
ABCD is a quadrilateral and AP and DP are bisectors of ∠A and ∠D. The value of \[{\text{x}}\] is
Answer
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Hint: In Euclidean plane geometry, a quadrilateral is a polygon with four edges (sides) and four vertices (corners). Other names for quadrilateral include quadrangle (in analogy to triangle), tetragon (in analogy to pentagon, \[{\text{5}}\]-sided polygon, and hexagon, \[{\text{6}}\]-sided polygon), and \[{\text{4}}\]-gon (in analogy to k - gons for arbitrary values of k). A quadrilateral with vertices A, B, C and D is sometimes denoted as \[\square {\text{ABCD}}\].
The interior angles of a simple (and planar) quadrilateral ABCD add up to \[{\text{360}}\] degrees of arc, that is.
\[\angle {\text{A}} + \angle {\text{B}} + \angle {\text{C}} + \angle \;{\text{D }} = {\text{ 36}}0\]
This is a special case of the n-gon interior angle sum formula: (n − 2) × 180°
In a convex quadrilateral, all interior angles are less than 180°, and the two diagonals both lie inside the quadrilateral. The two diagonals of a convex quadrilateral are the line segments that connect opposite vertices.
In a concave quadrilateral, one interior angle is bigger than 180°, and one of the two diagonals lies outside the quadrilateral.
Complete answer:
As we know that the sum of all angles of a quadrilateral is \[{\text{36}}{{\text{0}}^o}\]
So we get
\[\angle {\text{A}} + \angle {\text{B}} + \angle {\text{C}} + \angle \;{\text{D }} = {\text{ 36}}{0^o}\]
\[\angle {\text{A}} + \angle {\text{D}}\; = {\text{ 17}}{0^o}\]
\[\dfrac{{\angle {\text{A}}}}{{\text{2}}} + \dfrac{{\angle {\text{D}}}}{{\text{2}}}\; = {\text{ 8}}{{\text{5}}^o}\]
Now in the triangle ADP
\[\dfrac{{\angle {\text{A}}}}{{\text{2}}}{\text{ }} + \dfrac{{\angle {\text{D}}}}{{\text{2}}}{\text{ }} + \angle {\text{x}}\; = {\text{ 18}}{{\text{0}}^o}\]
=> \[\angle {\text{x}} = {\text{9}}{{\text{5}}^o}\]
Hence, the value of \[{\text{x}} = {\text{9}}{{\text{5}}^o}\]
Note: Quadrilaterals can be classified into Parallelograms, Squares, Rectangles, and Rhombuses. Square, Rectangle, and Rhombus are also Parallelograms.
The interior angles of a simple (and planar) quadrilateral ABCD add up to \[{\text{360}}\] degrees of arc, that is.
\[\angle {\text{A}} + \angle {\text{B}} + \angle {\text{C}} + \angle \;{\text{D }} = {\text{ 36}}0\]
This is a special case of the n-gon interior angle sum formula: (n − 2) × 180°
In a convex quadrilateral, all interior angles are less than 180°, and the two diagonals both lie inside the quadrilateral. The two diagonals of a convex quadrilateral are the line segments that connect opposite vertices.
In a concave quadrilateral, one interior angle is bigger than 180°, and one of the two diagonals lies outside the quadrilateral.
Complete answer:
As we know that the sum of all angles of a quadrilateral is \[{\text{36}}{{\text{0}}^o}\]
So we get
\[\angle {\text{A}} + \angle {\text{B}} + \angle {\text{C}} + \angle \;{\text{D }} = {\text{ 36}}{0^o}\]
\[\angle {\text{A}} + \angle {\text{D}}\; = {\text{ 17}}{0^o}\]
\[\dfrac{{\angle {\text{A}}}}{{\text{2}}} + \dfrac{{\angle {\text{D}}}}{{\text{2}}}\; = {\text{ 8}}{{\text{5}}^o}\]
Now in the triangle ADP
\[\dfrac{{\angle {\text{A}}}}{{\text{2}}}{\text{ }} + \dfrac{{\angle {\text{D}}}}{{\text{2}}}{\text{ }} + \angle {\text{x}}\; = {\text{ 18}}{{\text{0}}^o}\]
=> \[\angle {\text{x}} = {\text{9}}{{\text{5}}^o}\]
Hence, the value of \[{\text{x}} = {\text{9}}{{\text{5}}^o}\]
Note: Quadrilaterals can be classified into Parallelograms, Squares, Rectangles, and Rhombuses. Square, Rectangle, and Rhombus are also Parallelograms.
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