Free Constructions 03 Practice Test - 10th Grade

Question 1

$Δ A B C$ 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 $Δ A^{'} B C^{'}$ .

$4 : 1$

$4 : 5$

$1 : 4$

$1 : 5$

SOLUTION

Solution : 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 2

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

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

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

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

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

SOLUTION

Solution : 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.

Question 3

To divide a line segment PQ in a certain ratio, we draw a ray PM. Why don’t we draw it with an obtuse angle?

We cannot measure an obtuse angle with the given line segment.

Drawing an obtuse angle would make my constructions very congested.

The diagram would be very large.

The textbook says we should draw an acute angle.

SOLUTION

Solution : B

Suppose we draw $P M$ such that the angle $∠ Q P M$ is an obtuse angle. Next we mark $(m + n)$ arcs on the ray PM such that $P P_{1}$ = $P_{1} P_{2}$ = $P_{2} P_{3}$ = … = $P_{m + n - 1}$ $P_{m + n}$ .

For our convenience let us assume $m + n = 5$ so we can explain things much easier. (In our case the required ratio could be 1:4 or 2:3 or 3:2 or 4:1 etc...)

Now in the $△$ $Q P P_{5}$ ,

If, $∠$ $Q P P_{5}$ is obtuse then $∠$ $Q P_{5} P$ and $∠$ $P Q P_{5}$ would become very small (compared to the case where $∠$ $Q P P_{5}$ is acute) and it would be difficult to draw a line parallel to $Q P_{5}$ (Because small angles and hence small radii of arcs are involved in construction of parallel lines)

This will make drawing of parallel lines very congested compared to drawing parallel lines on line segment PN with acute angle $∠ Q P N$ .

You can see from the diagram that $P_{4} Q^{'} ∥ P_{5} Q$ is more closely spaced compared to $P_{4}^{'} Q^{'} ∥ P_{5}^{'} Q$ .

So we prefer to draw an acute angle to make it more clear and spacious i.e. to avoid congestion.

Question 4

An arc AB forms $∠ A O B$ at the centre of the circle. If I draw 2 arcs with A and B as centre and a fixed radius 'r' such that they intersect at point P. The line OP will ___________ the $∠ A O B$ .

be at right angles to

bisect

trisect

be at zero angle to

SOLUTION

Solution : B

If we draw two arcs from any random points A and B with radius larger than half the distance between AB, the perpendicular bisector of AB passes through the point of intersection of the two arcs. If you consider the line AB to be a chord then the perpendicular of AB passing through centre will bisect the chord. By combining the two concepts we can arrive at the conclusion that the perpendicular passing though P and O bisects the arc AB. Therefore, OP will bisect the $∠ A O B$ .

Question 5

A triangle $A B C$ with $B C = 6 c m, A B = 5 c m$ and $∠ A B C = 60^{\circ}$ . The image of constructing a similar triangle of $Δ A B C$ whose sides are $\frac{3}{4}$ times the corresponding sides of the $Δ A B C$ is given below.

Arrange the steps of construction in the correct 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 $B X$ 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. 2, 1, 3, 4

C. 2, 3, 1, 4

D. 2, 4, 1, 3

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 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'.

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

Question 6

A triangle $A B C$ with $B C = 6 c m, A B = 5 c m$ and $∠ A B C = 60^{\circ}$ The image of constructing a similar triangle of $Δ A B C$ whose sides are $\frac{3}{4}$ times the corresponding sides of the $Δ A B C$ 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 a similar triangle of $Δ A B C$ , we draw any ray BX making an acute angle with BC on the side opposite to the vertex A.

Question 7

A triangle ABC with sides BC =7cm, $∠ B = 45^{\circ}, ∠ A = 105^{\circ}$ is given. The image of constructing a similar triangle of $Δ A B C$ whose sides are $\frac{3}{4}$ times the corresponding sides of the $Δ A B C$ is given below. Arrange the steps of construction in the correct order.

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

2.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'.

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

4.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 $B B_{1} = B_{1} B_{2} = B_{2} B_{3} = B_{3} B_{4}$

A. 1, 3, 2, 4

B. 1, 4, 2, 3

C. 2, 1, 3, 4

D. 2, 3, 1, 4

SOLUTION

Solution : B

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 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'.

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

Question 8

\frac{3}{4}

\frac{4}{3}

\frac{3}{7}

\frac{4}{7}

SOLUTION

Solution : A

As we draw $Δ A^{'} C^{'} B$ similar to $Δ A C B$ with corresponding side ratio is $\frac{3}{4}$ :
$\frac{B A^{'}}{B A} = \frac{B C^{'}}{B C} = \frac{A^{'} C^{'}}{A C} = \frac{3}{4}$

Question 9

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.
$∠ A^{'} C^{'} B = ∠ A C B \to$ AC||A'C'.

Question 10

The postulate used in the division of line segment to draw $A_{3} C$ parallel to $A_{5} B$ is :

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 03 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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Arithmetic Progressions 03
Pair of Linear Equations in Two Variables 03
Probability 03