Math - Geometry Lesson Plans : Perimeter & Area of Quadrilaterals II

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Mensuration - Formulae for Perimeter and Area of Quadrilaterals II

Perimeter of a rectangle = 2 (Length + Breadth)

Diagonal of a rectangle = Ö

Length^{2} + Breadth^{2}

Area of a rectangle = Length × Breadth

Perimeter of a square = 4 × Side

Diagonal of a square = Ö2 Side

Area of a square = Side^{2}

Area of a square = ½ × Diagonal^{2}

Area of a quadrilateral when diagonals intersect at right angles = ½ × Product of diagonals

Area of a quadrilateral when one diagonal and the lengths of the perpendiculars from its opposite vertices to this diagonal are given = ½ × Diagonal × Sum of the lengths of the perpendiculars

Area of a parallelogram = Base × Height

Area of a rhombus = ½ × Product of diagonals

Area of a trapezoid = ½ × Sum of parallel sides × Distance between parallel sides

Example
Calculate the perimeter (in cm) of the quadrilateral EFGH in which EF = 42 cm, GH = 74 cm, EH = 56 cm, FH = 70 cm and ∠GFH = 90°. Solution.
In ΔFGH, by Pythagorean Theorem, GH^{2} = FG^{2} + FH^{2}
∴ FG^{2} = GH^{2} − FH^{2} = 74^{2} − 70^{2} = 576 or FG = 24 cm.
Perimeter of the quadrilateral = EF + FG + EH + GH = 42 + 24 + 56 + 74 = 196 cm.

Example
Find the area (in cm^{2}) of the quadrilateral PQRS in which PQ = 36 cm, QR = RS = 50 cm, PS = 48 cm and ∠QPS = 90°. Solution.
By Pythagorean Theorem, QS^{2} = PQ^{2} + PS^{2}
= 36^{2} + 48^{2} = 3600 or QS = 60 cm.
Area of ΔPQS = ½ × Base × Height
= ½ (36) (48) = 864 cm^{2}.
Area of isosceles ΔQRS = ¼ bÖ(4a^{2} − b^{2})
= ¼ (60) Ö(4 × 50^{2} − 60^{2}) = 1200 cm^{2}.
Area of the quadrilateral = Area of ΔPQS + Area of ΔQRS
= 864 + 1200 = 2064 cm^{2}.

Example
Find the area (in cm^{2}) of the trapezoid ABCD in which AB = 22 cm, BC = 50 cm, CD = 8 cm and ∠BAD = ∠ADC = 90°. Solution.
Construction: Draw CE ⊥ AB.
Then, AE = 8 cm.
∴ BE = 22 − 8 = 14 cm.
By Pythagorean Theorem, BC^{2} = BE^{2} + CE^{2}
∴ CE^{2} = BC^{2} − BE^{2} = 50^{2} − 14^{2} = 2304 or CE = 48 cm.
Area of ΔBCE = ½ × Base × Height
= ½ × 48 × 14 = 336 cm^{2}.
Area of rectangle ADCE = Length × Breadth
= 48 × 8 = 384 cm^{2}.
Area of the trapezoid = Area of ΔBCE + Area of rectangle ADCE
= 336 + 384 = 720 cm^{2}.

Example
The shaded region of the given figure is a villa 68 m long and 24 m broad. It is bordered by a lawn of uniform breadth 6 m on three sides. Find the total area (in m^{2}) occupied by the villa and the lawn. Solution.
Total area occupied by the villa and the lawn = Length × Breadth
= 80 × 30 = 2400 m^{2}.