Take any quadrilateral, say ABCD (Fig 3.4). Divide it into two triangles, by drawing a diagonal. You get six angles 1, 2, 3, 4, 5 and 6. Use the angle-sum property of a triangle and argue how the sum of the measures of ∠A, ∠B, ∠C and ∠D amounts to

Q1. Take any quadrilateral, say ABCD (Fig 3.4). Divide it into two triangles, by drawing a diagonal. You get six angles 1, 2, 3, 4, 5 and 6. Use the angle-sum property of a triangle and argue how the sum of the measures of and amounts to $180\degree + 180\degree = 360\degree$ .

Answers (1)

As shown in $\triangle$ ACD,

$\angle 1 + \angle 2+\angle 3 = 180\degree$

As shown in $\triangle$ ABC,

$\angle 4 + \angle 5+\angle 6 = 180\degree$

$\angle A + \angle B +\angle C + \angle D= (\angle 1+\angle 4) +\angle 6 + (\angle 2+\angle 5) + \angle 3$ ( Since, $\angle A = (\angle 1+\angle 4)$ , $\angle B = \angle 6$ , $\angle C = (\angle 2+\angle 5)$ ,

$\angle D = \angle 3$ )

$\angle A + \angle B +\angle C + \angle D= (\angle 1+\angle 2 +\angle 3) + (\angle 4+\angle 5 + \angle 6)$

$= 180\degree + 180\degree$

$= 360\degree$

Hence proved, the sum of the measures of $\angle A,\angle B,\angle C$ and
$\angle D$ amounts to $180\degree + 180\degree = 360\degree$ .

Posted by

seema garhwal

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Q1. Take any quadrilateral, say ABCD (Fig 3.4). Divide it into two triangles, by drawing a diagonal. You get six angles 1, 2, 3, 4, 5 and 6. Use the angle-sum property of a triangle and argue how the sum of the measures of and amounts to .

Answers (1)

Posted by

seema garhwal

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Q1. Take any quadrilateral, say ABCD (Fig 3.4). Divide it into two triangles, by drawing a diagonal. You get six angles 1, 2, 3, 4, 5 and 6. Use the angle-sum property of a triangle and argue how the sum of the measures of and amounts to $180\degree + 180\degree = 360\degree$ .