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Quadrilaterals - Rectangles and Squares

Grade 8CBSE

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

🔑Concepts

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A rectangle is a parallelogram where every interior angle is a right angle (90∘90^{\circ}). Its opposite sides are equal and parallel, and its diagonals are equal in length and bisect each other.

Rectangle ABCD with diagonals AC and BD intersecting at point O.
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A square is a special type of rectangle where all four sides are equal. It possesses all properties of a rectangle and a rhombus. Its diagonals are equal, bisect each other at right angles (90∘90^{\circ}), and bisect the vertex angles.

Square with equal sides and diagonals bisecting at 90 degrees.
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In a rectangle, the diagonals are equal. If ACAC and BDBD are diagonals of rectangle ABCDABCD, then AC=BDAC = BD.

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In a square, the angle between the diagonals is always 90∘90^{\circ}, making them perpendicular bisectors of each other.

📐Formulae

Perimeter of Rectangle=2(l+b)\text{Perimeter of Rectangle} = 2(l + b)

Area of Rectangle=l×b\text{Area of Rectangle} = l \times b

Diagonal of Rectangle=l2+b2\text{Diagonal of Rectangle} = \sqrt{l^2 + b^2}

Perimeter of Square=4×s\text{Perimeter of Square} = 4 \times s

Area of Square=s2\text{Area of Square} = s^2

Diagonal of Square=s2\text{Diagonal of Square} = s\sqrt{2}

💡Examples

Problem 1:

In a rectangle PQRSPQRS, the diagonals PRPR and QSQS intersect at OO. If OP=2x+4OP = 2x + 4 and OS=3x+1OS = 3x + 1, find the value of xx.

Solution:

In a rectangle, diagonals are equal and bisect each other. Therefore, PR=QSPR = QS and their halves are also equal. This means OP=OSOP = OS. 2x+4=3x+12x + 4 = 3x + 1 4−1=3x−2x4 - 1 = 3x - 2x x=3x = 3

Explanation:

Since diagonals of a rectangle bisect each other and are equal in length, the segments from the intersection point to any vertex are equal.

Problem 2:

Find the length of the diagonal of a square whose side is 55 cm.

Solution:

Given side s=5s = 5 cm. The formula for the diagonal of a square is d=s2d = s\sqrt{2}. d=52 cmd = 5\sqrt{2} \text{ cm} If we use 2≈1.414\sqrt{2} \approx 1.414: d≈5×1.414=7.07 cmd \approx 5 \times 1.414 = 7.07 \text{ cm}

Explanation:

The diagonal of a square forms a right-angled triangle with two sides. Using Pythagoras theorem: d2=s2+s2=2s2d^2 = s^2 + s^2 = 2s^2, hence d=s2d = s\sqrt{2}.

Problem 3:

The perimeter of a rectangle is 4040 cm. If its length is 1212 cm, calculate its area.

Solution:

Given Perimeter P=40P = 40 cm and length l=12l = 12 cm. P=2(l+b)P = 2(l + b) 40=2(12+b)40 = 2(12 + b) 20=12+b20 = 12 + b b=20−12=8 cmb = 20 - 12 = 8 \text{ cm} Now, Area A=l×bA = l \times b: A=12×8=96 cm2A = 12 \times 8 = 96 \text{ cm}^2

Explanation:

First, use the perimeter formula to find the unknown breadth, then use the area formula.

Problem 4:

In the given rectangle ABCDABCD, the length is 88 cm and the diagonal ACAC is 1010 cm. Find the breadth BCBC of the rectangle.

Rectangle ABCD with diagonal AC=10 and side AB=8

Solution:

In rectangle ABCDABCD, ∠ABC=90∘\angle ABC = 90^{\circ}. Thus, △ABC\triangle ABC is a right-angled triangle. Using Pythagoras theorem: AB2+BC2=AC2AB^2 + BC^2 = AC^2 82+BC2=1028^2 + BC^2 = 10^2 64+BC2=10064 + BC^2 = 100 BC2=100−64BC^2 = 100 - 64 BC2=36BC^2 = 36 BC=36=6 cmBC = \sqrt{36} = 6 \text{ cm}

Explanation:

Since all angles in a rectangle are 90∘90^{\circ}, we can use the Pythagorean theorem on the triangle formed by the length, breadth, and diagonal.

Problem 5:

Square PQRSPQRS has diagonals intersecting at OO. If ∠OPQ=(3x−15)∘\angle OPQ = (3x - 15)^{\circ}, find the value of xx.

Square PQRS with diagonals intersecting at O

Solution:

In a square, the diagonals bisect the vertex angles. Since each vertex angle is 90∘90^{\circ}, the diagonal bisects it into two 45∘45^{\circ} angles. Therefore, ∠OPQ=45∘\angle OPQ = 45^{\circ}. 3x−15=453x - 15 = 45 3x=45+153x = 45 + 15 3x=603x = 60 x=603x = \frac{60}{3} x=20x = 20

Explanation:

Diagonals of a square are angle bisectors of the interior 90∘90^{\circ} angles, so the angle between a side and a diagonal is always 45∘45^{\circ}.