# Free Constructions 03 Practice Test - 10th Grade

ΔABC of dimesions AB=4 cm,BC=5 cm and ∠B= 60is given.
A ray BX is drawn from B making an acute angle with AB.
5 points B1,B2,B3,B4 & B5 are located on the ray such that BB1=B1B2=B2B3=B3B4=B4B5.
B4 is joined to A and a line parallel to B4A is drawn through B5 to intersect the extended line AB at A.
Another line is drawn through A parallel to AC, intersecting the extended line BC at C.
Find the ratio of the corresponding sides of ΔABC and ΔABC.

A.

4:1

B.

4:5

C.

1:4

D.

1:5

#### SOLUTION

Solution : B According to the construction, ΔBB4AΔBB5A

And for similar triangles, the ratio of corresponding sides is ABAB.

The ratio of corresponding sides is 4:5.

Match the following based on the construction of similar triangles, if scale factor (mn) is
I.   >1a) The similar triangle is smaller than the original triangle.II.  <1b) The two triangles are congruent triangles.III. =1c) The similar triangle is larger than the original triangle.

A.

Ic,IIa,IIIb

B.

Ib,IIa,IIIc

C.

Ia,IIc,IIIb

D.

Ia,IIb,IIIc

#### 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 (mn)>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 (mn)<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 (mn)=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.

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?

A.

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

B.

Drawing an obtuse angle would make my constructions very congested.

C.

The diagram would be very large.

D.

The textbook says we should draw an acute angle.

#### SOLUTION

Solution : B

Suppose we draw PM such that the angle QPM is an obtuse angle. Next we mark (m+n) arcs on the ray PM such that PP1 = P1P2 = P2P3 = … = Pm+n1Pm+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 QPP5,

If, QPP5 is obtuse then QP5P and PQP5 would become very small (compared to the case where QPP5 is acute) and it would be difficult to draw a line parallel to QP5 (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 QPN. You can see from the diagram that P4QP5Q is more closely spaced compared to P4QP5Q.

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

An arc AB forms AOB 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 AOB.

A.

be at right angles to

B.

bisect

C.

trisect

D.

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 AOB. A triangle ABC with BC=6 cm,AB=5 cm and ABC=60. The image of constructing a similar triangle of ΔABC whose sides are 34 times the corresponding sides of the ΔABC is given below. Arrange the steps of construction in the correct order.
1. Join B4C and raw a line through B3
(the third point, 3 being smaller of 4 in 34) 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'.

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 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 raw 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'. A triangle ABC with BC=6 cm,AB=5 cm and ABC=60The image of constructing a similar triangle of ΔABC whose sides are 34 times the corresponding sides of the ΔABC is given below. CBX 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 ΔABC, we draw any ray BX making an acute angle with BC on the side opposite to the vertex A.

A triangle ABC with sides BC =7cm, B=45,A=105 is given. The image of constructing a similar triangle of ΔABC whose sides are 34 times the corresponding sides of the ΔABC 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 B4C and raw a line through B3( the third point, 3 being smaller of 4 in34) parallel to B4C  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 in34) points B1,B2,B3 and B4 on BX  so that BB1=B1B2=B2B3=B3B4 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 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 raw 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'. A triangle ABC with sides BC =7cm, B=45,A=105 is given. The image of constructing a similar triangle of ΔABC whose sides are 34 times the corresponding sides of the ΔABC is given below. Then ACAC is equal to : A. 34
B. 43
C. 37
D. 47

#### SOLUTION

Solution : A

As we draw ΔACB similar to ΔACB with corresponding side ratio is 34:
BABA=BCBC=ACAC=34

A triangle ABC with sides BC =7cm, B=45,A=105 is given. The image of constructing a similar triangle of ΔABC whose sides are 34 times the corresponding sides of the ΔABC is given below. The condition used to draw A'C' parallel to AC:

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.
ACB=ACB AC||A'C'. The postulate used in the division of line segment to draw A3C parallel to A5B 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.
AA3C=AA5C as A3C is parallel to A5B