Ninth Grade Mathematics: Rectangles, Rhombuses & Squares
While general parallelograms possess equal opposite sides and angles, introducing right angles or equal adjacent sides creates three special subclasses of parallelograms: rectangles, rhombuses, and squares. In this chapter, you will master the defining properties, diagonal behaviors, and hierarchical taxonomy of these classical geometric figures.
The Special Parallelogram Taxonomy
The relationships among special parallelograms form a strict hierarchical family tree:
PARALLELOGRAMS
(Opposite sides parallel & equal)
|
+------------------+------------------+
| |
v v
RECTANGLE RHOMBUS
(Equiangular: 4 Right Angles) (Equilateral: 4 Equal Sides)
| |
+------------------+------------------+
|
v
SQUARE
(Both Equiangular & Equilateral:
4 Right Angles AND 4 Equal Sides)
A rectangle is a parallelogram with four right angles. A rhombus is a parallelogram with four equal sides. A square is simultaneously a rectangle AND a rhombus, possessing all the properties of both figures.
Rectangles and Diagonal Equality
In an ordinary parallelogram, one diagonal is longer than the other. In a rectangle, the two diagonals are always equal in length:
The Rectangle Diagonal Theorem:
A parallelogram is a rectangle if and only if its diagonals are equal in length.
In Rectangle ABCD:
Diagonal AC = Diagonal BD
D C
*--------------------*
| \ / |
| \ / |
| \ / |
| \ / |
| \/ |
*---------/\---------*
A / \ B
AC == BD
Proof: In rectangle ABCD, consider Triangle ABC and Triangle DCB. AB = DC (opposite sides of a rectangle are equal). Angle ABC = Angle DCB = 90 degrees. BC = CB (common side). By the SAS Congruence Criterion, Triangle ABC =~ Triangle DCB. By CPCTC, hypotenuse AC = hypotenuse BD.
Rhombuses and Perpendicular Diagonals
In a rhombus, all four sides are equal, producing two unique diagonal behaviors:
The Rhombus Diagonal Theorems:
1. The diagonals of a rhombus are perpendicular to each other:
AC is perpendicular to BD (Angle AOD = 90 degrees)
2. The diagonals of a rhombus bisect the interior angles at the vertices.
A
*
/|\
/ | \
D *--+--* B
\ |O/
\|/
*
C
Diagonals intersect at 90 deg!
Proof: Let diagonals AC and BD intersect at O. In a rhombus, AB = BC = CD = DA. Consider Triangle AOD and Triangle COD: side AD = CD (all sides of rhombus equal), side OD = OD (common), and AO = CO (diagonals of a parallelogram bisect each other). By the SSS Congruence Criterion, Triangle AOD =~ Triangle COD. By CPCTC, Angle AOD = Angle COD. Because these two angles form a linear pair along diagonal AC, Angle AOD + Angle COD = 180 degrees, so 2*Angle AOD = 180, which proves Angle AOD = 90 degrees.
The Square: The Ultimate Regular Quadrilateral
A square combines every single property of parallelograms, rectangles, and rhombuses:
1. All four sides are equal (AB = BC = CD = DA).
2. All four angles are right angles (90 degrees each).
3. Opposite sides are parallel (AB || CD and AD || BC).
4. Diagonals are equal in length (AC = BD).
5. Diagonals bisect each other at right angles (perpendicular bisectors: AC is perpendicular to BD).
6. Diagonals bisect the 90-degree corner angles into two 45-degree angles.
Chapter Exercises
Exercise 1: In a rectangle ABCD, the diagonals intersect at point O. If Angle BOC = 50 degrees, calculate the measure of Angle OAD.
Exercise 2: The diagonals of a rhombus ABCD have lengths 16 cm and 12 cm. Find the perimeter of the rhombus.
Exercise 3: Prove that if the diagonals of a parallelogram are equal and intersect at right angles, the parallelogram must be a square.
Solutions and Explanations
Solution 1: In rectangle ABCD, diagonals AC and BD are equal in length and bisect each other at point O, meaning OA = OB = OC = OD. Since OB = OC, triangle OBC is an isosceles triangle with Angle OBC = Angle OCB. We are given vertex Angle BOC = 50 degrees. In triangle OBC, Angle OBC + Angle OCB + 50 = 180, so 2*Angle OBC = 130, which gives Angle OBC = 65 degrees. Since each corner angle of the rectangle is 90 degrees: Angle OBA = 90 - 65 = 25 degrees. Triangle OAB is also isosceles with OA = OB, so Angle OAB = Angle OBA = 25 degrees. Finally, Angle OAD = 90 - Angle OAB = 90 - 25 = 65 degrees.
Solution 2: Let diagonals AC and BD intersect at point O. The diagonals of a rhombus bisect each other perpendicularly: AO = AC / 2 = 16 / 2 = 8 cm, and BO = BD / 2 = 12 / 2 = 6 cm. Triangle AOB is a right-angled triangle with the right angle at O. By the Pythagorean theorem: AB^2 = AO^2 + BO^2 = 8^2 + 6^2 = 64 + 36 = 100. Taking the square root gives side AB = sqrt(100) = 10 cm. Since all four sides of a rhombus are equal, the perimeter is 4 * side length = 4 * 10 = 40 cm.
Solution 3: We are given a parallelogram ABCD whose diagonals AC and BD are equal (AC = BD) and perpendicular (AC is perpendicular to BD). Since the diagonals of the parallelogram are equal in length (AC = BD), by the Rectangle Diagonal Theorem, ABCD must be a rectangle (meaning all four angles are 90 degrees). Since the diagonals of the parallelogram intersect at right angles, by the Rhombus Diagonal Theorem, ABCD must be a rhombus (meaning all four sides are equal). A quadrilateral that is simultaneously a rectangle (four right angles) and a rhombus (four equal sides) satisfies the complete definition of a square. Therefore, ABCD is a square.