1. find the sum of the measures of exterior angles, one at each vertex, of an octagon.
2 answers:
Answer: First question: b. 360°
Second question : b. 7
Third question: Yes, it is a parallelogram.
Fourth question: It is a rhombus.
Step-by-step explanation:
1. Since, The sum of exterior angles in a polygon is always equal to 360 degrees.
And, an octagon is also a polygon.
Therefore the sum of the all exterior angle of an octagon = 360°
2. Since, the sum of the interior angles of a polygon of n sides is (n-2)×180°
But here (n-2)×180°=900
⇒ n-2 = 5 ( After dividing both sides by 180° )
⇒ n = 5 + 2 = 7
Thus, If the sum of the interior angles of a polygon is 900° then the polygon has 7 sides.
10. Since, In triangles XNY and WNZ,
XN≅NZ (Given)
NY≅NW ( Given)
∠XNY≅∠WNZ ( vertically opposite angles)
Thus, By SAS postulate of congruence,
ΔXNY≅ΔWNZ,
By CPCTC, XY≅WZ
Similarly, Δ XNW≅ Δ YNZ,
By CPCTC, XW≅YZ
Therefore, In quadrilateral XYZW, Opposite sides are equal.
⇒ XYZW is a parallelogram.
11. Let ABCD is a parallelogram,
In which AC is a diagonal.
Also, AB = CD and AD= BC ( By the property of parallelogram)
And, It is given that ∠DAC=∠DCA = 72°
Therefore ADC is an isosceles triangle ( By the property of isosceles triangle)
Thus, AD=DC
Similarly, ABC is an isosceles triangle.
Thus, AB= BC
Thus, AB=BC=CD=DA.
Also, ∠ADC=∠ABC ( By the property of parallelogram)
Therefore, In ABCD all sides are equal and Opposite angles are equal.
⇒ ABCD is a rhombus. ( The diagram is shown below)
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