Question

So,janta how would you attempt this question​

Answer 1

Given :-

• in ∆ACB CD = CB
• CE = 6 cm.
• ∠CEB = 60°
• ∠ACB = 120°

To Find :-

• Area ∆DEC.

Construction :-

• Draw CF ⟂ EB .

Solution :- (Excellent Question.)

❁❁ Refer To Image First .. ❁❁

From image in Right ∆CFE :-

→ ∠CEF = 60°

→ CE = Hypotenuse = 6cm.

So,

→ sin60° = (Perpendicular)/(Hypotenuse) .

→ (√3/2) = CF / CE

→ (√3/2) = CF / 6

→ CF = 3√3 cm.

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Now, in Right ∆CFB, we have :-

→ sinθ = CF / CB

→ CB = (3√3/sinθ) -------------------------- Equation ❶

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Now, in ∆DCE we have :-

→ CB = CD (Given).

So,

→ CD = (3√3/sinθ) { From Equation ❶ }.

→ CE = 6 cm. (Given).

→ ∠DCE = θ

So,

→ ∆DCE Area = (1/2) * CD * CE * ( Sine Angle b/w these two sides).

Putting values we get :-

→ Area [∆DCE] = (1/2) * (3√3/sinsinθ) * 6 * sinθ

→ Area [∆DCE] = 9√3 (unit)².

Hence, Area of ∆DCE with orange Shaded Region is 9√3 unit².

[ Best Question For Basic Geometry .]

Answer 2

$\huge{\fbox{\fbox{\bigstar{\mathfrak{\red{Solution-}}}}}}$

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Given -

∆ ABC CD = CB

CE = 6 cm

angle CEB = 60°

angle ACB = 120°

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To find -

Area of ∆ DEC

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Construct-

Drawing CF bicetor to EB

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♀️From image,♀️

______________________

In ∆CFE,

➠/_CEF = 60°

➠CE = Hypotenus = 6 centimetres

So,

➠sin60° = (perpendicular)/(Hypotenus)

➠(√3/2) = CF / CE

➠(√3/2) = CF / 6

➠CF = 3 √ 3

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In ∆CEB,

➩sinθ = CF / CB

➩ CB = (3√3/sinθ) ____(eq.1)

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In ∆DCE,

⇝CB = CD (given)

⇝CD = (3√3/ sinθ ) ___using (eq.1)

⇝CE = 6 centimetres

⇝angle DCE = thetha

so,

Area of ∆DCE = 1/2 × CD × CE

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Putting values altogether,

⟹Area[∆ACE] = 1/2 × (3√3/sinsinθ ) × 6 × sinθ

⟹ Area [∆DCE ] =9√3(units)²

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