Download PRACTICE QUESTIONS: UNIT 12 (Geometry II)

PRACTICE QUESTIONS: UNIT 12 (Geometry II)
1. AC is the diameter of the circle and the chord BD intersects AC at X. Given that
and
, calculate:
i.
ii.
B
A
D
Solution:
i.
In ABC, C=26˚ (given)
ABC=90˚ (The angle in a semi-circle is 90o)
We know that sum of interior angles of a triangle is 180˚
BAC+ ABC+ BCA=180˚
BAC+90˚+26˚=180˚
BAC +116˚=180˚
BAC =180˚-116˚
BAC =64˚.
ii.
ACB = ADB = 26˚
(Angles in the same segments are equal.)
XAD = 47˚ (Given)
AXD + XAD + ADX
180˚
(Sum of interior angles of an angle is 180˚)
AXD +
°+
X
˚
180°
AXD + 3° = 180°
AXD 180 – 73 = 107°
C
2.
In the diagram below, not drawn to scale, ABCDE is a pentagon inscribed in a circle with centre, O. The
diameter AD is produced to F. Angle
and angle
.
a. Calculate, giving reasons for your answer, the magnitude of angles:
i.
CDA
ii.
BCD
iii.
AED
b. Given that OA = 15 cm and angle EAD = 35o, calculate the length of AE.
B
C
A
D
F
D
E
Solution:
i.
CDF=136˚ (Given)
CDA+ CDF=180˚ (We know that)
CDA=180˚-136˚
CDA=44˚
ii.
BAD +
BCD
180
(Opposite angles in a cyclic quadrilateral add up to 180.)
72 + BCD
180
BCD
180 – 72
BCD
108˚
iii.
AED=90˚ (Angle in a semi-circle is 90˚)
AED= 90˚
(Angle in semi-circle is 90˚)
3. The diagram below, not drawn to scale, shows a circle, centre O. The lines BD and DCE
are tangents to the circle, and angle BCD = 70o.
B
A
D
E
C
Calculate, giving reasons for each step of your answer.
i.
OCE
ii.
BAC
iii.
BOC
iv.
BDC
Solution:
i.
OCE=90˚ (The angle between a tangent and a radius in a circle is 90o)
ii.
BAC=70˚ (The angle between the tangent and the chord at the point of contact is equal to the
angle in the alternate segment.)
iii.
BOC=2* BAC (The angle at the center is twice the angle at the circumference.)
BOC=140˚
iv.
OBD = 90
OCD
90
{Angles from the tangent to center of a circle is 90˚)
BCD
0˚
So CBD
0˚ (Similarly)
BCD+ CBD+ BDC 180
0+ 0+ BDC 180
BDC 180-140
BDC
0˚
4. Using ruler and compasses only, construct a triangle PQR with 𝑃𝑄 = 8.0𝑐𝑚, 𝑄𝑅= 6.5𝑐𝑚 and angle
PQR=75o. Measure PR and state its length. Show all construction lines clearly.
Solution:
I.
Draw a line PQ=8.0cm using ruler.
II.
Place the compass on Q and draw an arc which cuts PQ at A.
III.
To draw 75˚, draw two angles of 60˚ and 90˚ and divide them half as shown below. The resultant
will be our desired 75˚.
IV.
Set the compass to a separation of 6.5cm and mark an arc on line QB which cut QB at R.
V.
Join R to P; PQR is our desired triangle.
PR is 8.8cm long.
𝟓. Using ruler and compasses only, construct a triangle PQR in which
,
𝟓
𝟓
i.
Using the triangle PQR above, construct the parallelogram PQRS such that PR is a diagonal of
the parallelogram.
ii.
Measure and state the length of QS and the size of angle PQR.
Solution:
1.
2.
3.
4.
5.
Construct a line QR=7.2cm using ruler.
Set the compass to 6.5cm using ruler and mark an arc with center Q.
Now set the compass to 5cm and draw an arc with center R.
The two arcs will cut each other at point P.
Join point P to Q and R.
i.
We know that a parallelogram has opposite sides of same length. So we construct PS=7.2cm and
SQ=5cm.
1.
2.
3.
4.
5.
Set the compass at 7.2cm and draw an arc with the center P.
Now set the compass to 5cm and draw another arc with center Q.
These arcs will cut each other at point S.
Join S to Q and P.
PQRS is our desired parallelogram with PR as the diagonal of the parallelogram.
ii.
Measure of SQ is 5cm because it is alternate side of PR and PR=5cm.
Measure of PQR is 45˚. (Measured by protractor)
6. Using ruler and compasses only, construct a quadrilateral VWXY in which 𝑌 = 8𝑐𝑚, angle XYV =
80o, VY = 6cm, XW = 7cm and angle XYW = 35o. Measure and state the length of VW correct to 1
decimal place.
Ans:
XY = 8cm, YV=6cm, XW=7cm and VW=8.2cm
7. All construction lines must be shown clearly:
i.
Construct a triangle LMN with MN=10cm, LM=6.5cm and angle LMN = 50o.
ii.
Measure and state the length of LN.
iii.
Measure and state the size of angle LMN.
iv.
Using, ruler and compasses only, construct LX so that LX is perpendicular to MN and meets MN
at X.
Solution:
1. Draw the line MN=10cm using ruler.
2. Place the protractor on the point on M and mark a point L at the angle 50˚.
3. Now joint the point L with M and N.
4. LMN is our desired triangle.
ii. LN is 5.7cm
iii. LMN is 50˚
iv.
I.
II.
III.
IV.
V.
Open compass a bit more than 5 cm.
With center M mark an arc.
With center n mark another arc of same radius.
These arcs cut each other at point A.
Now join the point L with MN through A which join MN at point X.
8.
A straight line HK cuts the y-axis at (0, -1). The gradient of HK is 23. Show that the equation of
the line HK is 2𝑥−3𝑦=3.
Solution:
Equation of line is
Y= mx + c
Given equation:
2x - 3y = 3
Put the values
2(0) – 3(-1) = 3
0 +3 = 3
Equation satisfied.
9.
P is the point (4, 2), Q is the point (12, 5) and R is the point (1, 3).
i.
The length of PR.
ii.
The gradient of PQ.
iii.
The equation of the line passing through R and parallel to PQ.
Solution:
i.
The equation to find the length is
√(𝑥
𝑥 ) + (𝑦
𝑦 )
Putting values
Length or PR = (1
) +(
)
PR = 64 + 9
PR = 73
ii.
𝑚
The gradient of PQ.
𝑦
𝑥
𝑦
𝑥
m=
m=
iii.
Substitute the value of m into the general equation:
Y = mx + c
Using the values of (12, 5)
5 = (3/8)12 + c
5 = 36/8 + c
5 = 9/4 + c
C = 5 – 9/4 (Equation of line)
10.
The coordinates of A and B are (3, 5) and (7, 1), respectively. X is the midpoint of AB.
Determine
i.
The length of AB
ii.
The gradient of AB
iii.
The coordinates of X
The equation of the perpendicular bisector of AB and state the coordinates of the point at which the
perpendicular bisector meets the y-axis
Solution:
i.
Length of AB
3) + (1
AB = √(
)
AB = 1 + 1
AB = 6.92cm
ii.
The gradient of AB
𝑚
𝑦
𝑥
𝑦
𝑥
m=
m=m = -1
iii.
x=
x=5
y=
y=3
The coordinates of X