Questions
 The size of an interior angle of a rectangular polygon is 6 ½ times that of its exterior angle.
Determine the number of sides of the polygon.  The sum of interior angles of two regular polygons of sides n and n + 2 are in the ratio 3:4.
Calculate the sum of the interior angles of the polygons with n sides  The area of a rhombus is 60cm2. Given that one of its diagonals is 15cm long. Calculate the perimeter of the rhombus.
 In the figure below AE is parallel to BD. BC = BD, AB = 7.25cm, AE = 15.25cm and ED = 5.25 cm
Find the perimeter of the figure .  The figure below shows a trapezium ABCD in which side AB is perpendicular to both AD and BC. Side AD=17cm, DC=10cm
 What is the length of side AB
 Find the value of cos(90^{o }– x^{o}) in the form ^{a}/_{b} where a and b are integers
 The size of an interior angle of a regular polygon is 3x^{o }while its exterior angle is (x20)^{o}.
Find the number of sides of the polygon 
In the figure above, angle a is half the sum of the other angles. Evaluate the triangle  The sum of the interior angles of an nsided polygon is 1260^{o}. Find the value of n and hence deduce the polygon
 Giving reason, find the angle marked n
 Solve for y in the equation 125^{y+1 }+ 5^{3y }= 63^{0}
 The interior angle of a regular polygon is 108o larger than the exterior angle. How many sides has the polygon?
 The interior angle of a regular polygon is 4 times the exterior angle. How many sides has the polygon
 In the figure below ABCD is a trapezium with DC parallel to AB. DC = 5cm, CB = 4cm, BD = 8cm and AB = 10cm
Calculate: the size of angle BD
 the area of triangle ABD
 In the figure below, DE bisects angle BDG and AB is parallel to DE. Angle DCF = 60^{o }and angle CFG = 100^{o}Find the value of angle:
 CDF
 ABD
 The size of an interior angle of a regular polygon is 4x^{o}, while its exterior angle is (x – 30)^{o}
Find the number of sides of the polygon  The sum of interior angles of a polygon is 1440^{o}. Find the number of sides of the polygon hence name the polygon
 In the figure below PQ is parallel to RS. Calculate the value of x and y
 The interior angle of a nsided regular polygon exceeds its exterior angle by 132^{o}.Find the value of n
 The sum of angles of a triangle is given by the expression (2a+b)^{o} while that of a quadrilateral is given by (13a  b)^{o}. Calculate the values of a and b (4 mks)
 The figure below represents a quadrilateral ABCD. Triangle ABX is an equilateral triangle. If <ADX = 50^{o} , find <AXD with <BAD = 90^{o} (2 mks)
 Wanjiku is standing at a point P, 160m south of a hill H on a level ground. From point P she observes the angle of elevation of the top of the hill to be 67^{o}
 Calculate the height of the hill (3 mks)
 After walking 420m due east to the point Q, Wanjiku proceeds to point R due east of Q, where the angle of elevation of the top of the hill is 35^{o}. Calculate the angle of elevation of the top of the hill from Q (3 mks)
 Calculate the distance from P to R (4 mks)
Answers
 Let the exterior ∠be x
6.5x + x = 180
7.5x = 1800
x = 24
No. of sides = ^{360}/_{24}= 15 sides.  (2n – 4) 90 = 3
(2(n+2)  4)90 4
2n – 4 = 3
2n 4
8n – 16 = 6n
2n = 16
n = 8
(2(8) – 4) 90
= 12 x 90 = 1080  15 b = 60
2 2
15b = 60 x 4
b = 16cm (diagonal)
= √8^{2 }+ 7.5^{2}∴per = 4 √8^{2 }+ 7.5^{2}
= 43.86cm  x^{2 }= 7.25^{2 }– 5.25^{2}x = √7.25^{2 }– 5.25^{2}= 52.5625
27.5625 
√25
= 5cm
BC = 15.25 + 5 = 22.25cm
Arc CD = ^{90}/_{360 }X 3.142 X 2 X 22.25
= 34.65475
Perimeter = AB + BC + CD + DE + EA
= 15.25 +7.25 + 22.25 + 34.95 + 5.25
= 84.95cm  AB^{2} = 102 – 82= 100 – 64
AB^{2} = 36
AB = 6cm
Cos (90^{o }– x^{o}) 8/10 = 4/5  x 20 + 3x = 180^{o}C
4x = 200
x = 50^{o}  2x + 40 + x – 25
3x + 15 + 9 = 180
3x + 15 = 29
9 = ½ (3x + 15)
3x + ^{3x}/_{2} = 180 15^{15}/_{2}x = 35^{o}
x = 35 = 10^{o}½ ( 10 + 110) = 60^{o}  1260 = 14rt ∠s
90
Sum of interior ∠s
(2n 4) rt ∠s
2n4 = 14
n = 9
9 sided polygon  N = 50 + 40 = 90^{o}
Alternative angles  53(y+1) + 53y = 630
Let x = 53y
53 x 53y + 53y =630
125x + x = 630
x = 5
53y = 51
3y = 1
y =^{1}/_{3}  _{}360/_{n} + ^{108}/_{n} = 180  360
360 + 108n = 180n – 360
72n = 720
n = 10  Let exterior angle be x
^{4x}/_{4} = 180^{o}/_{4}x = 45^{o}
n=360
Exterior angle
n = ^{360}/_{45}= 8 sides 
 Let ∠BDC = ø
A^{2} = 5^{2} + 8^{2 }– 2 x 5 x 8 cos ø  Area of ABD
= ½ x 8 x 10 sin 24°9^{1}
= 40 x 0.4091
= 16.36cm^{3}
 Let ∠BDC = ø

 ∠CDF = 10060=40^{o }(exterior angle of a Δ)
 ∠BDE = 20^{o }(DE is bisector of BDG)
∠ABD = 20^{o }(alternate angles)
 4x + x – 30 = 180
5x = 210°
x = 42
(x  30)n = 360°
12n = 360°
n = 360°/_{12}n = 30  180(n20) = 1440
n  2 = ^{1440}/_{180} = 8
n = 10
Decagon  ∠PQR =∠SRT = x (Alt ∠SPQ //RS)
∴5x + 3x + x = 180° <’s of
9x = 180°
X = 20°
∴5 x 20 + y = 180
y = 180 – 120 = 60  Let the interior ∠be x and exterior be y
∴ x + y = 180
+
x – y = 132
2x = 312
x = 156
y = 180 – 156 = 24^{o}No. of sides (n) = 360^{o}/_{24}= 15
= 15 sides
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