# Free Constructions 02 Practice Test - 10th Grade

### Question 1

The ratio of BCCC′ is 3:5. What will the ratio of corresponding sides be if triangles ABC and A'BC' are similar? Given that BC and BC' lie

on the same ray BY.

5:8

3:8

8:5

8:3

#### SOLUTION

Solution :B

Since BC and BC' lie on the same line, the points B, C and C' are collinear.

BCCC′=35Therefore BC′BC = BC+CC′BC

BC′BC= 1+ CC′BC

CC′BC = 53.

So BC′BC=83

∴BCBC′ =38

The ratio of corresponding sides of triangles ABC and A′BC′ =38

### Question 2

AA postulate is used to prove the similarity of the constructed triangle.

True

False

#### SOLUTION

Solution :A

AA postulate is used to prove the similarity of the constructed triangle.

In the above diagram, ΔA′BC′ is similar to ΔABC. A'C' has been constructed parallel to AC. Hence, the similarity of the triangles can be proved by the AA similarity method.

### Question 3

Image of the division of line segment AB in the ratio 3:2 is given below.

∠BAX is a/an :

#### SOLUTION

Solution :A

During the construction to divide line segment AB in the given ratio, we draw any ray AX, making an acute angle with AB.

### Question 4

A triangle similar to given ΔABCwith sides equal to 34 of the sides of ΔABC is to be constructed as the given image. Arrange the following steps of construction in order:

1. Join B4C and raw a line through B3( the third point, 3 being smaller of 4 in34) parallel to B4C to intersect BC at C'.

2. Draw any ray BX making an acute angle with BC on the side opposite to the vertex A.

3. Locate 4(the greater of 3 and 4 in34) points B1,B2,B3 and B4 on BX so that BB1=B1B2=B2B3=B3B4

4. Draw a line through C' parallel to the line CA to intersect BA at A'.

#### SOLUTION

Solution :C

Steps to be followed to for the construction of a similar triangle of ΔABC is given below.

1. Draw any ray BX making an acute angle with BC on the side opposite to the vertex A.

2.Locate 4(the greater of 3 and 4 in34) points B1,B2,B3 and B4 on BX so that BB1=B1B2=B2B3=B3B4

3.Join B4C and draw a line through B3( the third point, 3 being smaller of 4 in34) parallel to B4C to intersect BC at C'.

4.Draw a line through C' parallel to the line CA to intersect BA at A'.

### Question 5

A triangle similar to given ΔABCwith sides equal to 34 of the sides of ΔABC is constructed.

BC′BC is equal to____.

#### SOLUTION

Solution :B

As ΔABC∼ΔA′BC′,

BC′BC=BA′BA=A′C′AC=34

### Question 6

Initial step for constructing a similar triangle of ΔABC is given below ∠CBX is a/an:

#### SOLUTION

Solution :A

For the construction of similar triangle, we draw a ray BX making anacute anglewith BCon the side opposite to to the vertex A.

### Question 7

Construction of a ΔA′C′B similar to ΔACB is shown in the figure below, which condition is used to construct A′C′ ||AC?

#### SOLUTION

Solution :A

Here we construct A′C′||AC by making ∠A′C′B=∠ACB and we used the corresponding angle property (if corresponding angles of two lines are equal, then the lines are parallel).

### Question 8

If a triangle similar to given ΔABC with sides equal to 34 of the sides of ΔABC is to be constructed, then the number of points to be marked on ray BX is __.

#### SOLUTION

Solution :B

In the ratio between sides 34 , 4>3

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

### Question 9

Steps to divide a line segment AB in the given ratio m : n by drawing alternated angles is given. Choose the correct order.

1.Draw any ray AX making acute angle with AB and ray BY such that ∠BAX=∠ABY.

2.Locate the points A1,A2,A3...Am on AX and B1,B2,B3...Bn on BY such that AA1=A1A2=BB1=B1B2

3.Draw a ray BY parallel to AX by making ∠ABY=∠BAX

4.Join AmBn.

#### SOLUTION

Solution :B

Steps to divide a line segment AB in the given ratio m:n by corresponding angles method is given. Choose the correct order.

1.Draw any ray AX making acuter angle with AB.

2Draw a ray BY parallel to AX by making ∠ABY=∠BAX

3..Locate the points A1,A2,A3...Am on AX and B1,B2,B3...Bn on BY such that AA1=A1A2=BB1=B1B2

4.Join AmBn. Let it intersect AB at point C.

Then we get,

### Question 10

What is the ratio ACBC for the line segment AB following the construction method below?

Step 1. A ray is extended from A and 30 arcs of equal lengths are cut, cutting the ray at A1,A2,...,A30

Step 2. A line is drawn from A30 to B and a line parallel to A30B is drawn, passing through the point A17 and meet AB at C.

17:30

17:13

13:17

13:30

#### SOLUTION

Solution :B

Here the total number of arcs is equal to m+n in the ratio m:n.

The triangles △ AA17C and △ AA30B are similar.

Hence, ACAB = AA17AA30=1730

BCAB = AB−ACAB

BCAB = 1 - 1730 = 1330Hence, ACBC = 1713→17:13