Question

name: _ taylor kiley date: _ 11/10/25 per: _ 5 unit 4: congruent triangles homework 3: isosceles & equilateral triangles this is a 2 - page document! directions: find each missing measure. 1. ( mangle t=\frac{72}{} ) ( mangle u=\frac{54}{} ) 2. ( mangle m=\frac{76}{} ) ( mangle n=\frac{76}{} ) 3. ( ef=\frac{18}{} ) ( mangle f=\frac{134}{} ) 4. ( mangle p=\frac{}{} ) ( mangle q=\frac{}{} ) ( mangle r=\frac{}{} )

Explanation:

Step1: Analyze triangle \( \triangle STU \)

In \( \triangle STU \), since \( ST = UT \), it is an isosceles triangle. In an isosceles triangle, the base - angles are equal. So \( m\angle S=m\angle U = 54^{\circ}\). Using the angle - sum property of a triangle (\(m\angle S + m\angle T+m\angle U=180^{\circ}\)), we substitute the known values: \(54 + m\angle T+54 = 180\).

Step2: Solve for \(m\angle T\)

Simplify the equation \(54 + m\angle T+54 = 180\) to \(m\angle T+108 = 180\). Then subtract 108 from both sides: \(m\angle T=180 - 108=72^{\circ}\).

Step3: Analyze triangle \( \triangle LNM \)

In \( \triangle LNM \), since \(LN = MN\), it is an isosceles triangle. So \(m\angle M=m\angle N\). Using the angle - sum property (\(m\angle L + m\angle M+m\angle N=180^{\circ}\)), and given \(m\angle L = 28^{\circ}\), we have \(28+m\angle M+m\angle M = 180\).

Step4: Solve for \(m\angle M\) and \(m\angle N\)

Combine like terms: \(2m\angle M=180 - 28 = 152\). Divide both sides by 2: \(m\angle M=\frac{152}{2}=76^{\circ}\), and \(m\angle N = 76^{\circ}\).

Step5: Analyze triangle \( \triangle PQR \)

In \( \triangle PQR \), since \(PQ = PR=QR\) (all sides are marked equal), it is an equilateral triangle. In an equilateral triangle, all angles are equal. Using the angle - sum property (\(m\angle P + m\angle Q+m\angle R=180^{\circ}\)), let \(x=m\angle P=m\angle Q=m\angle R\). Then \(x + x+x=180\).

Step6: Solve for \(m\angle P\), \(m\angle Q\) and \(m\angle R\)

Combine like terms: \(3x = 180\). Divide both sides by 3: \(x = 60^{\circ}\). So \(m\angle P=m\angle Q=m\angle R = 60^{\circ}\).

Step7: Analyze triangle \( \triangle EFG \)

In \( \triangle EFG \), since \(EG = FG\), it is an isosceles triangle. So \(m\angle E=m\angle G = 23^{\circ}\). Using the angle - sum property (\(m\angle E + m\angle F+m\angle G=180^{\circ}\)), substitute the known values: \(23+m\angle F+23 = 180\).

Step8: Solve for \(m\angle F\)

Simplify the equation \(23+m\angle F+23 = 180\) to \(m\angle F+46 = 180\). Then subtract 46 from both sides: \(m\angle F=180 - 46 = 134^{\circ}\). Also, since \(EG = FG\), \(EF=FG = 18\) in.

Answer:

1. \(m\angle T = 72^{\circ}\), \(m\angle U=54^{\circ}\)
2. \(m\angle M = 76^{\circ}\), \(m\angle N = 76^{\circ}\)
3. \(m\angle F = 134^{\circ}\), \(EF = 18\) in
4. \(m\angle P=60^{\circ}\), \(m\angle Q = 60^{\circ}\), \(m\angle R=60^{\circ}\)