- Find the value of x in the following.
X+45+50=1800 (Angles on a straight lines are supplementary i.e. add up to 180º )
The value of x =85º
- Find the sum of angle “a” and angle “b” in the figure below.
Lines AB and C D are transversals are Therefore 90+b = 1800
Co-interior angles - supplementally
B = 90º
Angle a = 120º - (Corresponding angles)
Therefore a = 120º
Sum of a and b
=120 + 90
- Find the size of angle marked A B D in the figure below.
X+4x+x+30=180º (angles on a straight line are supplementary)
6x = 150
X = 25
Angle A B D =x + 4X
But x = 25
Therefore 25 + (4 x 25)
= 25 + 100
- Draw an equilateral triangle A B C where Line AB = 6cm.
Draw a circle touching the 3 vertices of the triangle. What is the radius of the circle?
- Draw line A B = 6cm
- With A as the Centre with the same radius 6cm, mark off an arc above line A B.
- With B as the Centre with the same radius 6cm, mark off an arc above line A B to meet the arc in (II) above. Call the point of intersection point C
- Join C to A and C to B
- Bisect line A B and B C and let the bisectors meet at point X.
- With X as the Centre, draw a circle passing through points A, B and C.
- Measure the radius of the circle.
- Construct a triangle P Q R in which Q P = 6cm. Q R = 4cm and P R =8cm. Draw a circle that touches the 3 sides of the triangle, measure the radius of the circle.
- Draw line Q P 6cm
- With Centre Q, make an arc 4cm above line Q P.
- With Centre P, make an arc 8cm above line Q P and let the arc meet the one in (II) above. Label the point of intersection as R.
- Join R to P and R to Q.
- Bisect any two angles and let the bisectors meet at point Y.
- With Y as the Centre, draw a circle that touches the 3 sides of the triangle.
R = 3.5cm
- A rectangle measures 6cm by 2½ cm. What is the length of the diagonal?
AC2 = AB2 + BC2 [ Pythagoras Theorem]
AC2 = 62 + 2 ½2
AC2 = 36 + 6.25
AC2 = 42.25
AC = √42.25
= 6.5 or 6 ½
NB: The Pythagoras theorem states
H2 =B2 +h2
h2 = H2 – b2
b2 = H2 –h2
- In the figure below, A B C is a straight line and B C D E is a quadrilateral. Angle CBD = 620 and lines EB = BD = DC. Line EB is parallel to DC.
What is the size of angle BDE?
Consider triangle BCD (isosceles triangle)
Therefore base angles are equal
CBD = 62º
BCD = 62º
Therefore, BDC = 180 – 124 = 56º
Angle CDB = angle EBD [Alternate triangle]
Therefore EBD = 56º
Angle BDE =180 - 56/2
Therefore, BDE = 62º
- Find the size of the largest angle from the following triangle.
4X – 10 + x – 20 + 3x + 10 = 180 [Angle sum of a triangle]
8x – 20 = 180
8x = 200
X = 25
4x – 10 = (100 – 10)º
= 90º largest angle.