Free Constructions 01 Practice Test - 10th Grade

Question 1

A line segment $A B$ is divided in the ratio $m : n$ where $m$ and $n$ are co-prime, using a single ray $A X$ . The number of arcs to be drawn on $A X$ is _____.

m+n

m-n

SOLUTION

Solution : C

The number of arcs to be drawn, when a single ray is used to divide a line segment in the ratio $m : n$ , where $m$ and $n$ are co-prime, is $m + n$ .

Question 2

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

SAS Similarity

AA Similarity

SSS Similarity

ASA Similarity

SOLUTION

Solution : B

AA(Angle-Angle) similarity is used to prove that the constructed triangles are similar.

Question 3

What will the ratio $A B : A C$ be if $C$ divides the line segment $A B$ in the ratio $5 : 12$ ?

$5 : 12$

$17 : 12$

$12 : 17$

$17 : 5$

SOLUTION

Solution : D

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 4

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.

1:2

1:1

2:1

3:1

SOLUTION

Solution : 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 5

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

SOLUTION

Solution : B

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

Question 6

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 = $120^{\circ}$ .

SOLUTION

Solution : 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 7

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

___

SOLUTION

Solution :
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 8

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.

A. 2,1,3,4

B. 1,2,3,4

C. 1,3,2,4

D. 2,3,1,4

SOLUTION

Solution : 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.

Question 9

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?

A. Corresponding angle are equal

B. Alternate interior angles are equal

C. Co-interior angle are supplementary

D. Perpendicular bisector theorem

SOLUTION

Solution : 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 10

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?

A. Corresponding angles are equal

B. Alternate interior angles are equal

C. Co-interior angles are supplementary

D. Perpendicular bisector theorem

SOLUTION

Solution : A

Here we use the principle that when corresponding angles are equal, the lines are parallel.
$∠ A A_{3} C = ∠ A A_{5} C$ as $A_{3} C$ is parallel to $A_{5} B .$

Free Constructions 01 Practice Test - 10th Grade

Question 1

SOLUTION

Question 2

SOLUTION

Question 3

SOLUTION

Question 4

SOLUTION

Question 5

SOLUTION

Question 6

SOLUTION

Question 7

SOLUTION

Question 8

SOLUTION

Question 9

SOLUTION

Question 10

SOLUTION

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