
In the figure AB = BC = CD = DE = EF = FG = GA, then angle DAE is equal to
$\left( a \right){24^o}$
$\left( b \right){25^o}$
$\left( c \right){27^o}$
$\left( d \right)\dfrac{{{{180}^o}}}{7}$
Answer
575.1k+ views
Hint: In this particular question we will use the concept of the theorem, that is, “angles opposite to the equal sides of the triangle are also equal” and also use the exterior angle property i.e. in any triangle ABC the exterior angle of angle C is equal to the sum of the remaining two angles A and B, so use these concepts to reach the solution of the question.
Complete step-by-step answer:
Now moving to the question,
Let us assume, $\angle $DAE = x
Triangle ABC is isosceles as AB = AC
Therefore, $\angle $BCA = $\angle $CAB = x [angles opposite to equal sides of a triangle are also equal]
Hence, $\angle $CBD = $\angle $CAB + $\angle $BCA = x + x = 2x [external angle of triangle ABC]
Triangle BCD is isosceles as BC = CD
$\angle $CBD = $\angle $CDB = 2x [angles opposite to equal sides of triangle are also equal]
Hence, $\angle $DCE = $\angle $DAE + $\angle $CDA = x + 2x = 3x [exterior angle of triangle ACD]
Triangle CDE is isosceles as CD = DE
$\angle $DCE = $\angle $DEC = $\angle $AED = 3x
Similarly,
$\angle $ADE = $\angle $EFD = $\angle $AEF + $\angle $DAE = $\angle $EGF + $\angle $DAE = ($\angle $DAE + $\angle $GFA) + $\angle $DAE = $\angle $DAE + $\angle $DAE + $\angle $DAE = 3x
Hence in triangle ADE,
$\angle $ADE + $\angle $DAE + $\angle $AED = ${180^o}$ [angle sum property of triangle]
Now substituting the values we have,
$ \Rightarrow $ 3x + x + 3x = 7x = ${180^o}$
Hence, 7x = ${180^o}$
$ \Rightarrow $ x = $\dfrac{{{{180}^o}}}{7}$
So this is the required answer.
Hence option (d) is the correct answer.
Note: Whenever we face such types of questions the key concept we have to remember is that an Isosceles triangle is a triangle which has two equal sides. Also, the angles opposite to the equal sides are equal. Suppose in a triangle ABC, if sides AB and AC are equal, then ABC is an isosceles triangle where angle B = angle C, so if we consider angle DAE is equal to x so by using this property first we have to calculate all the angles in terms of x as above then use the property that in any triangle the sum of all angles are equal to 180 degrees.
Complete step-by-step answer:
Now moving to the question,
Let us assume, $\angle $DAE = x
Triangle ABC is isosceles as AB = AC
Therefore, $\angle $BCA = $\angle $CAB = x [angles opposite to equal sides of a triangle are also equal]
Hence, $\angle $CBD = $\angle $CAB + $\angle $BCA = x + x = 2x [external angle of triangle ABC]
Triangle BCD is isosceles as BC = CD
$\angle $CBD = $\angle $CDB = 2x [angles opposite to equal sides of triangle are also equal]
Hence, $\angle $DCE = $\angle $DAE + $\angle $CDA = x + 2x = 3x [exterior angle of triangle ACD]
Triangle CDE is isosceles as CD = DE
$\angle $DCE = $\angle $DEC = $\angle $AED = 3x
Similarly,
$\angle $ADE = $\angle $EFD = $\angle $AEF + $\angle $DAE = $\angle $EGF + $\angle $DAE = ($\angle $DAE + $\angle $GFA) + $\angle $DAE = $\angle $DAE + $\angle $DAE + $\angle $DAE = 3x
Hence in triangle ADE,
$\angle $ADE + $\angle $DAE + $\angle $AED = ${180^o}$ [angle sum property of triangle]
Now substituting the values we have,
$ \Rightarrow $ 3x + x + 3x = 7x = ${180^o}$
Hence, 7x = ${180^o}$
$ \Rightarrow $ x = $\dfrac{{{{180}^o}}}{7}$
So this is the required answer.
Hence option (d) is the correct answer.
Note: Whenever we face such types of questions the key concept we have to remember is that an Isosceles triangle is a triangle which has two equal sides. Also, the angles opposite to the equal sides are equal. Suppose in a triangle ABC, if sides AB and AC are equal, then ABC is an isosceles triangle where angle B = angle C, so if we consider angle DAE is equal to x so by using this property first we have to calculate all the angles in terms of x as above then use the property that in any triangle the sum of all angles are equal to 180 degrees.
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