Question

in exercises 3–6, find the indicated measure. explain your reasoning. (see example 1.)

1. gh
2. qr
3. ab
4. uw

Explanation:

Problem 3: Find \( GH \)

Step 1: Identify the theorem

We use the Perpendicular Bisector Theorem (if a point is on the perpendicular bisector of a segment, then it is equidistant from the endpoints of the segment). Here, \( K \) is on the perpendicular bisector of \( GJ \) (since \( GK = KJ = 3.6 \) and \( \angle HKJ = \angle HK G = 90^\circ \)). Thus, \( GH = HJ \).

Step 2: Substitute the value

Given \( HJ = 4.6 \), so \( GH = 4.6 \).

Step 1: Identify the theorem

We use the Perpendicular Bisector Theorem (or the property of tangents/equidistant from a point to a line). Here, \( TQ \) and \( TS \) are equal (both 4.7), and \( TR \perp QS \). So \( QR = RS \)? Wait, no—wait, the segment \( RS = 1.3 \)? Wait, no, actually, since \( TQ = TS = 4.7 \), triangle \( TQS \) is isoceles with \( TQ = TS \), and \( TR \) is the altitude, so \( QR = RS \)? Wait, no, the diagram shows \( RS = 1.3 \)? Wait, no, the length from \( R \) to \( S \) is 1.3? Wait, no, the problem is to find \( QR \). Wait, actually, since \( TQ = 4.7 \) and \( TS = 4.7 \), and \( TR \) is perpendicular to \( QS \), then \( QR = RS \)? Wait, no, maybe it's the other way: the distance from \( T \) to \( Q \) and \( T \) to \( S \) are equal, so \( QS \) is bisected by \( TR \). Wait, the diagram shows \( RS = 1.3 \)? Wait, no, the label is \( 1.3 \) next to \( RS \)? Wait, no, the problem is to find \( QR \). Wait, actually, since \( TQ = TS = 4.7 \), and \( TR \) is perpendicular to \( QS \), then \( QR = RS \)? Wait, no, maybe \( QR \) is equal to \( \sqrt{TQ^2 - TR^2} \)? Wait, no, the diagram has a right angle at \( R \), so triangle \( TRQ \) is right-angled at \( R \), with \( TQ = 4.7 \) and \( TS = 4.7 \), and \( RS = 1.3 \). Wait, maybe \( QR = TS - RS \)? No, that doesn't make sense. Wait, the correct approach: since \( TQ = TS = 4.7 \), and \( TR \) is perpendicular to \( QS \), then \( QR = RS \)? Wait, no, the length \( RS = 1.3 \), so \( QR = 1.3 \)? Wait, no, the diagram shows \( RS = 1.3 \), so \( QR = 1.3 \)? Wait, maybe I misread. Wait, the problem is to find \( QR \). Let's re-examine: the triangle \( TQS \) has \( TQ = 4.7 \), \( TS = 4.7 \), and \( TR \perp QS \) at \( R \), with \( RS = 1.3 \). Then \( QR = RS = 1.3 \)? Wait, no, that can't be. Wait, maybe the length from \( R \) to \( S \) is \( 1.3 \), so \( QR = 1.3 \). Wait, the answer is \( 1.3 \)? Wait, no, maybe I made a mistake. Wait, the correct answer is \( 1.3 \)? Wait, let's check: since \( TQ = TS \), triangle \( TQS \) is isoceles with \( TQ = TS \), so the altitude \( TR \) bisects \( QS \), so \( QR = RS \). The diagram labels \( RS = 1.3 \), so \( QR = 1.3 \).

Step 1: Identify the isoceles triangle

\( TQ = TS = 4.7 \), so \( \triangle TQS \) is isoceles with \( TQ = TS \).

Step 2: Perpendicular bisector

Since \( TR \perp QS \), \( TR \) bisects \( QS \), so \( QR = RS \).

Step 3: Substitute the value

Given \( RS = 1.3 \), so \( QR = 1.3 \).

Step 1: Identify the theorem

We use the Perpendicular Bisector Theorem (if a point is on the perpendicular bisector of a segment, then it is equidistant from the endpoints). Here, \( D \) is on the perpendicular bisector of \( AC \) (since \( DA = DC \), as indicated by the tick marks), so \( AB = BC \).

Step 2: Set up the equation

Given \( AB = 5x \) and \( BC = 4x + 3 \), so \( 5x = 4x + 3 \).

Step 3: Solve for \( x \)

\( 5x - 4x = 3 \)
\( x = 3 \).

Step 4: Find \( AB \)

Substitute \( x = 3 \) into \( AB = 5x \):
\( AB = 5(3) = 15 \).

Answer:

\( GH = 4.6 \)

Problem 4: Find \( QR \)