Polygons \& Quadrilaterals Date: Per: Homework 5: Rhombi and Squares ** This is a 2-page document! *
The Correct Answer and Explanation is:
which has these important properties:
Key Properties of Rhombi:
- All four sides are congruent.
- Diagonals bisect each other at right angles (90°).
- Diagonals bisect the vertex angles.
- Opposite angles are equal.
Solutions:
1. UV = 8, WX = 5
- Since diagonals of a rhombus bisect each other:
- TU = WU = 4 (half of UV)
- TX = TV = 2.5 (half of WX)
Answer:
- TU = 4, WU = 4, TX = 2.5, TV = 2.5
2. BC = 28, BD = 32
- All sides are equal, so:
- CD = EF = FE = EC = 28
- Diagonals bisect:
- FD = 16 (half of BD)
Answer:
- CD = 28, FD = 16, EF = 28, EC = 28
3. MK = 24, JL = 20, ∠MJL = 50°
- Since diagonals bisect each other and sides are congruent:
- NK = 12, NL = 10
- All sides = 24 (since MK = 24 and rhombi have equal sides)
- ∠MJL is split in half by diagonal → other half = 50°
- Diagonals form 90° angles
Answer:
- NK = 12, NL = 10, ML = 24, JM = 24
- ∠LKN = 90°, ∠KJL = 50°, ∠LMK = 90°, ∠JKM = 50°, ∠JML = 40°
4. Find PQ (5x + 16 = 9x – 32)
Solve:
- 5x + 16 = 9x – 32
- 48 = 4x → x = 12
Now plug into PQ:
5(12) + 16 = 76
Answer: PQ = 76
5. Find ∠HGI
Use the equation:
- ∠HGI = 7x – 1, ∠GIJ = 4x + 3
- Diagonals intersect at 90°, so:
(7x – 1) + (4x + 3) = 90
→ 11x + 2 = 90
→ x = 8
Then ∠HGI = 7(8) – 1 = 55°
Answer: ∠HGI = 55°
6. Find ∠ADB:
Given:
- ∠ACD = 13x – 16
- ∠CBD = 9x + 4
Add them:
(13x – 16) + (9x + 4) = 22x – 12 = ∠ADB
Set = 90° since diagonals of a rhombus intersect at 90°:
→ 22x – 12 = 90
→ x = 21
Then ∠ADB = 90°
Answer: ∠ADB = 90°
7. If ∠XYZ = 136°, solve for x:
In rhombi, opposite angles are equal, and adjacent angles are supplementary.
∠XYZ = 136°, ∠WXY = 44°
Also, (10x – 8) = ∠WXY → 10x – 8 = 44 → x = 5.2
Answer: x = 5.2
Explanation
This worksheet focuses on the properties of rhombi and how they relate to side lengths, angles, and diagonals. A rhombus is a special type of parallelogram where all four sides are congruent, and the diagonals bisect each other at right angles. Because of this, many problems can be solved using simple geometry principles.
For example, in problems involving diagonals (like #1 and #2), we use the fact that diagonals bisect each other. If a diagonal measures 8 units, then each half is 4. This helps in quickly finding missing segments.
In problems like #3 and #5, understanding how diagonals create right triangles and bisect angles is key. In problem #3, recognizing that diagonals divide the rhombus into four right triangles helps us determine the angles. In #5, angles around the intersection point of diagonals sum to 90°, allowing us to set up equations and solve for x.
For algebraic problems like #4 and #7, equations are created by setting congruent sides or supplementary angles equal to expressions. Solving these equations gives us the value of x, which we then use to find side lengths or angles.
Overall, solving problems involving rhombi involves applying a consistent set of rules. Understanding these properties allows us to approach each question logically and with confidence, reinforcing key geometry and algebra skills.
