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,

AC^{2} = AE^{2} + EC^{2}

⇒ (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 cm^{2} = 4√3 cm^{2}

Area = 4√3 cm^{2}

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