Free Constructions 02 Practice Test - 10th Grade

Question 1

The ratio of $\frac{B C}{C C^{'}}$ 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.
$\frac{B C}{C C^{'}}$ = $\frac{3}{5}$

Therefore $\frac{B C^{'}}{B C}$ = $\frac{B C + C C^{'}}{B C}$

$\frac{B C^{'}}{B C}$ = 1+ $\frac{C C^{'}}{B C}$

$\frac{C C^{'}}{B C}$ = $\frac{5}{3}$ .

So $\frac{B C^{'}}{B C} = \frac{8}{3}$

$∴ \frac{B C}{B C^{'}}$ = $\frac{3}{8}$

The ratio of corresponding sides of triangles $A B C$ and $A^{'} B C^{'}$ = $\frac{3}{8}$

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, $Δ A^{'} B C^{'}$ is similar to $Δ A B C$ . 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.

$∠ B A X$ is a/an :

A. acute angle

B. right angle

C. obtuse angle

D. reflex angle

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 $Δ A B C$ with sides equal to $\frac{3}{4}$ of the sides of $Δ A B C$ is to be constructed as the given image. Arrange the following steps of construction in order:

1. Join $B_{4} C$ and raw a line through $B_{3}$ ( the third point, 3 being smaller of 4 in $\frac{3}{4}$ ) parallel to $B_{4} C$ 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 in $\frac{3}{4}$ ) points $B_{1}, B_{2}, B_{3} and B_{4}$ on BX so that $B B_{1} = B_{1} B_{2} = B_{2} B_{3} = B_{3} B_{4}$

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

A. 1,2,3,4

B. 4,3,2,1

C. 2,3,1,4

D. 2,1,3,4

SOLUTION

Solution : C

Steps to be followed to for the construction of a similar triangle of $Δ A B C$ 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 in $\frac{3}{4}$ ) points $B_{1}, B_{2}, B_{3} and B_{4}$ on BX so that $B B_{1} = B_{1} B_{2} = B_{2} B_{3} = B_{3} B_{4}$

3.Join $B_{4} C$ and draw a line through $B_{3}$ ( the third point, 3 being smaller of 4 in $\frac{3}{4}$ ) parallel to $B_{4} C$ 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 $Δ A B C$ with sides equal to $\frac{3}{4}$ of the sides of $Δ A B C$ is constructed.

$\frac{B C^{'}}{B C}$ is equal to____.

\frac{B A^{'}}{C A^{'}}

\frac{B A^{'}}{B A}

\frac{B C}{A C}

\frac{B C^{'}}{B_{3} C^{'}}

SOLUTION

Solution : B

As $Δ A B C \sim Δ A^{'} B C^{'}$ ,
$\frac{B C^{'}}{B C} = \frac{B A^{'}}{B A} = \frac{A^{'} C^{'}}{A C} = \frac{3}{4}$

Question 6

Initial step for constructing a similar triangle of $Δ A B C$ is given below $∠ C B X$ is a/an:

A. acute angle

B. right angle

C. obtuse angle

D. reflex angle

SOLUTION

Solution : A

For the construction of similar triangle, we draw a ray BX making an acute angle with BC on the side opposite to to the vertex A.

Question 7

Construction of a $Δ A^{'} C^{'} B$ similar to $Δ A C B$ is shown in the figure below, which condition is used to construct $A^{'} C^{'}$ || $A C$ ?

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 construct $A^{'} C^{'} | | A C$ by making $∠ A^{'} C^{'} B = ∠ A C B$ 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 $Δ A B C$ 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 __.

A. 3

B. 4

C. 7

D. 6

SOLUTION

Solution : 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 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 $∠ B A X = ∠ A B Y$ .
2.Locate the points $A_{1}, A_{2}, A_{3} . . . A_{m}$ on AX and $B_{1}, B_{2}, B_{3} . . . B_{n}$ on BY such that $A A_{1} = A_{1} A_{2} = B B_{1} = B_{1} B_{2}$
3.Draw a ray BY parallel to AX by making $∠ A B Y = ∠ B A X$
4.Join $A_{m} B_{n}$ .

A. 1,2,3,4

B. 1,3,2,4

C. 2,1,3,4

D. 2,3,1,4

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 $∠ A B Y = ∠ B A X$
3..Locate the points $A_{1}, A_{2}, A_{3} . . . A_{m}$ on AX and $B_{1}, B_{2}, B_{3} . . . B_{n}$ on BY such that $A A_{1} = A_{1} A_{2} = B B_{1} = B_{1} B_{2}$
4.Join $A_{m} B_{n}$ . Let it intersect AB at point C.
Then we get,

Question 10

What is the ratio $\frac{A C}{B C}$ 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 $A_{1}, A_{2}, . . ., A_{30}$
Step 2. A line is drawn from $A_{30}$ to $B$ and a line parallel to $A_{30} B$ is drawn, passing through the point $A_{17}$ 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 $△$ $A A_{17} C$ and $△$ $A A_{30} B$ are similar.

Hence, $\frac{A C}{A B}$ = $\frac{A A_{17}}{A A_{30}} = \frac{17}{30}$

$\frac{B C}{A B}$ = $\frac{A B - A C}{A B}$
$\frac{B C}{A B}$ = 1 - $\frac{17}{30}$ = $\frac{13}{30}$

Hence, $\frac{A C}{B C}$ = $\frac{17}{13} \to 17 : 13$

Free Constructions 02 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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