Question

in the given figure find the value of a​

Answer 1

Answer:

Solution :

Here we have given that the exterior angles of quadrilateral are 66a⁰, 56⁰, 31⁰, 27⁰. We need to find the vale of a.

As we know that sum of exterior angles of quadrilateral is 360⁰.

So, now, finding the value of a.

$\begin{gathered} \quad{\dashrightarrow{\sf{Sum \: of \: exterior \: angles = {360}^{\circ}}}} \\\\\quad{\dashrightarrow{\sf{66{a}^{\circ} + 56{a}^{\circ} + 31{a}^{\circ} + 27{a}^{\circ} = {360}^{ \circ}}}}\\\\\quad{\dashrightarrow{\sf{122{a}^{\circ} + 31{a}^{\circ} + 27{a}^{\circ} = {360}^{ \circ}}}}\\\\\quad{\dashrightarrow{\sf{153{a}^{\circ} + 27{a}^{\circ} = {360}^{ \circ}}}}\\\\\quad{\dashrightarrow{\sf{180{a}^{\circ} = {360}^{\circ}}}}\\\\{\dashrightarrow{\sf{a = \frac{360}{180} }}}\\\\{\dashrightarrow{\sf{a = \cancel{\frac{360}{180}}}}}\\\\{\dashrightarrow{\sf{a = 2}}} \\ \\ \bigstar \: {\underline{\boxed{\frak{\pink{a = 2}}}}}\end{gathered}$

• Hence, the value of a is 2.

So, now after knowing the value of a. Finding the angles by multiplying by 2.

• ➝ 66a⁰ = 66 × 2 = 132⁰
• ➝ 56a⁰ = 56 × 2 = 112⁰
• ➝ 31a⁰ = 31 × 2 = 62⁰
• ➝ 27a⁰ = 27 × 2 = 54⁰

Hence, the angles of quadrilateral are 132⁰, 112⁰, 62⁰, 54⁰.

$\begin{gathered}\end{gathered}$

Vefication :

$\begin{gathered} \quad{\dashrightarrow{\sf{Sum \: of \: exterior \: angles = {360}^{\circ}}}} \\\\\quad{\dashrightarrow{\sf{66{a}^{\circ} + 56{a}^{\circ} + 31{a}^{\circ} + 27{a}^{\circ} = {360}^{ \circ}}}}\\ \\ \quad{\dashrightarrow{\sf{{132}^{\circ} + {112}^{\circ} + {62}^{\circ} + {54}^{\circ} = {360}^{ \circ}}}} \\ \\ \quad{\dashrightarrow{\sf{{244}^{\circ} + {62}^{\circ} + {54}^{\circ} = {360}^{ \circ}}}} \\ \\ \quad{\dashrightarrow{\sf{{306}^{\circ} + {54}^{\circ} = {360}^{ \circ}}}} \\ \\ {\dashrightarrow{\sf{{360}^{\circ} = {360}^{ \circ}}}} \\ \\ \bigstar \: {\underline{\boxed{\frak{\pink{LHS = RHS}}}}}\end{gathered}$

• Hence, Verified!

$\begin{gathered}\end{gathered}$

Learn More :

$\begin{gathered}\begin{gathered} \boxed{\begin{array}{l}\\ \large\dag\quad\underline{\bf Formulas\:of\:Areas:-}\\ \\ \star \: \: \sf Circle = \pi r^2 \\ \\ \star \: \; \sf Square=(side)^2\\ \\ \star\; \; \sf Rectangle=Length\times Breadth \\\\ \star \: \: \sf Triangle=\dfrac{1}{2}\times Breadth\times Height \\\\ \star \: \: \sf Scalene\triangle=\sqrt {s (s-a)(s-b)(s-c)}\\ \\ \star \: \: \sf Rhombus =\dfrac {1}{2}\times d_1\times d_2 \\\\ \star \: \: \sf Rhombus =\:\dfrac {1}{2}p\sqrt {4a^2-p^2}\\ \\ \star \: \: \sf Parallelogram =Breadth\times Height\\\\ \star \: \: \sf Trapezium =\dfrac {1}{2}(a+b)\times Height \\ \\ \star \: \: \sf Equilateral\:Triangle=\dfrac {\sqrt{3}}{4}(side)^2\end {array}}\end{gathered}\end{gathered}$

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Answer 2

$27 + a = 180$

$a = 180 - 27$

$a = 143$