Question

class nine ncert ch 10A qus no 4
solve this
give correct ans
its urgent
wrong ans will be reported​

Answer 1

Answer:

Given: ABCD is a square in which AB = BC = CD = DA. ∆EDC is an equilateral triangle in which ED = EC = DC and

∠ EDC = ∠ DEC = ∠DCE = 60∘.

To prove: AE = BE and ∠DAE = 15∘

Proof: In ∆ADE and ∆BCE, we have:

AD = BC [Sides of a square]

DE = EC [Sides of an equilateral triangle]

∠ADE = ∠BCE = 90∘ + 60∘ = 150∘

∴ ∆ADE ≅ ∆BCE

i.e., AE = BE

Now, ∠ADE = 150∘

DA = DC [Sides of a square]

DC = DE [Sides of an equilateral triangle]

So, DA = DE

∆ADE and ∆BCE are isosceles triangles.

i.e., ∠DAE = ∠DEA =

1/2(180° - 150°) = 30°/2 = 15°

Hope this is helpful for you.