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NCERT Exemplar Solution for Class 10 Mathematics: Constructions (Part-IV)

Aug 17, 2017 15:38 IST

    Class 10 Mathematics NCERT Exemplar, NCERT Exemplar Solution, Constructions Class 10 NCERT ExemplarHere 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, AA1 = A1A2 = A2A3 = A3A4. 

    6. Join A4B.

    7. Draw A3P || A4B 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 AB1 = B1B2 = B2B3 = B3B4.

    11. Join B4C.

    11. Draw B1Q || B4C 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 DBDC with scale factor

    4/3. Draw the line segment D’A’ parallel to DA, where A’ lies on extended side BA. Is ABC’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 ΔBD’C’ similar to DBDC 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, BB1 = B1B2 = B2B3 = B3B4.

    8. Join B3C and draw a line B4C’ || B3C 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, ΔBD’C’ is the required triangle similar to DBDC 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 DOPQ,

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

    Thus, using Pythagoras theorem in right angled DOPQ, we get:

    ∴         PQ2 = OP2 - OQ2                                

    ⟹       PQ2 = (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 ΔABC in which PQ = 8 cm. Also justify the construction.

    Solution.              

    To construct a ΔPQR similar to triangle ΔABC, we have to find the scale factor.

    Steps of construction:

    First we draw ΔABC 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, ΔABC is the required isosceles triangle.

    Now we draw ΔPQR similar to ΔABC 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 BB1 = B1B2 = B2B3 = B3B4.

    7. Join B3C and from B4 draw a line B4R || B3C 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, ΔPQR is the required triangle.

    Justification:

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

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

    Solution.

    Steps of construction:

    First we draw ΔABC 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.

    ΔABC is the required triangle.

    To draw another triangle similar to ΔABC 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 AA1 = A1A2 = A2A3 = A3A4 = A4A5 = A5A6 = A6A7.  

    6. Join A7B and from A5 draw A5B’ || A7B intersecting AB at B’.

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

    Then, ΔAB’C’ is the required triangle whose sides are equal to 5/7 of the corresponding sides of the ΔABC.

    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 ΔABC 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, ΔABC is the required triangle.

    Now we draw another triangle similar to ΔABC 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, BB1 = B1B2 = B2B3.

    6. Join B2C and from B3 draw B3C’ || B2C 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, ΔA’BC’ is the required triangle whose sides are equal to 3/2 of the corresponding sides of the ΔABC.

    Justification

    Here, the two triangles, ABC and ABC’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

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