Q. 1 B5.0( 5 Votes )
Calculate the areas of the parallelograms shown below:
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
We can consider Δ AEC as a half of equilateral Δ ACL.
Since, Δ ACL is an equilateral triangle.
AL = AC
⇒ (x + x) = 2 cm
⇒ 2x = 2 cm
⇒ x = 1 cm
Using the Pythagoras theorem in Δ AEC ,
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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