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Playing with Constructions - Exploring Diagonals of Rectangles and Squares

Grade 6CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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A rectangle is a quadrilateral where all four angles are right angles. One of its unique properties is that its diagonals are equal in length (d1=d2d_1 = d_2) and they bisect each other.

Rectangle ABCD showing equal diagonals AC and BD
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In a square, the diagonals are not only equal in length but also bisect each other at right angles (90∘90^\circ). This means the diagonals are perpendicular to each other.

Square PQRS showing perpendicular bisecting diagonals
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The point where the diagonals intersect is the midpoint of both diagonals. For both rectangles and squares, the distance from the center (intersection point) to any vertex is equal.

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A square is a special type of rectangle where all sides are equal. Therefore, it inherits the property of equal diagonals from the rectangle and adds the property of perpendicularity.

📐Formulae

For a Rectangle: d1=d2\text{For a Rectangle: } d_1 = d_2

For a Square: d1=d2 and d1⊥d2\text{For a Square: } d_1 = d_2 \text{ and } d_1 \perp d_2

Angle between diagonals of a square=90∘\text{Angle between diagonals of a square} = 90^\circ

💡Examples

Problem 1:

In a rectangle ABCDABCD, if the length of the diagonal ACAC is 7 cm7\text{ cm}, what is the length of the diagonal BDBD?

Solution:

In a rectangle, the diagonals are always equal in length. Therefore, BD=AC=7 cmBD = AC = 7\text{ cm}.

Explanation:

Since ABCDABCD is a rectangle, the property of equal diagonals applies.

Problem 2:

In a square PQRSPQRS, the diagonals PRPR and QSQS intersect at point OO. What is the measure of ∠POS\angle POS?

Solution:

∠POS=90∘\angle POS = 90^\circ

Explanation:

In a square, the diagonals are perpendicular bisectors of each other. This means the angle formed at the point of intersection is always a right angle (90∘90^\circ).

Problem 3:

If the diagonals of a quadrilateral are equal and bisect each other at right angles, identify the shape.

Solution:

The shape is a Square.

Explanation:

A rectangle has equal diagonals that bisect each other, but they do not necessarily meet at 90∘90^\circ. A rhombus has diagonals that meet at 90∘90^\circ, but they are not equal. Only a square satisfies all three conditions: equal length, bisecting, and perpendicularity.

Problem 4:

In a rectangle KLMNKLMN, the diagonals KMKM and LNLN intersect at point OO. If KO=4 cmKO = 4\text{ cm}, find the length of the diagonal LNLN.

Rectangle KLMN with diagonals intersecting at O and KO marked as 4cm

Solution:

KO=4 cmKO = 4\text{ cm} KM=2×KO=2×4=8 cmKM = 2 \times KO = 2 \times 4 = 8\text{ cm} In a rectangle, LN=KM\text{In a rectangle, } LN = KM LN=8 cmLN = 8\text{ cm}

Explanation:

Since the diagonals of a rectangle bisect each other, the full diagonal KMKM is twice the length of the segment KOKO. Because diagonals of a rectangle are equal, LNLN must be the same length as KMKM.

Problem 5:

In a square WXYZWXYZ, the diagonals WYWY and XZXZ meet at OO. If ∠WOX=(3x+30)∘\angle WOX = (3x + 30)^\circ, find the value of xx.

Square WXYZ with angle WOX labeled

Solution:

∠WOX=90∘ (Diagonals of a square are perpendicular)\angle WOX = 90^\circ \text{ (Diagonals of a square are perpendicular)} 3x+30=903x + 30 = 90 3x=90−303x = 90 - 30 3x=603x = 60 x=603x = \frac{60}{3} x=20x = 20

Explanation:

The diagonals of a square always intersect at right angles (90∘90^\circ). By setting the given expression for the angle equal to 90, we can solve for the unknown variable xx.