Here you get the CBSE Class 10 Mathematics chapter 10, Constructions: NCERT Exemplar Problems and Solutions (Part-IV). This part of the chapter includes solutions for questions from Exercise 10.4 of NCERT Exemplar Problems for Class 10 Mathematics Chapter: Constructions. This exercise comprises of only the Long Answer Type Questions framed from various important topics in the chapter. Each question is provided with a detailed solution.

NCERT Exemplar problems are a very good resource for preparing the critical questions like Higher Order Thinking Skill (HOTS) questions. All these questions are very important to prepare for CBSE Class 10 Mathematics Board Examination 2017-2018 as well as other competitive exams.

**Find below the NCERT Exemplar problems and their solutions for Class 10 Mathematics Chapter, Constructions:**

**Exercise 10.4 **

**Long Answer Type Questions:**

**Solution.**

**Steps of construction:**

1. Draw a line segment *AB = *5 cm.

2. Draw ∠*BAZ* = 60°.

3. With *A* as centre and radius equal to 7 cm draw an arc cutting the line *AZ *at *C*.

4. Draw a ray *AX*, making an acute ∠*BAX*.

5. Divide *AX *into four equal parts, *AA*_{1 }= *A*_{1}*A*_{2} = *A*_{2}*A*_{3} = *A*_{3}*A*_{4}._{ }

6. Join *A*_{4}*B*.

7. Draw *A*_{3}*P* || *A*_{4}*B* meeting *AB *at *P*.

8. Thus, *P *is the point on *AB *such that *AP* = ¾ *AB.*

9. Next, draw a ray *AY*, making an acute ∠*CAY.*

10. Divide *AY *into four parts *AB*_{1} = *B*_{1}*B*_{2 }= *B*_{2}*B*_{3} = *B*_{3}*B*_{4}.

11. Join *B*_{4}*C*.

11. Draw *B*_{1}*Q* || *B*_{4}*C *meeting *AC *at *Q*.

Then, *Q* is the point on *AC *such that *AQ* = ¼ *AC*.

12. Join *PQ *and measure it.

13. *PQ *= 3.25 cm.

**Question. 2 **Draw a parallelogram *ABCD *in which *BC = *5 cm, *AB* = 3 cm and *∠ABC* = 60°, divide it into triangles *BCD *and *ABD *by the diagonal *BD. *Construct the triangle *BD’C’ *similar to D*BDC* with scale factor

4/3. Draw the line segment *D’A’ *parallel to *DA, *where *A*’ lies on extended side *BA.* Is *A*’*BC’D’ *a parallelogram?

**Solution.**

**Steps of construction:**

First we draw the parallelogram *ABCD *by following the steps given below:

1. Draw a line segment *AB = *3 cm.

2. Draw *∠ABY = *60°.

3. Taking *B* as centre and radius equal to 5 cm draw an arc cutting *BY* at *C.*

4. Draw CD || *AB *and *AD* || *BC.* Thus *ABCD* is the required parallelogram.

5. Join *BD* to get two triangles *BCD *and *ABD*.

**Now to construct the Δ BD’C’ similar to **

**D**

*BDC***with scale factor 4/3, follow the steps given below:**

6. From *B* draw a ray *BX* making an acute *∠**CBX.*

7. Divide *BX* into four equal parts, *BB*_{1} = *B*_{1}*B*_{2} = *B*_{2}*B*_{3} = *B*_{3}*B*_{4}.

8. Join *B*_{3}*C* and draw a line *B*_{4}*C*’ || *B*_{3}*C *intersecting the extended line segment *BC *at *C*’.

9.** **From point *C’ *draw *C’D’ || CD *intersecting the extended line segment *BD *at *D’. *

Thus, Δ*BD’C’ *is the required triangle similar to D*BDC* with scale factor 4/3*.*

10.* *Now draw a line segment *D’A’ *parallel to *DA, *where *A’ *lies on line segment *BA *at* A*’.

11. Thus, we obtain *A’BC’D’*, which* *is a parallelogram*.*

**Question.**** 3 **Draw two concentric circles of radii 3 cm and 5 cm. Taking a point on outer circle construct the

pair of tangents to the other. Measure the length of a tangent and verify it by actual calculation.

**Solution.**

**Steps of construction:**

1. Draw two concentric circles with centre *O* and radii 3 cm and 5 cm.

2. Take a point *P *on outer circle and join *OP.*

3. Draw a perpendicular bisector of *OP *intersecting OP at *M*.* *Thus *M *is the mid-point of *OP*.

4. Now, taking *M *as centre and *OM *as radius draw a circle meeting the inner circle at points *Q *and *R.*

5. Join *PQ *and *PR. *Thus, *PQ *and *PR *are the required tangents.

6. On measuring *PQ *and *PR, *we get *PQ *= *PR = *4 cm.

**Actual calculation:**

In *D**OPQ, *

∠*OQP = *90° (Tangent at a point is perpendicular to the radius through that point)

Thus, using Pythagoras theorem in right angled *D**OPQ,* we get:

∴ *PQ ^{2} = OP^{2} *

*-*

*OQ*

^{2 }⟹ *PQ*^{2} = (5)^{2} - (3)^{2} = 25 - 9 = 16

⟹ *PQ = *4cm

Hence, verified.

**Question. 4 **Draw an isosceles triangle *ABC *in which *AB = AC = *6 cm and *BC = *5 cm. Construct a triangle *PQR *similar to Δ*ABC *in which *PQ = *8 cm. Also justify the construction.

**Solution. **

To construct a Δ*PQR* similar to triangle Δ*ABC*, we have to find the scale factor.

**Steps of construction:**

First we draw Δ*ABC* by following the steps given below:

1.** **Draw a line segment *BC = *5 cm.

2. Taking *B *and *C *as centers draw two arcs each of 6 cm, intersecting each other at point *A*.

4. Join *AB *and *AC*.

Thus, Δ*ABC* is the required isosceles triangle.

Now we draw Δ*PQR *similar to Δ*ABC *with scale factor 4/3 by following the steps given below:

5. From *B, *draw a ray *BX* making an acute ∠*CBX *opposite to point* A* with respect to side *BC*.

6. Divide *BX* into four equal parts *BB*_{1} = *B*_{1}*B*_{2} = *B*_{2}*B*_{3} = *B*_{3}*B*_{4.}

7. Join *B*_{3}C and from *B*_{4} draw a line *B*_{4}*R* || *B _{3}C *intersecting the extended line segment

*BC*at

*R*.

8. From point *R, *draw *RP || CA *meeting the extended line segment *BA *at *P*.

9. Take point *B* as *Q.*

Then, Δ*PQR* is the required triangle.

**Justification:**

Hence, the *ΔPQR* is similar to Δ*ABC*, with its sides equal to 4/3 times of the corresponding sides of the Δ*ABC*.

**Question. 5** Draw a *ΔABC *in which *AB = *5 cm, *BC = *6 cm and ∠*ABC = *60°. Construct a triangle similar to Δ*ABC* with scale factor 5/7. Justify the construction.

**Solution.**

**Steps of construction: **

First we draw Δ*ABC* by following the steps given below:

1. Draw a line segment *AB* = 5 cm.

2. Draw ∠*ABY = *60°

3. Taking *B* as centre and radius = 6cm draw an arc cutting *BY* at *C*. Join *AC*.

Δ*ABC* is the required triangle.

To draw another triangle similar to Δ*ABC* with scale factor 5/7, we follow the steps given below:

4. From *A*, draw any ray *AX* making an acute angle *BAX* in the opposide side to vertex *C* with respect to side *AB*.

5. Divide *AX* into 7 equal parts *AA*_{1} = *A*_{1}*A*_{2} = *A*_{2}*A*_{3 }= *A*_{3}*A*_{4} = *A*_{4}*A*_{5 }=* A*_{5}*A*_{6} =* A*_{6}*A*_{7}.

6. Join *A*_{7}*B *and from *A*_{5} draw *A*_{5}*B’ || A*_{7}*B* intersecting *AB *at *B’*.

7. Now from point *B’ *draw *B’C’* || *BC* intersecting *AC *at *C’*.

Then, Δ*AB’C’ *is* *the required triangle whose sides are equal to 5/7 of the corresponding sides of the Δ*ABC*.

**Justification:**

Hence justified.

**Question. 6 **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**

1. Draw a circle of with centre as *O* and radius *OA = *4 cm.

2. Produce *OA *to *B* such that *OA = AB = *4 cm.

3. Taking *A *as the centre and *AO = *4 cm as radius draw a ciscle cutting the previous circle at points *P* and *Q*.

4. Join *BP *and *BQ*.

Thus *BP *and *BQ *are the desired tangents.

**Justification: **

Join *OP*.

**Question. 7 **Draw a *ΔABC *in which *AB *= 4 cm, *BC = *6 cm and *AC* = 9 cm. Construct a triangle similar to Δ*ABC *with scale factor 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:**

First we draw* ΔABC* by following the steps given below:

1. Draw a line segment *BC = *6 cm.

2. Taking *B* and *C* as centres, draw two arcs of radii 4 cm and 9 cm respectively, which intersect each other at point *A*.

3. Join *AB *and *AC. *

Thus, Δ*ABC *is the required triangle.

Now we draw another triangle similar to Δ*ABC *with scale factor 3/2.

4. From *B*, draw a ray *BX* downwards making an acute angle *CBX*

5. Divide *BX* into 3 equal parts, *BB*_{1} = *B*_{1}*B*_{2} = *B*_{2}*B*_{3}.

6. Join *B*_{2}*C *and from *B*_{3} draw *B*_{3}*C’* ||* B*_{2}*C *intersecting the extended line segment *BC *at *C’.*

7. From point *C’, *draw *C’A’* || *CA *intersecting the extended line segment *BA *at *A’.*

Then, Δ*A’BC’ *is the required triangle whose sides are equal to 3/2 of the corresponding sides of the Δ*ABC.*

**Justification**

Here, the two triangles, *ABC* and *A*’*BC’*are not congruent because, for the two triangles to be congruent, their corresponding sides and interior angles must be same. But here, all the three angles are same but three sides are not same.

**CBSE Class 10 Mathematics Syllabus 2017-2018**

**CBSE Class 10 NCERT Textbooks & NCERT Solutions**

**NCERT Solutions for CBSE Class 10 Maths**

**NCERT Exemplar Problems and Solutions Class 10 Science: All Chapters**