Here we are providing Constructions Class 10 Extra Questions Maths Chapter 11 with Answers Solutions, Extra Questions for Class 10 Maths was designed by subject expert teachers.

## Extra Questions for Class 10 Maths Constructions with Answers Solutions

**Extra Questions for Class 10 Maths Chapter 11 Constructions with Solutions Answers**

### Constructions Class 10 Extra Questions Very Short Answer Type

Question 1.

Is construction of a triangle with sides 8 cm, 4 cm, 4 cm possible?

Solution:

No, we know that in a triangle sum of two sides of a triangle is greater than the third side. So the condition is not satisfied.

Question 2.

To divide the line segment AB in the ratio 5 : 6, draw a ray AX such that ∠BAX is an acute angle, then draw a ray BY parallel to AX and the point A_{1}, A_{2}, A_{3}… and B_{1}, B_{2}, B_{3}… are located at equal distances on ray AX and BY respectively. Then which points should be joined?

Solution:

A_{5} and B_{6}.

Question 3.

To draw a pair of tangents to a circle which are inclined to each other at an angle of 60°, it is required to draw tangents at end points of those two radii of the circle. What should be the angle between them?

Solution:

120°

Question 4.

In Fig. 9.1 by what ratio does P divide AB internally.

Solution:

From Fig. 9.1, it is clear that there are 3 points at equal distances on AX and 4 points at equal distances on BY. Here P divides AB on joining A_{3} B_{4}. So P divides internally by 3 : 4.

Question 5.

Given a triangle with side AB = 8 cm. To get a line segment AB’ = 2 of AB, in what ratio will line segment AB be divided?

Solution:

Given AB = 8 cm

AB’ = \(\frac{3}{4}\) of AB

= \(\frac{3}{4}\) × 8 = 6 cm

BB’ = AB – AB’ = 8 – 6 = 2 cm.

⇒ AB’: BB’ = 6 : 2 = 3 : 1

Hence the required ratio is 3 : 1.

### Constructions Class 10 Extra Questions Short Answer Type I and II

Question 1.

Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are \(\frac{2}{3}\) of the corresponding sides of the first triangle.

OR

Draw a triangle with sides 4 cm, 5 cm and 6 cm. Then construct another triangle whose sides are \(\frac{2}{3}\) of the corresponding sides of first triangle.

Solution:

Steps of Construction:

Step I: Draw a line segment BC = 6 cm

Step II: Draw an arc with B as centre and radius equal to 5 cm.

Step III: Draw an arc, with C as centre and radius equal to 4 cm intersecting the previous drawn arc at A.

Step IV: Join AB and AC, then ∆ABC is the required triangle.

Step V: Below BC make an acute angle CBX

Step VI: Along BX mark off three points at equal distance: B_{1}, B_{2}, B_{3}, such that BB_{1} = B_{1}B_{2}, = B_{2}B_{3}.

Step VII: Join BC_{3}.

Step VIII: From B_{2}, draw B_{2}, D || B_{3},C, meeting BC at D.

Step IX: From D draw ED || AC meeting BA at E. Then we have ∆EDB which is the required triangle.

Justification:

Since DE || CA

Hence, we have the new AEBD similar to the given ∆ABC, whose sides are equal to \(\frac{2}{3}\) of the corresponding sides of ∆ABC.

Question 2.

Draw a line segment of length 7.6 cm and divide it in the ratio 5 : 8. Measure the two parts.

Solution:

Steps of Construction:

Step I: Draw a line segment AB = 7.6 cm

Step II: Draw any ray AX making an acute angle ∠BAX with AB.

Step III: On ray AX starting from A, mark 5 + 8 = 13 equal arcs. AA_{1}, A_{1}A_{2}, A_{2}A_{3}, A_{3}A_{4}, … A_{11}A_{12}, and A_{12}A_{13}.

Step IV: Join A_{13}B.

Step V: From A_{5}, draw A_{5}P || A_{13}B, meeting AB at P. Thus, P divides AB in the ratio 5 : 8. On measuring the two parts. We find AP = 2.9 cm and PB = 4.7 cm (approx).

Justification:

In ∆ABA_{13}, PA_{5} || BA_{13} .

∴ By Basic Proportionality Theorem

Question 3.

Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then draw another triangle whose sides are 1\(\frac{1}{2}\) times the corresponding sides of the isosceles triangle.

Solution:

Steps of Construction:

Step 1: Draw BC = 8 cm.

Step II: Construct XY, the perpendicular bisector of line segment BC, meeting BC at M.

Step III: Along MP, cut-off MA = 4 cm.

Step IV: Join BA and CA. Then ∆ABC so obtained is the required ∆ABC.

Step V: Extend BC to D, such that BD = 12 cm

Step VI: Draw DE || CA meeting BA produced at E. Then AEBD is the required triangle.

Justification:

Since, DE || CA .

Hence, we have the new triangle similar to the given triangle whose sides are 1 \(\frac{1}{2}\) i.e, \(\frac{3}{2}\) times the corresponding sides of the isosceles ABC.

Question 4.

Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct a triangle whose sides of the corresponding sides of the triangle ABC.

Solution:

Steps of Construction:

Step 1: Construct a ∆ABC in which BC = 6 cm and, AB = 5 cm and ∠ABC = 60°.

Step II: Below BC make an acute ∠CBX.

Step III: Along BX mark off four arcs: B_{1}, B_{2}, B_{3} such that BB_{1} = B_{1}B_{2} = B_{2}B_{3} = B_{3}B_{4}.

Step IV: Join B_{4}C.

Step V: From B_{3}, draw B_{3}D || B_{4}C, meeting BC at D.

Step VI: From D, draw ED || AC, meeting BA at E.

Now, we have AEBD which is the required triangle whose sides are \(\frac{3}{4}\)th of the corresponding sides of ∆ABC.

Justification:

Here, DE || CA

∴ ∆ABC ~ ∆EBD.

Hence, we get the new triangle similar to the given triangle whose sides are equal to \(\frac{3}{4}\)th of the corresponding sides of ∆ABC.

Question 5.

Draw a circle of radius 6 cm. From a point 10 cm away from its centre, construct the pair of tangents to the circle and measure their lengths.

Solution:

Steps of Construction:

Step 1: Take a point O and draw a circle of radius 6 cm.

Step II: Take a point P at a distance of 10 cm from the centre 0.

Step III: Join OP and bisect it. Let M be the mid-point.

Step IV: With M as centre and MP as radius, draw a circle to intersect the circle at Q and R.

Step V: Join PQ and PR. Then, PQ and PR are the required tangents. On measuring, we find, PQ = PR = 8cm.

Justification:

On joining OQ, we find that ∠PQO = 90°, as ∠PQO is the angle in the Semicircle.

∴ PQ ⊥ OQ

Since OQ is the radius of the given circle, PQ has to be a tangent to the circle. Similarly, PR is

also a tangent to the circle.

Question 6.

Construct a tangent to a circle of radius 4 cm from a point on the concentric circle of radius 6 cm and measure its length. Also, verify the measurement by actual calculation.

Solution:

Steps of Construction:

Step 1: Take a point O and draw a circle of radius OA = 4 cm. Also, draw a concentric circle of radius OB = 6 cm

Step II: Find the mid-point C of OB and draw a circle of radius OC = BC. Suppose this circle intersects the circle of radius 4 cm at P and Q.

Step III: Join BP and BQ to get the desired tangents from a point B on the circle of radius 6 cm. By actual measurement, we find BP = BQ = 4.5 cm.

Justification:

In ∆BPO, we have

∠BPO = 90°, OB = 6 cm and OP = 4 cm

∴ OB^{2} = BP^{2} + OP^{2} [Using Pythagoras theorem]

Similarly, BQ = 4.47 cm

Question 7.

Draw a line segment AB of length 8 cm. Taking A as centre, draw a circle of radius 4 cm and

taking B as centre, draw another circle of radius 3 cm. Construct tangents to each circle from the centre of the other circle.

Solution:

Steps of Construction:

Step I: Draw a line segment AB = 8 cm.

Step II: With A as centre, draw a circle of radius 4 cm and let it intersect the line segment AB in M.

Step III: With B as centre, draw a circle of radius 3 cm.

Step IV: With M as centre, draw a circle of radius AM and let it intersect the given two circles in P, e and R, S.

Step V: Join AP, AQ, BR and BS.

These are the required tangents.

Justification:

On joining BP, we have ∠BPA = 90°, as ∠BPA is the angle in the semicircle.

∴ AP ⊥ PB

Since BP is the radius of given circle, so AP has to be a tangent to the circle. Similarly, AQ, BR and BS are the tangents.

### Constructions Class 10 Extra Questions Long Answer Type

Question 1.

Construct a triangle similar to a given triangle ABC with its sides equal to \(\frac{5}{3}\) of the corresponding sides of the triangle ABC (i.e., of scale factor ).

Solution:

Steps of Construction:

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

Step II: From B cut off 5 arcs

B_{1}, B_{2}, B_{3}, B_{4} and B_{5} on BX so that

BB_{1} = B_{1}B_{2} = B_{2}B_{3} = B_{3}B_{4} = B_{4}B_{5}.

Step III: Join B_{3} to C and draw a line through B_{5}, parallel to B_{3}C intersecting the extended line segment BC at C’.

Step IV: Draw a line through C’ parallel to CA intersecting the

extended line segment BA at A’ (see figure). Then, A’ BC’ is the required triangle.

Justification:

Note that ∆ABC ~ ∆A’BC” (Since AC || A’C’)

Question 2.

Draw a circle of radius of 3 cm. Take two points P and Q on one of its diameters extended on both sides, each at a distance of 7 cm on opposite sides of its centre. Draw tangents to the circle from these two points P and Q.

Solution:

Steps of Construction:

Step 1: Taking a point ( as centre, draw a circle of radius 3 cm.

Step II: Take two points P and Q on one of its extended diameter such that OP = OQ = 7 cm.

Step III: Bisect OP and OQ and let M_{1} and M_{2} be the mid-points of OP and OQ respectively.

Step IV: Draw a circle with M_{1} as centre and M_{1} P as radius to intersect the circle at T_{1}, and T_{2}.

Step V: Join PT_{1} and PT_{2}.

Then, PT_{1} and PT_{2} are the required tangents. Similarly, the tangents QT_{3} and QT_{4} can be obtained

Justification:

On joining OT_{1}, we find ∠PT_{1}O = 90°, as it is an angle in the semicircle.

PT_{1} ⊥ OT_{1}

Since OT_{1} is a radius of the given circle, so PT_{1} has to be a tangent to the circle.

Similarly, PT_{2}, QT_{3} and QT_{4} are also tangents to the circle.

Question 3.

Let ABC be a right triangle in which AB = 6 cm, BC = 8 cm and ∠B = 90°. BD is the perpendicular from B on AC. The circle through B, C, D is drawn. Construct the tangents from A to this circle.

Solution:

Steps of Construction:

Step I: Draw ∆ABC and perpendicular BD from B on AC.

Step II: Draw a circle with BC as a diameter. This circle will pass through D.

Step III: Let O be the mid-point of BC. Join A0.

Step IV: Draw a circle with AO as diameter. This circle cuts the circle drawn in step II at B and E.

Step V: Join AE. AE and AB are desired tangents drawn from A to the circle passing through B, C and D.

Question 4.

Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. Then construct another triangle whose sides are \(\frac{5}{3}\) times the corresponding sides of the given triangle.

Solution:

Steps of Construction:

Step I: Construct a SABC in which BC = 4 cm, ∠B = 90° and BA = 3 cm.

Step II: Below BC, make an acute ∠CBX.

Step III: Along BX mark off five arcs: B_{1}, B_{2}, B_{3}, B_{4} and B_{5} such that

BB_{1} = B_{1}B_{2} = B_{2}B_{3} = B_{3}B_{4} = B_{4}B_{5}.

Step IV: Join B_{3}C.

Step V: From B_{5}, draw B_{5}D || B_{3}C, meeting BC produced at D.

Step VI: From D, draw ED || AC, meeting BA produced at E. Then EBD is the required triangle whose sides are \(\frac{5}{3}\) times the corresponding sides of ∆ABC.

Justification:

Since, DE || CA

Hence, we have the new triangle similar to the given triangle whose sides are equal to \(\frac{5}{3}\) times the corresponding sides of ∆ABC.

Question 5.

Construct a triangle similar to a given triangle ABC with its sides equal to \(\frac{3}{4}\) of the corresponding sides of the triangle ABC ( i.e., of scale factor \(\frac{3}{4}\)).

Solution:

Steps of Construction:

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

Step II: Locate 4 arcs B_{1}, B_{2}, B_{3}, and B_{4} on BX so that

BB_{1} = B_{1}B_{2} = B_{2}B_{3} = B_{3}B_{4}.

Step III: Join B_{4}C and draw a line through B_{3} parallel to B_{4}C to intersect BC at C’.

Step VI: Draw a line through C’ parallel to the line CA to intersect BA at A’ (Fig. 9.14).

Then, ∆A’ BC’ is the required triangle.

Let us now see how this construction gives the required triangle.

Justification:

Question 6.

Draw a triangle ABC in which AB = 4 cm, BC = 6 cm and AC = 9 cm. Construct a triangle similar to ∆ABC with scale factor \(\frac{3}{2}\). Justify the construction. Are the two triangles congruent?

Note that all the three angles and two sides of the two triangles are equal.

Solution:

Steps of Construction:

Step I: Draw a line segment BC = 6 cm.

Step II: With centre B and radius 4 cm draw an arc.

Step III: With centre C and radius 9 cm draw another arc which intersects the previous arc at A.

Step IV: Join BA and CA. ABC is the required triangle.

Step V: Through B, draw an acute angle CBX on the side opposite to vertex A.

Step VI: Locate three arcs B_{1}, B_{2}, and B_{3} on BX such that BB_{1} = B_{1}B_{2} = B_{2}B_{3}.

Step VII: Join B_{2}C.

Step VIII: Draw B_{3}C’ || B_{2}C intersecting the extended line segment BC at ∠C’.

Step IX: Draw C’A’ || CA intersecting the extended line segment BA to A’.

Thus, ∆A’BC’ is the required triangle (∆A’BC’ ~ ∆ABC).

Justification:

∵ B_{2}C || B_{3}C’

### Constructions Class 10 Extra Questions HOTS

Question 1.

Draw a circle with the help of a bangle. Take a point outside the circle. Construct the pair of tangents from this point to the circle.

Solution:

Steps of Construction:

Step I: Draw a circle with the help of a bangle.

Step II: Let P be the external point from where the tangents are to be drawn to the given circle. Through P, draw a secant PAB to intersect the circle at A and B (say).

Step III: Produce AP to a point C, such that AP = PC, i.e., P, is the mid-point of AC.

Step IV: Draw a semicircle with BC as diameter.

Step V: Draw PD ⊥ CB, intersecting the semicircle at D.

Step VI: With P as centre and PD as radius, draw arcs to intersect the given circle at T and T_{1}.

Step VII: Join PT and PT_{1}. Then, PT and PT_{1} are the required tangents.

Question 2.

Draw a ∆ABC with side BC = 7 cm, ∠B = 45°, ∠A = 105°. Then, construct a triangle whose sides are \(\frac{3}{4}\) times the corresponding sides of ∆ABC.

Solution:

Steps of Construction:

Step 1: Construct a ∆ABC in which BC = 7 cm,

∠B = 45°, ∠C = 180° – (∠A + ∠B)

= 180° – (105° + 45°) = 180o – 150° = 30°.

Step II: Below BC, make an acute angle ∠CBX.

Step III: Along BX, mark off four arcs: B_{1}, B_{2}, B_{3}, and B_{4} such that BB_{1} = B_{1}B_{2} = B_{2}B_{3} =B_{3}B_{4}.

Step IV: Join B_{4}C

Step V: From B_{3}, draw B_{3}D || B_{4}C meeting BC at D.

Step VI: From D, draw ED || AC, meeting BA at E. Then EBD is the required triangle whose sides are \(\frac{3}{4}\) times the corresponding sides of ∆ABC.

Justification:

Hence, we have the new triangle similar to the given triangle whose sides are equal to \(\frac{3}{4}\) times the corresponding sides of ∆ABC.

Question 3.

Draw a pair of tangents to a circle of radius 4 cm which are inclined to each other at an angle of 60°

OR

Draw a circle of radius 4 cm. Construct a pair of tangents to it, the angle between which is 60°. Also justify the construction. Measure the distance between the centre of the circle and the point of intersection of tangents.

Solution:

Steps of Construction:

Step I: Draw a circle with centre 0 and radius 4 cm.

Step II: Draw any diameter AOB.

Step III: Draw a radius OC such that ∠BOC = 60°.

Step IV: At C, we draw CM ⊥ OC and at A, we draw AN ⊥ OA.

Step V: Let the two perpendiculars intersect each other at P. Then, PA and PC are required tangents.

Justification:

Since OA is the radius, so PA has to be a tangent to the circle. Similarly, PC is also tangent to the circle.

∠APC = 360° – (∠OAP + ∠OCP + ∠AOC)

= 360° – (90° + 90° + 120°) = 360° – 300° = 60°

Hence, tangents PA and PC are inclined to each other at an angle of 60°

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