Question

in a figure ABCD is a rectangle and DEC is an equilateral triangle . find the area of shaded region​

Answer 1

Answer:

Area of shaded region = (48 - 9√3) cm²

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Explanation :

G I V E N :

An equilateral triangle DEC with side = 6 cm

A rectangle ABCD with dimensions (L × B) = 8 cm × 6 m.

T OㅤF I N D :

Area of shaded region = ?

S O L U T I O N :$\:$

✇ Finding area of rectangle ABCD :

➵ Area (rectangle ABCD) = L × B

➵ Area (rectangle ABCD) = 8 × 6

➵ Area (rectangle ABCD) = 48 cm² ㅤ(ℹ)

✇ Finding area of triangle DEC :

➵ Area (triangle DEC) = √3/4(side)²

➵ Area (triangle DEC) = √3/4(6)²

➵ Area (triangle DEC) = √3/4 × 6 × 6

➵ Area (triangle DEC) = √3/2 × 3 × 6

➵ Area (triangle DEC) = √3 × 3 × 3

➵ Area (triangle DEC) = √3 × 9

➵ Area (triangle DEC) = 9√3 cm²ㅤㅤ (ℹℹ)

✇ Finding area of shaded region :

➵ Area (shaded region) = (ℹ) - (ℹℹ)

➵ Area (shaded region) = (48 - 9√3) cm²

∴ Hence, area of shaded region = (48 - ㅤ9√3) cm²

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Answer 2

Answer:

The required area is $(48-9\sqrt{3} )$sq cm.

Step-by-step explanation:

It is given that ABCD is a rectangle and DEC is an equilateral triangle.
We need to find the area of shaded region.

Area of rectangle is $A=L*B$

$L=6cm$

$B=8cm$

$A=6*8=48$sq cm.

Area of equilateral triangle is given by $\frac{\sqrt{3} }{4} side^{2}$

Side is $6cm$

Area is $\frac{\sqrt{3} }{4} 6^{2}$

$=\frac{\sqrt{3} }{4} 36=9\sqrt{3}$sq cm.

The shaded area is $(48-9\sqrt{3} )$sq cm.

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