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

## Questions

1. 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.
2. 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
3. The area of a rhombus is 60cm2. Given that one of its diagonals is 15cm long. Calculate the perimeter of the rhombus.
4. 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 .
5. 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(90o – xo) in the form a/b where a and b are integers
6. The size of an interior angle of a regular polygon is 3xo while its exterior angle is (x-20)o.
Find the number of sides of the polygon
7. In the figure above, angle a is half the sum of the other angles. Evaluate the triangle
8. The sum of the interior angles of an n-sided polygon is 1260o. Find the value of n and hence deduce the polygon
9. Giving reason, find the angle marked n 10. Solve for y in the equation 125y+1 + 53y = 630
11. The interior angle of a regular polygon is 108o larger than the exterior angle. How many sides has the polygon?
12. The interior angle of a regular polygon is 4 times the exterior angle. How many sides has the polygon
13. 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
14. In the figure below, DE bisects angle BDG and AB is parallel to DE. Angle DCF = 60and angle CFG = 100o
Find the value of angle:- 1. CDF
2. ABD
15. The size of an interior angle of a regular polygon is 4xo, while its exterior angle is (x – 30)o
Find the number of sides of the polygon
16. The sum of interior angles of a polygon is 1440o. Find the number of sides of the polygon hence name the polygon
17.  In the figure below PQ is parallel to RS. Calculate the value of x and y 18. The interior angle of a n-sided regular polygon exceeds its exterior angle by 132o.Find the value of n
19. 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)
20. The figure below represents a quadrilateral ABCD. Triangle ABX is an equilateral triangle. If <ADX = 50o , find <AXD with <BAD = 90o (2 mks) 21. 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 67o
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 35o. 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)

1. Let the exterior ∠be x
6.5x + x = 180
7.5x = 1800
x = 24
No. of sides = 360/24
= 15 sides.
2. (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
3. 15 b = 60
2   2
15b = 60 x 4
b = 16cm (diagonal)
= √82 + 7.52
∴per = 4 √82 + 7.52
= 43.86cm
4. x2 = 7.252 – 5.252
x = 7.252 – 5.252
= 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
5. AB2 = 10– 82= 100 – 64
AB2 = 36
AB = 6cm
Cos (90– xo) 8/10 4/5
6. x -20 + 3x = 180oC
4x = 200
x = 50
o
7. 2x + 40 + x – 25
3x + 15 + 9 = 180
3x + 15 = 29
9 = ½ (3x + 15)
3x + 3x/2 = 180 -15-15/2
x = 35o
x = 35 = 10o
½ ( 10 + 110) = 60o
8. 1260 = 14rt ∠s
90
Sum of interior ∠s
(2n -4) rt ∠s
2n-4 = 14
n = 9
9 sided polygon
9. N = 50 + 40 = 90o
Alternative angles
10. 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
11. 360/n + 108/n = 180 - 360
360 + 108n = 180n – 360
-72n = -720
n = 10
12. Let exterior angle be x
4x/4 = 180o/4
x = 45o
n=360
Exterior angle
n = 360/45
= 8 sides
13.
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
14.
1. ∠CDF = 100-60=40o (exterior angle of a Δ)
2. ∠BDE = 20o (DE is bisector of BDG)
ABD = 20o (alternate angles)
15. 4x + x – 30 = 180
5x = 210°
x = 42
(x - 30)n = 360°
12n = 360°
n = 360°/12
n = 30
16. 180(n-20) = 1440
n - 2 = 1440/180 = 8
n = 10
Decagon
17. ∠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
18.  Let the interior ∠be x and exterior be y
x + y = 180
+
x – y = 132
2x = 312

x = 156
y = 180 – 156 = 24o
No. of sides (n) = 360o/24= 15
= 15 sides

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