10th Grade > Mathematics
CONSTRUCTIONS MCQs
:
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+n−1Pm+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 P4Q′∥P5Q is more closely spaced compared to P′4Q′∥P′5Q.
So we prefer to draw an acute angle to make it more clear and spacious i.e. to avoid congestion.
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'.
:
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'.
:
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
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
:
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
As we draw ΔA′C′B similar to ΔACB with corresponding side ratio is 34:
BA′BA=BC′BC=A′C′AC=34