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Leto [7]
1 year ago
15

Please help with p and q maths geometry

Mathematics
1 answer:
baherus [9]1 year ago
4 0

Step-by-step explanation:

p)

in a very similar way, as the sum of all 4 angles in a triangle is always 180°, so is the sum of all angles in a quadrilateral (a polygon with 4 sides) 360°.

just look at rectangles at special representatives of such quadrilaterals.

now that we know this, we can simply say, the sum of all angles in our shape must be 360° :

(8x + 2) + (4x + 2) + (x - 2) + (5x -2) =

= 18x = 360

x = 20

A = 8×20 + 2 = 162°

B = 4×20 + 2 = 82°

C = 5×20 - 2 = 98°

D = 20 - 2 = 18°

a trapezium has the following properties (in the UK):

like other quadrilaterals, the sum of all the four angles of the trapezium is equal to 360°.

a trapezium has two parallel sides and two non-parallel sides.

the diagonals of a regular trapezium bisect each other.

in the USA for example the request for 2 parallel sides does not exist.

we have proven the point with the angle sum.

the other 2 are given by the diagram.

so, yes, it is a trapezium.

q)

the sum of all interior angles of a polygon is

(n - 2) × 180°

n being the number of d sides.

we have here a regular pentagon (5 equal sides).

the size of the angle at any vertex is then

(5 - 2) × 180 / 5 = 3 × 180 / 5 = 3×36 = 108°.

so, theta = 108°.

alpha + beta = 108°

ABE is an isoceles triangle.

so, beta = angle 1

beta + angle 1 + theta = 180

2×beta + 108 = 180

2×beta = 72

beta = 36°

therefore,

alpha = 108 - beta = 108 - 36 = 72°

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