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

10th Grade > Mathematics

CONSTRUCTIONS MCQs

Total Questions : 58 | Page 4 of 6 pages

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

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 32.

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 33.

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

Answer: Option B. -> AA Similarity
:

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 34.

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 = $60^{\circ}$ and $∠$ NOE = $60^{\circ}$ .
Mark M and N on the circle such that $∠$ MOE = $120^{\circ}$ and $∠$ NOE = $120^{\circ}$ .

Discuss Question

Answer: Option C. -> Mark M and N on the circle such that

∠

MOE =

60^{\circ}

and

∠

NOE =

60^{\circ}

.
:
C

Since the angle between the tangents is 60°, we get $∠ M O N = 120^{\circ}$
(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 ray at E. This is the point from where the tangents should be drawn and EM is one tangent.

4. Similarly, EN is another tangent.

Question 35.

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 to B

A_{m + n}

through

A_{m}

to AB.

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

Question 36.

State whether true or false:
The line segment AB can be divided in a ratio only if the given ratio is less than 1.

True
False
2:1
3:1

Discuss Question

Answer: Option B. -> False
:
B

False. As long as the ratio is any positive rational number, a line segment can be divided in that ratio.

Question 37.

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?

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

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

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 38.

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 39.

$Δ A B C$ of dimesions $A B = 4 c m, B C = 5 c m$ and ∠B= 60^ois given.
A ray $B X$ is drawn from $B$ making an acute angle with $A B$ .
5 points $B_{1}, B_{2}, B_{3}, B_{4} & B_{5}$ are located on the ray such that $B B 1 = B 1 B 2 = B 2 B 3 = B 3 B 4 = B 4 B 5$ .
$B_{4}$ is joined to $A$ and a line parallel to $B_{4} A$ is drawn through $B_{5}$ to intersect the extended line $A B$ at $A^{'}$ .
Another line is drawn through $A^{'}$ parallel to $A C$ , intersecting the extended line $B C$ at $C^{'}$ .
Find the ratio of the corresponding sides of $Δ A B C$ and $Δ A^{'} B C^{'}$ .

$4 : 1$
$4 : 5$
$1 : 4$
$1 : 5$

Discuss Question

Answer: Option B. ->

4 : 5

:
B

According to the construction, $Δ B B_{4} A \sim Δ B B_{5} A^{'}$

And for similar triangles, the ratio of corresponding sides is $\frac{A B}{A^{'} B}$ .

The ratio of corresponding sides is $4 : 5$ .

Question 40.

Match the following based on the construction of similar triangles, if scale factor $(\frac{m}{n})$ is
$\begin{matrix} I . > 1 & a) T h e s i m i l a r t r i a n g l e i s s m a l l e r t h a n t h e o r i g i n a l t r i a n g l e . I I . < 1 & b) T h e t w o t r i a n g l e s a r e c o n g r u e n t t r i a n g l e s . I I I . = 1 & c) T h e s i m i l a r t r i a n g l e i s l a r g e r t h a n t h e o r i g i n a l t r i a n g l e . \end{matrix}$

$I - c, I I - a, I I I - b$
$I - b, I I - a, I I I - c$
$I - a, I I - c, I I I - b$
$I - a, I I - b, I I I - c$

Discuss Question

Answer: Option A. ->

I - c, I I - a, I I I - b

:
A

Scale factor basically defines the ratio between the sides of the constructed triangle to that of the original triangle.
So when we see the scale factor $(\frac{m}{n}) > 1$ , it means the sides of the constructed triangle is larger than the original triangle i.e., the triangle constructed is larger than the original triangle.
Similarly, if scale factor $(\frac{m}{n}) < 1$ , then the sides of the constructed triangle is smaller than that of the original triangle i.e., the constructed triangle is smaller than the original triangle.
When we have scale factor $(\frac{m}{n}) = 1$ , then the sides of both the constructed triangle and that of the original triangle is equal.
When a pair of similar triangles have equal corresponding sides, then the pair of similar triangles can be called as congruent because then the triangles will have equal corresponding sides and equal corresponding angles.