Angles and Plane Figures Questions and Answers

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Angles and Plane Figures Questions and Answers - Form 1 Topical Mathematics

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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
1. What is the length of side AB
2. 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^owhile its exterior angle is (x-20)^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 n-sided 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= 63⁰
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:
1. the size of angle BD
2. the area of triangle ABD
In the figure below, DE bisects angle BDG and AB is parallel to DE. Angle DCF = 60^oand angle CFG = 100^oFind the value of angle:-
1. CDF
2. 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 n-sided 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
1. Calculate the height of the hill (3 mks)
2. 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)
3. 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)
= √8²+ 7.5²∴per = 4 √8²+ 7.5²
= 43.86cm
x²= 7.25²– 5.25²x = √7.25²– 5.25²= 52.5625
27.5625 -
√25
= 5cm

BC = 15.25 + 5 = 22.25cm
Arc CD = ⁹⁰/₃₆₀X 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
AB² = 102 – 82= 100 – 64
AB² = 36
AB = 6cm
Cos (90^o– x^o) 8/10 = 4/5
x -20 + 3x = 180^oC
4x = 200
x = 50^o
2x + 40 + x – 25
3x + 15 + 9 = 180
3x + 15 = 29
9 = ½ (3x + 15)
3x + ^3x/₂ = 180 -15-¹⁵/₂x = 35^o
x = 35 = 10^o½ ( 10 + 110) = 60^o
1260 = 14rt ∠s
90
Sum of interior ∠s
(2n -4) rt ∠s
2n-4 = 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/₄x = 45^o
n=360
Exterior angle
n = ³⁶⁰/₄₅= 8 sides
1. Let ∠BDC = ø
  A²= 5²+ 8²– 2 x 5 x 8 cos ø
2. Area of ABD
  = ½ x 8 x 10 sin 24°9¹
  = 40 x 0.4091
  = 16.36cm³
1. ∠CDF = 100-60=40^o(exterior angle of a Δ)
2. ∠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°/₁₂n = 30
180(n-20) = 1440
n - 2 = ¹⁴⁴⁰/₁₈₀ = 8
n = 10
Decagon
∠PQR =∠SRT = x (Alt ∠SPQ //RS)
∴5x + 3x + x = 180° <’s of ï
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^oNo. of sides (n) = 360^o/₂₄= 15
= 15 sides