles (adjacent), so supplementary. \( \angle N \) and \( \angle P \) ( \( \angle NPQ \)) are consecutive angles, so supplementary? Wait, no, \( \angle N \) is at \( N \), between \( M \)-\( N \)-\( P \), and \( \angle P \) ( \( \angle NPQ \)) is at \( P \), between \( N \)-\( P \)-\( Q \). So \( MN \parallel PQ \), so \( \angle N + \angle P = 180^\circ \)? Wait, no, \( MQ \parallel NP \), so \( \angle M + \angle P = 180^\circ \)? Wait, I think I made a mistake. Let's re-express:

In parallelogram \( MNPQ \), \( \angle M \) and \( \angle P \) are opposite angles, so they are equal. \( \angle N \) and \( \angle Q \) are opposite angles, equal. Consecutive angles (e.g., \( \angle M \) and \( \angle N \)) are supplementary.

So \( \angle M = (3x + 1)^\circ \), \( \angle N = (7x - 1)^\circ \), so \( \angle M + \angle N = 180^\circ \) (consecutive angles). So:

\( 3x + 1 + 7x - 1 = 180 \)
\( 10x = 180 \)
\( x = 18 \)

Then \( \angle M = 3(18) + 1 = 55^\circ \), so \( \angle P = \angle M = 55^\circ \)? Wait, no—wait, \( \angle NPQ \) is \( \angle P \), which is opposite to \( \angle M \), so equal. Wait, but the options are 18, 35, 55, 125. So \( 55^\circ \) is option C? Wait, no—wait, maybe \( \angle NPQ \) is supplementary to \( \angle N \)? Wait, let's check \( \angle N \): \( 7x - 1 = 7(18) - 1 = 125^\circ \). Then \( \angle NPQ \) is supplementary to \( \angle N \)? Wait, no, \( \angle N \) and \( \angle P \) ( \( \angle NPQ \)): if \( MN \parallel PQ \), then \( \angle N + \angle P = 180^\circ \)? Wait, no, \( MQ \parallel NP \), so \( \angle M + \angle N = 180^\circ \), and \( \angle M + \angle Q = 180^\circ \), \( \angle N + \angle P = 180^\circ \), \( \angle P + \angle Q = 180^\circ \). Wait, so \( \angle N = 125^\circ \), so \( \angle P = 180 - 125 = 55^\circ \). Yes, that makes sense. So \( \angle NPQ = 55^\circ \), option C.

Step1: Recall parallelogram and linear pair

In parallelogram \( MNPR \), \( MN \parallel PR \), so \( \angle M \) and the exterior angle at \( R \) are equal (corresponding angles). Wait, the exterior angle at \( R \) is \( (3x - 2)^\circ \), and \( \angle M = (x + 34)^\circ \). Since \( MN \parallel PR \), \( \angle M = \) exterior angle at \( R \) (corresponding angles), so \( x + 34 = 3x - 2 \)

Step2: Solve for \( x \)

\( x + 34 = 3x - 2 \)
Subtract \( x \): \( 34 = 2x - 2 \)
Add 2: \( 36 = 2x \)
\( x = 18 \)

Step3: Find \( \angle M \)

\( \angle M = 18 + 34 = 52^\circ \)

Step4: Find \( \angle NPR \)

In parallelogram, \( \angle M \) and \( \angle NPR \) are equal? Wait, no—wait, \( \angle NPR \) is adjacent to \( \angle M \)? Wait, \( MNPR \): \( M \)-\( N \)-\( P \)-\( R \)-\( M \). So \( \angle M \) and \( \angle NPR \): since \( MN \parallel PR \), \( \angle M + \angle NPR = 180^\circ \)? Wait, no—wait, \( \angle M \) and \( \angle R \) (interior angle at \( R \)) are supplementary. The exterior angle at \( R \) is \( (3x - 2)^\circ \), so interior angle at \( R \) is \( 180 - (3x - 2) = 182 - 3x \). But in parallelogram, \( \angle M = \angle R \) (opposite angles)? Wait, no, \( \angle M \) and \( \angle P \) are opposite, \( \angle N \) and \( \angle R \) are opposite. Wait, maybe I messed up. Let's re-express:

The exterior angle at \( R \) is \( (3x - 2)^\circ \), so the interior angle at \( R \) is \( 180 - (3x - 2) = 182 - 3x \). In parallelogram, \( \angle M = \angle R \) (opposite angles). Wait, no, \( \angle M \) is \( (x + 34)^\circ \), and \( \angle R \) (interior) is \( 182 - 3x \). But also, \( \angle M \) and the exterior angle at \( R \) are equal (corresponding angles, since \( MN \parallel PR \) and transversal \( MR \)). So \( x + 34 = 3x - 2 \), so \( x = 18 \), as before. Then \( \angle M = 52^\circ \), and \( \angle NPR \) is equal to \( \angle M \)? Wait, no—wait, \( \angle NPR \) is at \( P \), between \( N \)-\( P \)-\( R \). Since \( MN \parallel PR \), \( \angle M + \angle NPR = 180^\circ \)? No, wait, \( MP \) is a side? Wait, maybe \( \angle NPR \) is equal to \( \angle M \)? Wait, no, let's look at the options: A.18, B.38, C.52, D.128. Wait, \( \angle M = 52^\circ \), and \( \angle NPR \) is equal to \( \angle M \)? Wait, no—wait, in parallelogram, \( MN \parallel PR \), so \( \angle M \) and \( \angle NPR \) are alternate interior angles? Wait, \( MN \parallel PR \), and \( MP \) is a transversal? No, \( MR \) is the base. Wait, maybe \( \angle NPR = \angle M \), so \( 52^\circ \), option C.

Answer:

D. 38 units

Problem 9 (Parallelogram Angles)