### Geometry

Worked Exercise

- Find the value of x in the following.

Working

X+45+50=1800 (Angles on a straight lines are supplementary i.e. add up to 180º )

X+95=180º

X=85º

The value of x =85º - Find the sum of angle “a” and angle “b” in the figure below.

Working

Lines AB and C D are transversals are Therefore 90+b = 1800

Co-interior angles - supplementally

Therefore b=180-90

B = 90º

Angle a = 120º - (Corresponding angles)

Therefore a = 120º

Sum of a and b

=120 + 90

= 210º - 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+30=180

6x=180-30

6x = 150

X = 25

Angle A B D =x + 4X

But x = 25

Therefore 25 + (4 x 25)

= 25 + 100

= 125º - 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?

Working

Steps:- 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.

Working- 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.

Construction

R = 3.5cm

- A rectangle measures 6cm by 2½ cm. What is the length of the diagonal?

Working

AC^{2}= AB^{2}+ BC^{2}[ Pythagoras Theorem]

AC^{2}= 6^{2}+ 2 ½^{2}

AC^{2}= 36 + 6.25

AC^{2}= 42.25

AC = √42.25

= 6.5 or 6 ½

NB: The Pythagoras theorem states

H^{2}=B^{2}+h^{2}

h^{2}= H^{2}– b^{2}

b^{2}= H^{2}–h^{2} - 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?

Working

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}

= 62º

Therefore, BDE = 62º - Find the size of the largest angle from the following triangle.

Working

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.

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