Constructions(9th Grade > Mathematics ) Questions and answers for exam Preparation

9th Grade > Mathematics

CONSTRUCTIONS MCQs

Total Questions : 56 | Page 3 of 6 pages

Question 21. We can draw a triangle ABC in which BC is 12cm,

∠ B

75^{\circ}

and AB + AC = 6.5 cm.

True
False
Data insufficient
Cannot be determined

Discuss Question

Answer: Option B. -> False
:
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 22. 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
Data insufficient
Cannot be determined

Discuss Question

Answer: Option A. -> True
:
A
The procedure is standard construction for the given type of triangle. The image of construction is shown below.

The Steps To Construct A Triangle With Given Base Angles �...

Question 23. A triangle ABC in which BC = 3.4 cm, AB - AC = 1.5 cm and DB = 45

^{\circ}

can be constructed using compass and ruler.

True
False
Data insufficient
Cannot be determined

Discuss Question

Answer: Option A. -> True
:
A
A triangle in which the difference of two sides is lessthan 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

\circ

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 intersectthe Ray BX at point A.
Step 7 : Join A to C,

Δ

ABC is the required triangle.

A Triangle ABC In Which BC = 3.4 Cm, AB - AC = 1.5 Cm And DB...

Question 24. A 40° angle is obtained by drawing an angle bisector for a given 80° angle.

True
False
Cannot be determined
None of the above

Discuss Question

Answer: Option A. -> True
:
A
An angle bisector divides a given angle into two equal halves. So bisecting a given

80^{\circ}

angle will give two

40^{\circ}

angles. Hence, the given statement is true.

Question 25. We can use the concept of an equilateral triangle to construct a 60

^{\circ}

angle.

True
False
90∘
30∘

Discuss Question

Answer: Option A. -> True
:
A
A60

^{\circ}

angle can be made by first drawing a line segment AB of any length.
Construct anangle 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 behindan equilateral trianglecan be used in constructing a 60

^{\circ}

angle.

Question 26. In the following figure, if OQ is angle bisector of

∠

POA, then

∠

POQ will be equal to ___.

In The Following Figure, If OQ Is Angle Bisector Of ∠POA, ...

45∘
60∘
90∘
30∘

Discuss Question

Answer: Option A. -> 45∘
:
A
As OQ is angle bisector of

∠

POA, so

∠

POQ =

∠

QOA=

45^{\circ}

Question 27.

In which of the following situations an angle bisector cannot be constructed to bisect the angle formed between two given lines?

Two intersecting lines
Two parallel lines
Two perpendicular lines
None of the above

Discuss Question

Answer: Option B. -> Two parallel lines
:
B
Since two parallel lines do not intersect each other, an angle cannot be formed between them. Thus in this case, angle bisection is not possible.

Question 28.

For which of the following can a perpendicular bisector be drawn?

Line
Ray
Line segment
Both Line and Ray

Discuss Question

Answer: Option C. -> Line segment
:
C

A perpendicular bisector can be drawn only if a figure has end points. Only a line segment has a definite length and hence it can be bisected by a perpendicular bisector.

Question 29.

In which of the following quadrilaterals, a diagonal is not an angle bisector?

Square
Rectangle
Rhombus
Parallelogram

Discuss Question

Answer: Option B. -> Rectangle
:
B

In the case of a rectangle, the diagonal is not an angle bisector.
Consider a rectangle ABCD.

Here, $∠$ ADB = $∠$ DBC (Alternate angles)
But, $∠$ DBC is not equal to $∠$ BDC (Angles opposite to unequal sides of a triangle are unequal)
$∴$ $∠$ ADB is not equal to $∠$ BDC
So, diagonal DB does not bisect $∠$ D.

Question 30.

Each point on a/an _________ is such that it forms an isosceles triangle with the end points of the given line segment.

perpendicular bisector
angular bisector
altitude
median

Discuss Question

Answer: Option A. -> perpendicular bisector
:
A

Consider the above figure.
Here, XY is the perpendicular bisector of a line AB.
Let P be any random point on XY.
In $△$ PMA $≅$ $△$ PMB
AM = BM (Perpendicular bisector divides a line segment into two equal halves)
$∠$ PMA = $∠$ PMB = 90 $^{\circ}$
Also PM is common side
So $△$ PMA $≅$ $△$ PMB (SAS Rule)
$∴$ PA = PB (CPCT)
Hence, $△$ PAB is an isosceles triangle.
So, the given statement is true.