Mathematics
Find the area enclosed by the figure (i) given below, where ABC is an equilateral triangle and DEFG is an isosceles trapezium. All measurements are in centimeters.
Mensuration
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Answer
In right angle triangle ECF,
Using pythagoras theorem,
⇒ EF2 = EC2 + CF2
⇒ 52 = EC2 + 32
⇒ EC2 = 52 - 32
⇒ EC2 = 25 - 9 = 16
⇒ EC = = 4 cm.
Since, DEFG is an isosceles trapezium.
∴ GD = EF= 5 cm.
Since, BDEC is a rectangle,
∴ BD = EC = 4 cm and BC = DE = 6 cm.
In right angle triangle DBG,
Using pythagoras theorem,
⇒ GD2 = BD2 + GB2
⇒ 52 = 42 + GB2
⇒ GB2 = 52 - 42
⇒ GB2 = 25 - 16 = 9
⇒ GB = = 3 cm.
In trapezium,
GF = GB + BC + CF = 3 + 6 + 3 = 12 cm.
Area of trapezium DEFG = (sum of parallel sides) × distance between them
=
= 18 × 2
= 36 cm2.
Area of equilateral triangle ABC =
=
=
= 1.732 × 9
= 15.59 cm2
Area of figure = Area of trapezium DEFG + Area of equilateral triangle ABC
= 36 + 15.59 = 51.59 cm2.
Hence, area of figure = 51.59 cm2.
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