y. - Mags Maths

Report
Geometric Reasoning
Mahobe
The facts to learn
Ð sum D
• Angles in a triangle
add up to 180o
Ðs on a line
• Angles on a straight line
add up to 180o
Ðs at a pt
• Angles around a
point add up to 360o
ext Ð D = sum opp int Ðs
base ∠’s isos ∆ =
Ð sum isos D
∠ sum quad
• Interior angles of a
quadrilateral sum to
360o
opp ∠, cyclic quad
• Opposite angles of a
cyclic quadrilateral
sum to 180o
Ext ∠, of cyclic quad
• Exterior angle of a
cyclic quad equals
the opposite interior
angle
vert opp ∠s =
• Vertically opposite
angles are equal
alt ∠s // lines
• Alternate angles are equal if the
lines are parallel
corr ∠s // lines
• Corresponding angles are equal if
the lines are parallel
Co-int ∠s // lines
• Co-interior angles sum to 180o if
the lines are parallel
• Exterior angles of a
polygon add up to
360o
If the polygon is regular all exterior
angles are the same size
• Exterior angles of a
polygon add up to
360o
ext ∠ regular pentagon
• Exterior angles of a
regular pentagon is
72o
Interior angles of a polygon with n
sides sum to 180(n-2)
Find all the other angles in this figure
ÐFAB = ÐFBA = 75°
Ð sum isos D
ÐBFC = 75°
co-int Ðs / / lines
ÐCFE = 60°
Ðs on a line
ÐCEF = 90°
Ð sum D
ÐBCF = 60°
co-int Ðs, / / lines
ÐCED = 90°
Ðs on a line
ÐCBF = 45°
Ð sum D
Vert opp angles
Ð in a semi-circle
rad ^ tang
A triangle formed with 2 radii is
isosceles
Ðs same arc
Ðs same chord
Ðs same segment
Ð at centre
• The angle at the
centre is twice the
angle at the
circumference.
When writing angles use 3 letters.
The middle letter is where the angle is.
Find the angles u, v and w
O Is the centre
w = 140°
O Is the centre
Ð at centre
u = 70°
O Is the centre
ext Ð isos D
ÐADO = 70° base Ðs isos D
ÐDAB = 90° Ð in semi-circle
O Is the centre
v = 20°
Ð sum D
Find angle d
5x + 4x = 180° co-int Ðs // lines
x = 20°
5x + 4x = 180° co-int Ðs // lines
x = 20°
5x + d = 180 cyclic quad
d = 80°
Calculate the size of angle ABC
ÐACB = 55°, Ðs on a line
ÐBAC = 55°, base Ðs isos D
ÐABC = 70°,
Ð sum D
ÐACE = 54°, Ðs on a line
ÐCBA = 20°, Ðs on a line
ÐCAB = 106°, Ð sum D
Calculate angle CDE
ÐDEC = 67°,
vert opp Ðs
ÐCDE = 71°,
Ð sum D
Calculate the size of angle HIJ
ÐJHI = 52°, alt Ðs // lines
ÐHIJ = 64°, Ð sum isos D
Calculate the size of angle c
C = 96o,
• Co-int angle // lines
• Corr angles // lines
• OR
Angles on a line
Opp angles parallelogram
Calculate the size of angle VUW
ÐUVW = 55°, corr Ðs // lines
ÐVUW = 70°, Ð sum isos D
Shape P is a regular pentagon and
shape H is a regular heptagon.
Calculate the size of angle CBD
ÐCBA =P108°,
int Ð reg
pentagonand
Shape
is a regular
pentagon
shape
H is a regular
ÐABD
= 128.6°,
int Ð regheptagon.
heptagon
Calculate the size of angle CBD
ÐCBD = 123.4°, Ðs at pt
The diagram gives the angles of a
playground slide. Find angle XYZ.
ÐWYZ = 75°,
Ð sum rt Ð D
ÐUYW = 90°,
rectangle
ÐXYZ = 160°, Ðs at point
Find the size of angles NMO and OPN
ÐNMO
95°,
ext ÐNMO
cyclicand
quad
Find the=size
of angles
OPN
ÐOPN = 85°,
cyclic quad
Find the size of angles PRQ and PSQ
O
ÐPRQ = 60°,
angle at centre
ÐPSQ = 120°,
cyclic quad
O
XBC
90
Rad perp.
tang
CBA
20
Angles on a
line
CAB
20
Base angles
isos triangle
XCB
40
Ext angle =
sum opp int
angles
CXB
50
Angle sum
triangle
ÐNMQ = 90°,
Ð in a semi-circle
ÐMOQ = 132°, ext Ð D = sum opp int Ðs
ÐNQM = 24°, Ð sum isos D
O Is the centre of the circle.
Calculate the size of angle BCD
ÐBAO = 47°, base Ðs isos D
ÐAOB = 86°,
Ð sum D
ÐDOB = 144°
ÐBCD = 72° Ð at centre = 2Ð at circ
O Is the centre of the circle.
Calculate the size of angle WYX
ÐOWY = ÐOXY = 90°,
rad ^ tang
ÐWYX = 30°,
Ð sum quad
Similar triangles
EB 10
= Þ EB = 8 Þ BD = 4
12 15
ABC is an isosceles triangle.
O is the centre of the circle.
Calculate the size of angles BAC and ODC
ÐODB = 31°, base Ðs isos D
ÐBOC = 118°,
Ð sum D
ÐBAC = 59°, Ð at centre
ÐBDC = 121°,
cyclic quad
ÐODC = 60.5°,
symmetry
Calculate the size of angles x and y.
ÐPKQ = 54°, angle at centre
x = 126°,
.
cyclic quad
y = 35°, Ð at tang = opp Ð D
O is the centre of the circle.
Calculate the size of angle BAC.
.
ÐOAC = 32°, base Ðs isos D
ÐAOC = 116°,
Ð sum D
reflex ÐAOC = 244°,
Ðs at pt
ÐABC = 122°,
Ð at centre
ÐBCA = 32°, alt Ðs // lines
ÐBAC = 26°,
Ð sum D
ÐMNQ = 30°,
Ð sum D
ÐKQN = 60°, Ðs on line
ÐQNK = 60°,
base Ðs isos D
ÐKNM = 90° and ÐKLM = 90°,
symmetry
KNML is cyclic as
the opposite
angles sum to 180.
O is the centre of the circle.
Prove angle CBY = angle CAB
angle CBY = angle CAB
Let ÐCBY = x
ÐOBC = 90 - x,
rad ^ tang
ÐOCB = 90 - x, base Ðs isos D
ÐBOC = 2x, Ð sum D
ÐCAB = x, Ð at centre

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