Coordinates and Graphics Questions and Answers

Coordinates and Graphics Questions

1.  A triangle ABC is formed by the points A (3,4), B (-7,2), and C (1,-2).

(a) Find the coordinates of the mid-points k of AB and p of AC (1 mk)

(b) Find the equation of the perpendicular bisector of the line kp (2 mks)

2.  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.

3.  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

4 .  The area of a rhombus is 60cm². Given that one of its diagonals is 15cm long. Calculate

the perimeter of the rhombus.

5.  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 .

6.  The figure below shows a trapezium ABCD in which side AB is perpendicular to both AD

and BC. Side AD=17cm, DC=10cm

(i) What is the length of side AB

(ii) Find the value of cos(90^o – x^o) in the form a where a
and b are integers

b

7.  The size of an interior angle of a regular polygon is 3x^o while its exterior angle is (x-20)^o.

Find the number of sides of the polygon  

8.

 In the figure above, angle a is half the sum of the other angles. Evaluate the triangle  

9.  The sum of the interior angles of an n-sided polygon is 1260^o. Find the value of n and

 hence deduce the polygon

10.  Giving reason, find the angle marked n  

11.  Solve for y in the equation 125 ^y+1 + 5^3y = 630

12.  The interior angle of a regular polygon is 108^o larger than the exterior angle. How many

sides has the polygon?

13.  The interior angle of a regular polygon is 4 times the exterior angle. How many sides has

the polygon

14.  In the figure below ABCD is a trapezium with DC parallel to AB. DC = 5cm, CB = 4cm,

BD = 8cm and AB = 10cm

 Calculate:

 (a) the size of angle BDC

 (b) the area of triangle ABD

15.  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:-  

(a) CDF

 (b) ABD

16.  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  

17.  The sum of interior angles of a polygon is 1440^o. Find the number of sides of the polygon

hence name the polygon

18.  In the figure below PQ is parallel to RS. Calculate the value of x and y

19.   The interior angle of a n-sided regular polygon exceeds its exterior angle by 132^o.

Find the value of n  

Coordinates and graphics Answers

2.  Let the exterior  be x

  6.5x + x = 180

 7.5x = 180⁰

  x = 24

No. of sides = 360

24

 = 15 sides.

3.  (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

4.  15
b = 60

2 2

15b = 60 x 4

b = 16cm (diagonal)

   = 8² + 7.5²

per = 4 8² + 7.5²

= 43.86cm

5.   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

6.  AB² = 10² – 8²= 100 – 64

AB² = 36

AB = 6cm

Cos (90^o – x^{o) 8}/₁₀ = ⁴/₅

7.  x -20 + 3x = 180^oC

4x = 200

x = 50^o

8.  2x + 40 + x – 25

3x + 15 + 9 = 180

3x + 15 = 29

9 = ½ (3x + 15)

3x + 3x = 180 -15-15

  2 2

x = 35^o

x = 35 = 10^o

½ ( 10 + 110) = 60^o

9.  1260 = 14rt s

 90

Sum of interior s

(2n -4) rt s

2n-4 = 14

  n = 9 9 sided polygon

10.  N = 50 + 40 = 90^o

Alternative angles

11.  5^3(y+1) + 5^3y = 630

Let x = 5^3y

5³ x 5^3y + 5^3y =630

125x + x = 630

x = 5

5^3y = 5¹

3y = 1

y = ¹/₃

12.  360 + 108 = 180 – 360

 n n

 360 + 108n = 180n – 360

  -72n = -720

  n = 10

13.  Let exterior angle be x

4x = 180^o

4 4

x = 45^o

n=360

Exterior angle

n = 360

  45

= 8sides

14. a)   Let < BDC = ø

  A² = 5² + 8² – 2 x 5 x 8 cos ø

  cos ø = 89 – 16 = 73 = 0.9125

80 80

 Ø = 24°9   ⁼ 24° 8

  b) Area of ABD

  = ½ x 8 x 10 sin 24°9 ¹

  = 40 x 0.4091

 = 16.36cm ³  16.37  16.38

15.  (a) CDF = 100-60=40^o (exterior angle of a  )

(b) BDE = 20^o (DE is bisector of BDG)

ABD = 20^o (alternate angles)

16.  4x + x – 30 = 180

  5x = 210°

  x = 42

  (x -30)n = 360°

12n = 360°

n = 360°

12

n = 30

17.  180(n-20) = 1440

n- 2 = 1440 = 8

180

n = 10

 Decagon

18.  

  5x + 3x + x = 180° <'s of 

9x = 180°

X = 20°

5 x 20 + y = 180

y = 180 – 120 = 60

19.  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 = 15

24

= 15 sides