Q. 1 B5.0( 5 Votes )

Calculate the areas of the parallelograms shown below:


Answer :

Consider a parallelogram ABDC.
Let us draw a perpendicular on base AB from C say, CE.



Let length of AE = x cm and CE = y cm.


In Δ AEC:


CAE = 60° and CEA = 90°
ACE = 30° ( C + E + A = 180° )



We know that sides of any triangle of angles 30°, 60° and 90°
are in the ratio 1:√3:2.


x:y:2 = 1:√3:2
x = 1 cm and y = √3 cm


OR
We can consider Δ AEC as a half of equilateral Δ ACL.



Since, Δ ACL is an equilateral triangle.
We get,


AL = AC
(x + x) = 2 cm
2x = 2 cm
x = 1 cm


Using the Pythagoras theorem in Δ AEC ,
we get,


AC2 = AE2 + EC2
(2)2 = (1)2 + (y)2
4 - 1 = (y)2
(3) = (y)2
y = √3 cm


CE = √3 cm and also line segment CE is a height of parallelogram ABDC.
Also base AB = 4 cm.


Thus area of parallelogram = base × height


area (ABDC) = 4 × √3 cm2 = 4√3 cm2


Area = 4√3 cm2


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Kerala Board Mathematics Part-1