10th Grade > Mathematics
CONSTRUCTIONS MCQs
Total Questions : 58
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Question 1. 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.
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.
Answer: Option B. -> 1,3,2,4
:
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,
:
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,
Answer: Option C. -> m+n
:
C
The number ofarcs to be drawn, when a single ray is used to divide a line segment in the ratio m:n, where m and n are co-prime, is m+n.
:
C
The number ofarcs to be drawn, when a single ray is used to divide a line segment in the ratio m:n, where m and n are co-prime, is m+n.
Answer: Option A. -> I−c,II−a,III−b
:
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 haveequal corresponding sides, then the pair of similar triangles can be called as congruent because then thetriangles will have equal corresponding sides and equal corresponding angles.
:
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 haveequal corresponding sides, then the pair of similar triangles can be called as congruent because then thetriangles will have equal corresponding sides and equal corresponding angles.
Question 4. 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'.
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'.
Answer: Option C. -> 2, 3, 1, 4
:
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 and4 in34) points B1,B2,B3andB4 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 toB4C to intersect BC at C'.
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 and4 in34) points B1,B2,B3andB4 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 toB4C to intersect BC at C'.
4.Draw a line through C' parallel to the line CA to intersect BA at A'.
Answer: Option A. -> Corresponding angles are equal
:
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).
:
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 6. 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'.
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'.
Answer: Option C. -> 2,3,1,4
:
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 and4 in34) points B1,B2,B3andB4 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 toB4C to intersect BC at C'.
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 and4 in34) points B1,B2,B3andB4 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 toB4C to intersect BC at C'.
4.Draw a line through C' parallel to the line CA to intersect BA at A'.
Answer: Option A. -> acute angle
:
A
During the construction to divide line segment AB in the given ratio, we draw any ray AX, making an acute angle with AB.
:
A
During the construction to divide line segment AB in the given ratio, we draw any ray AX, making an acute angle with AB.
Answer: Option B. -> BA′BA
:
B
As ΔABC∼ΔA′BC′,
BC′BC=BA′BA=A′C′AC=34
:
B
As ΔABC∼ΔA′BC′,
BC′BC=BA′BA=A′C′AC=34