Question

answer the question with explanation

Answer 1

Question 1 :

In the the given figure, there are two triangles.

1st triangle = Δ ACE

2nd triangle = Δ DBF

By using formula, ( n - 2 ) × 180 we can get the sum of all angles of any quadrilateral, where n is the number of sides present in the quadrilateral. As Δ ACE and Δ DBF are triangle ( quadrilateral ), using formula,

        Sum of all angles of a triangle = ( n - 1 ) × 180

triangle has three sides, therefore n = 3

   

        Sum of all angles of a triangle  = ( 3 - 2 ) × 180

                                                              = 1 × 180

                                                              = 180°

  sum of all angles  = ∠A + ∠B + ∠C + ∠D + ∠E + ∠F

                                 = ∠A + ∠C + ∠E + ∠D + ∠F + ∠B

                                 = all angles of Δ ACE + all angles of ΔDBF

                                 = 180 + 180       { Sum of all ∠s of Δ = 180 }

                                  = 360°

Therefore, ∠A + ∠B + ∠C + ∠D + ∠E + ∠F = 360°

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Answer 2

In the given diagram <A , <C and <E are the angles of ∆ AEC which are at vertices. We know that the sum of all angles of triangle present at the vertix is 180°. Therefore <A + <C + <E = 180°



<A + <C + <E = 180° --- 1



In the given diagram <D , <B and <F are the angles of ∆ DBF which are at vertices. We know that the sum of all angles of triangle present at the vertix is 180°. Therefore <D + <B + <F = 180°


<D + <B + <F = 180° ---- 2




Adding 1 & 2 ,


<A + <C + <E + <D + <B + <F = 180° + 180°


<A + <B + <C + <D + <E + <F = 360°





Hence,
<A + <B + <C + <D + <E + <F is 360°

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