Question

in the below figure find x​

Answer 1

To find

• Value of x

Solution

→ 140° + ∠DAB = 180° [ Linear pair ]

→ ∠DAB = 180° - 140°

→ ∠DAB = 40°

Again

→ 130° + ∠DCB = 180° [ Linear pair ]

→ ∠DCB = 180° - 130°

→ ∠DCB = 50°

As we know that

• Sum of angles of a quadrilateral is equal to 360°

→ 140° + ∠DAB + ∠DCB + x = 360°

→ 140° + 40° + 50° + x = 360°

→ 180° + 50° + x = 360°

→ 230° + x = 360°

→ x = 360° - 230°

•°• x = 130°

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Answer 2

Note : For Understanding The Answer Very Clearly, We need Some Editing In Picture, The Edited Picture is in Attachment and Whole Answer is Designed Based on that Editing !! Thankyou !

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Firstly, Understand Question. We are Asked to Find the Value of x in :

$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\put(0,0){\line(-1,1){3}}\put(0,0){\line(1,1){3}}\put(-3,3){\line(1,1){3}}\put(3,3){\line(-1,1){3}}\put(-3,3){\vector(-1,1){1}}\put(3,3){\vector(1,-1){1}}\qbezier(-0.4,0.4)(0,0.7)(0.4,0.4)\qbezier(-0.4,5.6)(0,5.3)(0.4,5.6)\qbezier(-2.6,3.4)(-3,3.7)(-3.4,3.3)\qbezier(2.6,2.6)(3,2.3)(3.4,2.6)\put(-0.1,5){\bf\large x}\put(-0.4,0.8){ \bf 140^{\circ} }\put(2.6,2){ \bf 130^{\circ} }\put(-3.4,3.8){ \bf 140^{\circ} }\end{picture}$

It's Very Simple !! To Do This Problem, We need to find the Value of x, For Finding x you first need to Find y and z using Linear Pair Property in which Sum of Adjacent Angles is 180° . When You Equate The Angles y and z with Adjacent Angle to 180°, Then We Are There with y and z. Using Quadrilateral Property : Sum of interior angles of a quadrilateral is equal to 360°, We get The Value of x !!

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Before Going into Answer, It's Important to know about some Basic Terms related to the Answer !!

• Angle : An angle is a combination of two rays with a common endpoint (Vertex) . An angle is represented by the symbol "∠"

• Linear Pair : When The Measures Adjacent Angles adds upto 180°, Then the angles form Linear Pair. Linear pairs are always supplementary. The angles are said to be linear if they are adjacent to each other after the intersection of the two lines.

$\setlength{\unitlength}{1 cm}\begin{picture}(0,0)\thicklines\put(5,1){\vector(1,0){4}}\put(5,1){\vector(-1,0){4}}\put(5,1){\vector(1,1){3}}\put(1,3){ \boxed{\large\sf a + b = 180^{\circ}} }\put(4.5,1.3){ \sf a^{\circ} }\put(5.7,1.3){ \sf b^{\circ} }\end{picture}$

• Quadrilateral : Quadrilateral is a closed, two-dimensional shape which has four straight sides. It has 4 vertices and 4 edges. Sum of interior angles of a quadrilateral is equal to 360°

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From the Attachment

• ∠EAC = 140°

• ∠BDF = 130°

• ∠ABD = 140°

• We can observe We formed y or (∠CAB) and z or (∠CDB) as Other Two Angles in the Quadrilateral

✮ Solving For y :

⇒ y + 140° = 180°     (Linear Pair)

⇒ y + 140° - 140° = 180° - 140°

⇒ y = 40°

✮ Solving For z :

⇒ z + 130° = 180°     (Linear Pair)

⇒ z + 130° - 130° = 180° - 130°  

⇒ z = 50°

➠ We know : Sum of interior angles of a quadrilateral is equal to 360°

The Interior Angles Present In Our Quadrilateral :

• x , y , z , ∠ABD

⇒ x + y + z + ∠ABD = 360°

⇒ x + 40° + 50° + 140° = 360°

⇒ x + 230° = 360°

⇒ x + 230° - 230°  = 360° - 230°

⇒ x = 130°

$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\put(0,0){\line(-1,1){3}}\put(0,0){\line(1,1){3}}\put(-3,3){\line(1,1){3}}\put(3,3){\line(-1,1){3}}\put(-3,3){\vector(-1,1){1}}\put(3,3){\vector(1,-1){1}}\qbezier(-0.4,0.4)(0,0.7)(0.4,0.4)\qbezier(-0.4,5.6)(0,5.3)(0.4,5.6)\qbezier(-2.6,3.4)(-3,3.7)(-3.4,3.3)\qbezier(2.6,2.6)(3,2.3)(3.4,2.6)\put(-0.1,5){ \bf 130^{\circ} }\put(-0.4,0.8){ \bf 140^{\circ} }\put(2.6,2){ \bf 130^{\circ} }\put(-3.4,3.8){ \bf 140^{\circ} }\end{picture}$

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