Free Constructions 03 Practice Test - 9th Grade
Question 1
A 40° angle is obtained by drawing an angle bisector for a given 80° angle.
SOLUTION
Solution : A
An angle bisector divides a given angle into two equal halves. So bisecting a given 80∘ angle will give two 40∘ angles. Hence, the given statement is true.
Question 2
In the below figure if AP=BP, AQ=BQ, then AM=BM?
SOLUTION
Solution : A
AP = BP (Given)
AQ = BQ (Given)
PQ is common.
So, Δ PAQ ≅ Δ PBQ (SSS rule).
⇒∠ APQ=∠ BPQ (CPCT).
Also, AP = BP (Given) and PM is common
Hence, △ PMA ≅ △PMB (SAS Rule)So, AM = BM
Question 3
For constructing an angle of 60∘, we need to draw
SOLUTION
Solution :We need to draw just 2 arcs to draw a 60∘ angle. As shown in the figure below, the first arc should have P as a centre. The second arc should have the same radius as the first and should have B as a centre.
Question 4
We can draw a triangle ABC in which BC is 12cm, ∠B is 75∘ and AB + AC = 6.5 cm.
SOLUTION
Solution : B
We know that sum of the length of any two sides of a triangle must be greater than the length of the third side, but the condition in this question defies this law. So the given triangle can not be formed.
Question 5
A triangle ABC in which BC = 3.4 cm, AB - AC = 1.5 cm and DB = 45∘ can be constructed using compass and ruler.
SOLUTION
Solution : A
A triangle in which the difference of two sides is less than the third side can be made. The steps are as follows:
Step 1 : Draw a line segment BC of length 3.4cm.
Step 2 : Draw an angle of 45∘ from point B.
Step 3 : From Ray BX cut off the line segment BD = 1.5cm.
Step 4 : Join D to C.
Step 5 : Draw bisector of DC.
Step 6 : Extend bisector of DC to intersect the Ray BX at point A.
Step 7 : Join A to C, Δ ABC is the required triangle.
Question 6
The steps to construct a triangle with given base angles ∠ B and ∠ C and BC + CA + AB, are:.
A. Draw a line segment, say XY equal to BC + CA + AB.
B. Make ∠ LXY equal to ∠ B and MYX equal to ∠ C.
C. Bisect ∠ LXY and ∠ MYX. Let these bisectors intersect at a point A.
D. Draw perpendicular bisectors PQ of AX and RS of AY.
E. Let PQ intersect XY at B and RS intersect XY at C. Join AB and AC.
True
False
SOLUTION
Solution : A
The procedure is standard construction for the given type of triangle. The image of construction is shown below.
Question 7
In the following figure, if OQ is angle bisector of ∠POA, then ∠POQ will be equal to ___.
45∘
60∘
90∘
30∘
SOLUTION
Solution : A
As OQ is angle bisector of ∠POA, so ∠POQ = ∠QOA= 45∘
Question 8
We can use the concept of an equilateral triangle to construct a 60∘ angle.
True
False
SOLUTION
Solution : A
A 60∘ angle can be made by first drawing a line segment AB of any length.
Construct an angle of 60 degrees at both A and B using a compass.
Let the rays intersect each other at P.
The triangle ABP formed is an equilateral triangle. So we can see that the concept behind an equilateral triangle can be used in constructing a 60∘ angle.
Question 9
Which of the following angles can be constructed without using a protractor?
30∘
40∘
SOLUTION
Solution : A and D
Angles 60o and 30o can be constructed using a scale and compass.
1. Draw a line l.
2. With O as centre and with any convenient radius, draw an arc to cut line l at A.
3. With A as centre and with the same radius, draw an arc to cut the previous arc at B.
4. Join OB and extend. ∠AOB= 60o
5. To construct 30o we bisect 60o.
The remaining two angles cannot be constructed without using a protractor.
Question 10
In the given construction of a ΔABC, ∠CAB will be
SOLUTION
Solution : B
In ΔAXB
PQ is the perpendicular bisector of XA, the two triangles formed are congruent by SAS rule.
∴ AB = XB
⇒ ∠BAX=∠AXB
∠ABC=∠BAX+∠AXB
= 2 ∠AXB=∠LXY
[Since ∠AXB is angle bisector of ∠LXY]
Similarly, ∠ACB=∠MYXTherefore, ∠ABC=75∘ and ∠BCA=60∘In ΔABC, we have∠ABC+∠BCA+∠CAB=180∘75∘+60∘+∠CAB=180∘∠CAB=180∘−135∘∠CAB=45∘