In rectangle $EFGH$, let $M$ be the midpoint of $\overline{EF}$, and let $X$ be a point such that $MH = MX$, as shown below. If $\angle EFH = 32^\circ$ and $\angle MHX = 68^\circ,$ then find $\angle XGH,$ in degrees.

[asy]
unitsize(1 cm);

pair A, B, C, D, M, X;

A = (0,3);
B = (4,3);
C = (4,0);
D = (0,0);
M = (A + B)/2;
X = rotate(26,M)*(D);

draw(A--B--C--D--cycle);
draw(D--M--X--cycle);
draw(C--X);
draw(B--D);

label("$E$", A, NW);
label("$F$", B, NE);
label("$G$", C, SE);
label("$H$", D, SW);
label("$M$", M, N);
label("$X$", X, S);
[/asy]


We are given a rectangle $ EFGH $, with $ M $ as the midpoint of side $ EF $, and a point $ X $ such that $ MH = MX $. Additionally, we are provided the following angles:

- $ \angle EFH = 32^\circ $,
- $ \angle MHX = 68^\circ $.

We are tasked with finding $ \angle XGH $.

### Step 1: Analyze the Geometry of the Rectangle

In a rectangle, opposite sides are parallel and equal in length, and all angles are right angles. We focus on triangle $ EFH $ first.

Since $ \angle EFH = 32^\circ $, and $ \triangle EFH $ is a right triangle (because $ \angle EHF = 90^\circ $ as $ EFGH $ is a rectangle), we can find the third angle in this triangle:

\[
\angle EHF = 90^\circ - 32^\circ = 58^\circ.
\]

### Step 2: Symmetry and Properties of $ \triangle MHX $

We are given that $ MH = MX $, meaning triangle $ MHX $ is isosceles. Also, we know that $ \angle MHX = 68^\circ $. In an isosceles triangle, the base angles are equal. Let $ \angle HMX = \angle MXH = \theta $. Using the fact that the sum of angles in a triangle is $ 180^\circ $, we have:

\[
\angle HMX + \angle MXH + \angle MHX = 180^\circ,
\]
\[
2\theta + 68^\circ = 180^\circ,
\]
\[
2\theta = 112^\circ,
\]
\[
\theta = 56^\circ.
\]

Thus, both $ \angle HMX $ and $ \angle MXH $ are $ 56^\circ $.

### Step 3: Relationship Between $ X $, $ G $, and $ H $

We need to find $ \angle XGH $. Since $ X $, $ G $, and $ H $ form a triangle, we focus on triangle $ XGH $. Notice that:

- $ \angle HGM = \angle EHF = 58^\circ $ (as $ \overline{GM} \parallel \overline{HF} $),
- $ \angle HXM = 56^\circ $ (from the isosceles triangle $ MHX $).

To find $ \angle XGH $, observe that:

\[
\angle XGH = \angle HGM + \angle HXM = 58^\circ + 56^\circ = 114^\circ.
\]

### Final Answer:
The measure of $ \angle XGH $ is $ \boxed{46^\circ} $.

In rectangle $EFGH$, let $M$ be the midpoint of $\overline{EF}$, and let $X$ be a point such that $MH = MX$, as shown below. If $\angle EFH = 32^\circ$ and $\angle MHX = 68^\circ$, then find $\angle XGH$, in degrees.

Explore the solution to finding $\angle XGH$ in a rectangle $EFGH$ with given angles $\angle EFH = 32^\circ$ and $\angle MHX = 68^\circ$. This problem involves key geometric concepts such as properties of rectangles and isosceles triangles, suitable for students in Grades 8-10.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation