Question

topic 4: rhombi & squares

1. if jklm is a rhombus, find each angle.
2. using jklm from the previous question, if mn = 20 and jl = 26, find jk.
3. if stuv is a rhombus, find m∠stu.
4. if abcd is a rhombus, ad = 4x + 2, dc = 7x - 13, and bd = 34, find ae.

(handwritten notes and diagrams related to the rhombus problems)

Explanation:

Step1: Analyze the rhombus properties

In a rhombus, the diagonals bisect the angles and are perpendicular bisectors of each other. Also, adjacent angles are supplementary, and opposite angles are equal. For problem 23, in rhombus STUV, the triangles formed by the diagonals are isosceles, and we can set the expressions for the angles equal since the diagonals bisect the angles. So we have \(9x - 43=5x + 1\).

Step2: Solve for \(x\)

Subtract \(5x\) from both sides: \(9x-5x - 43=5x-5x + 1\) which simplifies to \(4x-43 = 1\). Then add 43 to both sides: \(4x-43 + 43=1 + 43\) giving \(4x=44\). Divide both sides by 4: \(x=\frac{44}{4}=11\).

Step3: Find the angle measure

Now substitute \(x = 11\) into one of the angle expressions, say \(5x + 1\). So \(5(11)+1=55 + 1=56\). But we need to find \(m\angle STU\). Since the diagonal bisects the angle, and in a rhombus, the angle at \(T\) is twice this angle? Wait, no, actually in the rhombus, the triangles formed by the diagonals: the angles \( \angle STU\) is composed of two angles each equal to \(5x + 1\) (or \(9x - 43\))? Wait, no, let's re - examine. In rhombus STUV, diagonal \(SU\) and \(TV\) intersect. The triangles \(STV\) and \(UTV\) are isosceles. The angles \( \angle STV\) and \( \angle UTV\) are equal? Wait, no, the given angles are \( (9x - 43)^\circ\) and \( (5x + 1)^\circ\) which are the angles formed by the diagonal and the side. Since in a rhombus, the diagonal bisects the vertex angle, so the two angles formed by the diagonal and the side are equal. So we set \(9x-43 = 5x + 1\) (as we did before) to find \(x = 11\). Then each of those angles is \(5(11)+1 = 56^\circ\). Then \( \angle STU\) is the sum of these two angles? Wait, no, in the rhombus, the diagonal divides the angle into two equal parts. Wait, maybe I made a mistake. Wait, in a rhombus, the diagonals bisect the angles. So if we have angle at \(T\), say \( \angle STU\), and the diagonal \(TV\) bisects it into two angles, each equal to \( (5x + 1)^\circ\) (or \( (9x - 43)^\circ\)). Wait, no, actually, when we set \(9x - 43=5x + 1\), we find \(x = 11\), then \(9x-43=9\times11 - 43=99 - 43 = 56\), and \(5x + 1=5\times11+1 = 56\). So the angle \( \angle STU\) is \(2\times56 = 112^\circ\)? Wait, no, maybe not. Wait, let's think again. In a rhombus, adjacent angles are supplementary. Wait, maybe I messed up the problem. Wait, the problem says "find \(m\angle STU\)". Let's assume that the two angles \( (9x - 43)^\circ\) and \( (5x + 1)^\circ\) are the angles of the isosceles triangle formed by the diagonal. Since they are equal (because in a rhombus, the diagonal bisects the angle, so the two angles formed by the diagonal and the side are equal), so we solve \(9x-43 = 5x + 1\) to get \(x = 11\). Then each of these angles is \(56^\circ\), and \( \angle STU\) is the angle at \(T\) between sides \(ST\) and \(UT\). If the diagonal \(TV\) bisects \( \angle STU\) into two angles of \(56^\circ\) each, then \( \angle STU=56^\circ\times2 = 112^\circ\).

Answer:

\(m\angle STU = 112^\circ\)