Patterns (Tokens)

Patterns (Rods)

Angles (Parallelogram and Triangle)

In the figure below, not drawn to scale, ABCD is a parallelogram, DE and AE are straight lines. Find the ∠AED.

Label the point with a 90 angle as F.

Method 1
CFE = 90 (adjacent angles on straight line BFC)
BCE = 65 (corresponding angles)
AED = 180 - 90- 65 = 25 (angle sum of triangle CFE)

Method 2
AFB = 90 (adjacent angles on straight line BFC)
ABC = 65 (diagonally opposite angles of parallelogram ABCD)
AED = BAF = 180 - 90 - 65 = 25 (angle sum of triangle BAF, alternate angles)

Angles (Isosceles Triangle)

The figure below is not drawn to scale. Given that TP = SP, find ∠RTS

TRS = 40 + 24 = 64 (exterior angle = interior opposite angles of triangle TPR)

TSP = (180 - 24) / 2 = 78 (base angles of isosceles triangle TSP)

RTS = 180 - 78 - 64 = 38 (angle sum of triangle RTS)

TRP = 180 - 40 - 24 = 116 (angle sum of triangle TRP)

TSP = (180 - 24) / 2 = 78 (base angles of isosceles triangle TSP)

RTS = 116 - 78 = 38 (exterior angle = interior opposite angles of triangle RTS)

Angles (2 Isosceles Triangles)

First

The figure below is not drawn to scale. ABC is an isosceles triangle where BA = BC. Given that D is the midpoint of BC and AC is 1/2 of BA, find ∠ADB.

Method 1

∠DCB = (180 - 20) / 2 = 80 (base angles of isosceles triangle ABC)

∠ADC = (180 - 80) / 2 = 50 (base angles of isosceles triangle ACD)

∠ADB = 180 - 50 = 130 (adjacent angles on straight line BDC)

Angles (Isosceles Triangle and Parallelogram)

First

The following figure is not drawn to scale. Given that ABCD is a parallelogram and ∠ADE is an isosceles triangle, find ∠y.

ADC = 180 - 118 = 62 (co-interior angles of parallelogram ABCD)

Y = 62 / 2 (exterior angle = interior opposite angles, base angles of isosceles triangle ADE)

DAB = 118 (diagonally opposite angles of parallelogram ABCD)

y = (180 - 118) / 2 = 31 (co-interior angles of trapezium ABCE)

Second

Study the figure below and find ∠FCB.

Method 1

FCD = EFA = (180 - 84) / 2 = 48 (base angles of isosceles triangle EAF, corresponding angles)

DCB = 180 - 68 = 112 (co-interior angles of parallelogram ABCD)

FCB = 112 - 48 = 64

Method 2

CFB = (180 - 84) / 2 = 48 (base angles of isosceles triangle EAF, vertically opposite angles)

FBC = 68 (diagonally opposite angles)

FCB = 180 - 48 - 68 = 64 (angle sum of triangle FCB)

Angles (Rhombus & Parallelogram)

In the figure below, ABCD is a rhombus and CDEF is a parallelogram, ∠ADE is 150° and ∠CFE is 115°. Find ∠x.

Method 1
∠FCB = 150° (corresponding angles)
∠FCD = 180° - 115° = 65° (co-interior angles of a parallelogram)
∠DCB = 150° - 65° = 85°
∠DAB = 85° (diagonally opposite angles of a rhombus)
∠x = 85° / 2 = 42.5° (angle bisector of a rhombus)

Method 2
∠EDC = 115° (diagonally opposite angles of a parallelogram)
∠CDA = 360° - 150° - 115° = 95° (angles at a point)
∠DAB = 180° - 95° = 85° (co-interior angles of a parallelogram)
∠x = 85° / 2 = 42.5° (angle bisector of a rhombus)