Constructions(10th Grade > Mathematics ) Questions and answers for exam Preparation

10th Grade > Mathematics

CONSTRUCTIONS MCQs

Total Questions : 58 | Page 2 of 6 pages

Question 11.
If a triangle similar to given

Δ A B C

with sides equal to

\frac{3}{4}

of the sides of

Δ A B C

is to be constructed, then the number of points to be marked on ray BX is __.

If A Triangle Similar To Given ΔABC With Sides Equal To 34 ...

Discuss Question

Answer: Option B. -> 4
:
B
In the ratio between sides

\frac{3}{4}

4 > 3

\Rightarrow

The number of points to be marked on BX to construct similar triangles is 4.

Question 12. Initial step for constructing a similar triangle of

Δ A B C

is given below

∠ C B X

is a/an:

Initial Step For Constructing A Similar Triangle Of ΔABC Is...

acute angle
right angle
obtuse angle
reflex angle

Discuss Question

Answer: Option A. -> acute angle
:
A
For the construction of similar triangle, we draw a ray BX making anacute angle with BCon the side opposite to to the vertex A.

Question 13. Which similarity is used to prove that the constructed triangles are similar?

SAS Similarity
AA Similarity
SSS Similarity
ASA Similarity

Discuss Question

Answer: Option B. -> AA Similarity
:
B
AA(Angle-Angle) similarity is used to prove that the constructed triangles are similar.

Question 14. In the given image, segment

A B

has been divided in the ratio

3 : 2

. This is done by
1) Drawing

∠ B A X

2) marking equal lengths

A A_{1}, A_{1} A_{2}, A_{2} A_{3}, A_{3} A_{4} & A_{4} A_{5}

3) Point

A_{5}

is joint to point B
4)

A_{3} C

is drawn parallel to

A_{5} B

by using which of the properties of parallel lines?

In The Given Image, Segment AB Has Been Divided In The Ratio...

Corresponding angles are equal
Alternate interior angles are equal
Co-interior angles are supplementary
Perpendicular bisector theorem

Discuss Question

Answer: Option A. -> Corresponding angles are equal
:
A

Here we use the principle that when corresponding angles are equal, the lines are parallel.

∠ A A_{3} C = ∠ A A_{5} C

A_{3} C

is parallel to

A_{5} B .

Question 15. The line segment AB was divided in the ratio 4:7 by taking 2 rays. The number of arcs to be made on the ray AX is
___

Discuss Question

:
The line segment AB is divided in the ratio 4:7. The number of divisions to be made on the ray AX is 4 + 7 = 11.

Question 16. Steps to divide a line segment AB in the given ratio 3 : 2 by corresponding angles method is given. Choose the correct order.
1. Draw any ray AX making an acute angle with AB
2.Locate 5 points

A_{1}, A_{2}, A_{3} . . . . A_{5}

on ray AX
3. Join

B A_{5}

4.Draw a line parallel to

B A_{5}

through

A_{3}

to AB.

2,1,3,4
1,2,3,4
1,3,2,4
2,3,1,4

Discuss Question

Answer: Option B. -> 1,2,3,4
:
B
To divide a line segment by corresponding angle method, we have to follow the steps
1. Draw any ray AX making an acute angle with AB
2.Locate (m+n)

A_{1}, A_{2}, A_{3} . . A_{m + n}

points in AX
3. Join B

A_{m + n}

4.Draw a line parallel toB

A_{m + n}

through

A_{m}

to AB.

Steps To Divide A Line Segment AB In The Given Ratio 3 : 2...

Question 17. In the given image, segment AB has been divided in the ratio 3:2. This is done by
1. Draw any ray AX making acute angle with AB.
2. Draw a ray BY parallel to AX by making

∠ A B Y = ∠ B A X

3. Locate the points

A_{1}, A_{2}, A_{3} . . . A_{3}

on AX and

B_{1}, B_{2}

on BY such that

A A_{1} = A_{1} A_{2} = B B_{1} = B_{1} B_{2}

4. Join

A_{3} B_{2}

by using which of the properties of parallel lines?

Corresponding angle are equal
Alternate interior angles are equal
Co-interior angle are supplementary
Perpendicular bisector theorem

Discuss Question

Answer: Option B. -> Alternate interior angles are equal
:
B

Here we use alternate interior angles are equal then the lin4es are parallel and

∠ X A B = ∠ A B Y

as AX parallel to BY.

Question 18. What is the ratio

\frac{A C}{B C}

for the following construction:
A line segment AB is drawn.
A single ray is extended from A and 12 arcs of equal lengths are cut, cutting the ray at

A_{1}, A_{2} \dots A_{12}

.
A line is drawn from

A_{12}

to B and a line parallel to

A_{12} B

is drawn, passing through the point

A_{6}

and cutting AB at C.

Discuss Question

Answer: Option B. -> 1:1
:
B

What Is The Ratio ACBC For The Following Construction:A li...

In the construction process given, triangles

△ A A_{12} B

and

△ A A_{6} C

are similar.
Hence, we get

\frac{A C}{A B} = \frac{6}{12} = \frac{1}{2}

.
By construction

\frac{B C}{A B} = \frac{6}{12} = \frac{1}{2}

\frac{A C}{B C} = \frac{\frac{A C}{A B}}{\frac{B C}{A B}}

= \frac{\frac{1}{2}}{\frac{1}{2}} = 1.

Question 19. What will the ratio

A B : A C

be if

C

divides the line segment

A B

in the ratio

5 : 12

5:12
17:12
12:17
17:5

Discuss Question

Answer: Option D. -> 17:5
:
D

What Will The Ratio AB:AC Be If C Divides The Line Segment A...

Given

\frac{A C}{B C} = \frac{5}{12}

Therefore,

\frac{B C}{A C} = \frac{12}{5}

\frac{A B}{A C} = \frac{A C + B C}{A C} = 1 + \frac{B C}{A C} = 1 + \frac{12}{5} = \frac{17}{5}

So, the required ratio

= 17 : 5

Question 20. You are given a circle with radius 'r' and centre 'O'. You are asked to draw a pair of tangents which are inclined at an angle of 60° with each other, from a point E.
Refer to the figure and select the option which would lead you to the required construction. The distance d is the distance OE.

Using trigonometry, arrive at d = r and mark E.
Construct the △MNO as it is equilateral triangle.
Mark M and N on the circle such that ∠MOE = 60∘ and ∠NOE = 60∘.
Mark M and N on the circle such that ∠MOE = 120∘ and ∠NOE = 120∘.

Discuss Question

Answer: Option C. -> Mark M and N on the circle such that ∠MOE = 60∘ and ∠NOE = 60∘.
:
C
Since the angle between the tangents is 60°, weget

∠ M O N = 120^{\circ}

(As MONE is a quadrilateral and sum of angles of a quadrilateral is

360^{\circ}

).
Hence, ΔMNO is NOT equilateral.
Since E is outside the circle, d can not be equal to r.
We know that ∠MOE = 60°, following are the steps of construction:
1. Draw a ray from the centre O.
2. With O as centre, construct ∠MOE = 60° .
3. Now extend OM and from M, draw a line perpendicular to OM. This intersects the rayat E. This is the point from where the tangents should be drawn andEM is one tangent.
4. Similarly, EN is another tangent.