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### CBSE Class 10 Maths Chapter 11- Construction Objective Questions

CBSE Class 10 Maths Chapter 11- Construction Objective Questions

CBSE Class 10 Maths Chapter 11 – Constructions include topics such as Division of a Line Segment, Constructions of Tangents to a Circle, Line Segment Bisector and so on. Class 10, Chapter 11-Constructions is one of the scoring chapters that come under the unit Geometry. Also, based on the latest modification of the exam pattern, we have compiled here CBSE Class 10 Maths Chapter 11- Construction Objective Questions for the students to get practice.

Find the sub-topics covered in the chapter and based on which the CBSE Class 10 Maths objective questions are formed.

Find from the link above the list of MCQs organized topic wise from the chapter 11 Constructions of CBSE Class 10 Maths.

11.1 Constructing Similar Triangles (8 MCQs From This Topic)

11.2 Construction of Tangents to a Circle (3 MCQs Listed From The Topic)

11.3 Drawing Tangents to a Circle (3 MCQs From The Given Topic)

11.4 Dividing a Line Segment (6 MCQs From The Topic)

A ray BX is drawn from B making an acute angle with AB.

5 points B1,B2,B3,B4 and 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 ΔA′BC′.

According to the construction, ΔBB 4 A∼ΔBB 5 A′

And for similar triangles, the ratio of corresponding sides is AB/A′B.

The ratio of corresponding sides is 4:5.

Answer: (C) Perpendicular distance between AC and Point P and angle between AC and PR.

Solution: Consider we need to draw a triangle △PQR ~ △ABC. But △PQR is at an isolated position w.r.t. △ABC. Therefore we will need to know the  perpendicular distance between AC and point P  and the  angle between AC and PR.

Now let us reframe the question with the required information.

Construct a triangle △PQR similar to △ABC, such that point P is at a perpendicular distance 5 cm from the line AC, PR makes an angle of 60 o  and which is 23rd of △ABC. In triangle △ABC, AB = 6 cm, AC = 7 cm, and ∠ABC = 30 o

2. Measure an angle of 30 o  w.r.t AC and draw a straight line

3. Set the compass at 6 cm length and with A as centre mark the length on the line, name the point of intersection as B.

Reason: This is done to draw the line AB. i.e. we are marking the length of 6 cm on the line 30 o  inclined to AC

4. Connect the points to BC to complete the triangle.

5. Draw a line perpendicular to AC at A.

Reason: We are drawing this step to identify the line on which the point P might lie.

6. For the given length of AP, cut the perpendicular at P Reason: We are drawing the arc to identify the exact position of point P on the line perpendicular to AC

7. Extrapolate line segment AC. Reason:

8. Draw a Ray XY 60 o  to AC, such that the line passes through the point P.

9. Measure the distance AC with a compass and draw an arc with point P as centre and radius equal to AC, the point of intersection is R’.

10. Measure z.BAC and draw a similar angle on PR’.

11. Measure AB and mark on PZ as PQ. Join QR’.

12. Draw a ray PM, mark 3 points P 1 P 2 P 3  such that PP 1 =P 1 P 2 =P 2 P 3

14. Through P2 draw a line parallel to P3R′, let the point of intersection of the line with PR’ is R.

15. Through the point R, draw a line parallel to R’Q’. Let the point of intersection of the line with line PO! is Q. The triangle △PQR which is similar to &△ABC such that, the angle between PR and AC is 120 o  and perpendicular distance of point P from AC is 5 cm.

Solution: If the perpendicular distance of AP is given then, we would start the construction of similar triangle by finding the position of the point P.

Solution: Once we know the position of Point P, we need to find the orientation of the △PQR w.r.t △ABC. For that purpose we need to know the angle of any one side that makes with the original triangle. Therefore, among the given options we will be able to determine the orientation of the △PQR if we know the angle which side PR makes with side AC.

Solution: 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 (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 (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 (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.

∠BA′C′=∠BAC (corresponding angles of similar triangles)

Solution: In the ratio between sides 3/4, 4>3

⇒ The number of points to be marked on BX to construct similar triangles is 4.

Answer: (B) Find two points one on the larger side and other on the smaller side using the given scale factor and use these scaled lengths to construct a similar parallelogram.

The following steps will give you the information on how to construct a similar parallelogram to ABCD.

Step 1: Find points E and F on longer and smaller sides respectively using the given scale factor.

Step 2: Draw a line from E parallel to smaller side AD.

Step 3: Taking the length of AF and E as center cut an arc on the line parallel to AD and let this new point be G.

Step 5: AEFG is the required parallelogram.

Now we have constructed the parallelogram AEFG ∼ ABCD.

Answer: (C) Mark M and N on the circle such that ∠MOE = 60 °  and ∠NOE = 60 °

Solution: Since the angle between the tangents is 60 °  and OE bisects ∠MEN, ∠MEO = 30 ° .

Now, since △OME is a right angled triangle, right angled at M, we realise that the ∠MOE = 60 ° . Since ∠MOE = 60 ° , we must have ∠NOE =60 °  and hence ∠MON = 120 ° . Hence △MNO is NOT equilateral.

Next, since in △OME, sin30 ° = 1/2 = OM/OE = r/d, we have d = 2r.

Recalling that ∠MOE = 60 ° , following are the steps of construction:

2. With O as centre, construct ∠MOE = 60 °  [constructing angle 60 °  is easy]

3. Now extend OM and from M, draw a line perpendicular to OM. This intersects the ray at E. This is the point from where the tangents should be drawn, EM is one tangent.

4. Similarly, EN is another tangent.

Solution: Since we need to draw a circle with radius ‘r’, we need the following points:

We have the radius ‘r’ but we need to adjust the centre on the line OA.

Realising that the tangents are common to both the circles, the radius of each circle at their point of contact being perpendicular to the common tangent, we can say that the radii are parallel. So we also have the ratio PQ: QO because we have a pair of similar triangles.

But before we can do all this, we first need to have the line AO and even before that, the point O. Only then we can draw tangents and then the inner circle. So the next step would be determining the point ‘O’.

Answer: (D) Centre of the circle ‘C’

Solution: Since we need to finally construct a circle of radius ‘R’ concentric to the previous circle, we need to determine the centre of these circles first, before proceeding with anything else.

Radius of the circle ‘D’ can be figured out after we get the radius of the first circle and using the centre.

Radius of the circle ‘C’ can be found after finding the centre of this circle.

Answer : (C) There are 2 tangents to the circle from point P

Solution: Only one tangent can be drawn from a point on the circle and the tangent is always perpendicular to the radius.

Solution: Since only one tangent can be drawn, this point P should be present on the circle.

Any point on the circle is at a distance equal to the radius of the circle. So OP is equal to the radius of the circle.

Solution: P is a point on the circle. We know that only one tangent can be drawn and it is perpendicular to the line joining the centre of the circle O and the point of contact P.

So our first step would be to join OP.

AB/BC= (AC+BC)/BC = (AC/BC) +1 = 5/6+1 = 11/6

XY/ XW= (XW/XW) + (WY/XW) = 1 + (n/m) = (m+n)/m = m+n: m

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.

Solution: Here the total number of arcs is equal to m+n in the ratio m: n.

The triangles △ AA 17 C and △ AA 30 B are similar.

Hence, AC/AB = AA 17 /AA 30  = 17/30

A single ray is extended from A and 12 arcs of equal lengths are cut, cutting the ray at A1, A2… A12.

A line is drawn from A12 to B and a line parallel to A12B is drawn, passing through the point A6 and cutting AB at C.

In the construction process given, triangles △AA 12 B   △AA 6 C are similar.

By construction BC/AB = 6/12 = 1/2.