Download PRACTICE QUESTIONS: UNIT 11(Geometry I)

PRACTICE QUESTIONS: UNIT 11(Geometry I)
1. Use your protractor to draw the following angles:
a)
b)
c)
d)
e)
30
150
225
54
128
Solution a)
I.
Draw a line segment of any length.
A
II.
B
Place the protractor on the line segment as showed below.
III.
Now mark a point “C” on the desired angle which is 30 .
IV.
Now remove the protractor and join the point C with A.
V.
∠CAB is the desired angle.
Solution b)
I.
Repeat the same process as above and mark a point “E” at 150 .
II.
Join E with A and ∠EAB is our desired angle.
Solution c)
I.
Since a protector cover 180 , so we will draw an angle of
135 and the remaining angle will be 225 .
II.
So we will draw 135 .
Solution d)
I.
Draw an angle of 54 by same procedure.
II.
∠CAB is our desired angle.
Solution e)
I.
Use the same process for this also.
2. Find angles x and y giving reasons for your answer.
y
60o
x
130o
Solution:
The sum of internal angles of a triangle is = 180˚
And a line segment is also an angle of 180˚.
X=180˚-130˚
X=50˚.
For Y:
Known inner angles = 60˚+50˚=110˚
Y=180˚ -110˚
Y=70˚
3. Estimate the size of angles a and b. Give reasons for your answers.
35o
15o
b
a 120o
Solution:
Note that angle 15˚ is not 15, its 150˚. And 120˚ is external angle.
∠A
8 -120 (Sum of all angles on a line is 8 ˚)
∠A=60˚
360= 60+150+35+∠B
360=245+∠B
∠B=360-245
(Sum of interior angles of 4 sided figures is 360˚)
∠B=115˚
4. In the diagram, not drawn to scale, BG is parallel to DE; and AF and CH are straight lines. Calculate
the values of x and y, showing clearly the steps in your calculation.
Solution:
x + y + 30 = 180 (Angles on a straight line)
x + y = 150 _________(1)
And
x + y = 150˚
Sum of internal angles of triangle is 180˚
x + y + 180 – 24 = 180
y – x = 0__________(2)
Adding equation 1 and 2 we get
2y = 150
y = 75 (put in equation 1)
x + 75 = 150
x = 75
5. How many sides have a regular polygon if each interior angle is 156˚?
Solution:
We know that
156n=180(n-2)
(From the formula 180(n-2))
156n=180n-360
360=24n
n=
n=15
So the polygon has 15 sides.
6. In a regular polygon, each interior angle is greater than each exterior angle by 90˚. Calculate the
number of sides of the polygon.
Solution:
i+e=180
(Interior angle plus exterior angle =180˚)
i=90+i
(Every interior angle is 90 greater than interior)
i+90+i=180
2i+90=180
2i=90
i=180/2
i=90
Interior angle = 90˚
We can now find the number of sides of polygon by the formula 180(n-2)
90n=180(n-2)
90n=180n-360
-90n=-360
n=-360/-90
n=4
So the sides of the polygon are 4.
7. Each of the interior angle of a regular polygon is 20 o. How many sides does the polygon have?
Solution:
We use the same formula:
180(n-2)
Interior angle=20˚
20n=180(n-2)
20n=180n-360
360=160n
n=360/160
n=2.25
8. ABCDE is a pentagon. The angles A, B, C, D, and E are
. Find the value of .
E
D
C
A
B
Solution:
The sum of interior angles in a polygon is
180(5-2)
=180(3)
(
).
=540
If one angle is 120˚ (given)
540-120
=420
The sum of remaining 4 angles is 420˚.
So
X+2X+3X+4X=420
10X=420
X=420/10
X=42˚
2X=2*42
2x=84˚
3X=3*42
3X=126
4X=4*42
4X=168
So ∠A=42˚, ∠B=84˚, ∠C=126˚, ∠D=168˚, ∠E=120˚.
9. In the diagram below, not drawn to scale, KM is parallel to HL, angle HKI = 97 o and angle MLI = 32o.
i.
ii.
Show that triangle JKI is similar to triangle JHL
Gi ven that KI = 4cm and HL = 10cm, write down the
v al ue of the ratio
Solution:
a)
I.
II.
III.
∠K=∠H
(Corresponding angles)
∠I=∠L
(Corresponding angles)
∠J=∠J
If 3 angles of any triangles are equal, the triangles are equal.
is simila o iangle
So
L.
b)
Ratio of KI to HL:
m
2:5
So the ratio of JKI to JHL is 2:5
10. In the diagram below, not drawn to scale, triangle ACD is an enlargement of triangle ABE, with A as the
centre of enlargement, AB = 4 cm and BC = 6cm.
D
E
A 4cm B
6cm
C
Calculate:
i.
The scale factor of the enlargement which maps triangle ABE onto triangle ACD.
The area of triangle ACD, in cm 2, given that the area of triangle ABE is 18 cm 2.
Solution:
Length from center A to B = 4
Length from center A to C = 10
Let scale factor = x
So 4 * x = 10
x = 10/4
x = 2.5 cm
b)
Area of ABE = 18 m
Area of ACD = y (let)
Scale factor = 2.5
So area of ACD = area of ABE * scale factor
y = 18 * 2.5
y = 45