You are here
Home > CBSE > NCERT Solutions For Class 10th Maths Chapter 11 : Constructions

NCERT Solutions For Class 10th Maths Chapter 11 : Constructions

CBSE NCERT Solutions For Class 10th Maths Chapter 11 : Constructions. NCERT Solutins For Class 10 Mathematics. Exercise 11.1, Exercise 11.2.


NCERT Solutions for Class X Maths Chapter 10 Circles – Mathematics CBSE

Page No: 219

Exercise 11.1

In each of the following, give the justification of the construction also:

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

Answer

http://1.bp.blogspot.com/-kvM9TPphSZ4/VYGyFveRklI/AAAAAAAAA2A/wvdJR6Hj6Sw/s320/ch11-maths-class10-construction-1.PNG

Steps of Construction:
Step I: AB = 7.6 cm is drawn.
Step II: A ray AX making an acute angle with                  AB is drawn.
Step III: After that, a ray BY is drawn parallel to               AX making equal acute angle as in the                 previous step.
Step IV: Point A1, A2, A3, A4 and A5 is marked               on AX and point B1, B2…. to B8 is                      marked on BY such that AA1 = A1A2 =              A2A3 =….BB1= B1B2 = …. B7B8
Step V: A5 and B8 is joined and it intersected AB at point C diving it in the ratio 5:8.
AC : CB = 5 : 8

Justification:
ΔAA5C ~ ΔBB8C
∴ AA5/BB8 = AC/BC = 5/8

2.  Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are 2/3 of the corresponding sides of the first triangle.

Answer
Steps of Construction:
http://1.bp.blogspot.com/-LMW7KWzPsJ4/VYHDvqKIyyI/AAAAAAAAA2Q/bH-OS4T3_yQ/s320/ch11-maths-class10-construction-2.PNG Step I: AB = 6 cm is drawn.
Step II: With A as a centre and radius equal to 4                cm, an arc is draw.
Step III: Again, with B as a centre and radius                     equal to 5 cm an arc is drawn on same                  side of AB intersecting previous arc at C.
Step IV: AC and BC are joined to form ΔABC.
Step V: A ray AX is drawn making an acute                      angle with AB below it.
Step VI: 5 equal points (sum of the ratio = 2 + 3                 =5) is marked on AX as A1 A2….A5
Step VII: A5B is joined. A2B’ is drawn parallel                   to A5B and B’C’ is drawn parallel to                    BC.
ΔAB’C’ is the required triangle
Justification:
∠A(Common)
∠C = ∠C’ and ∠B = ∠ B’ (corresponding angles)

Thus ΔAB’C’ ~  ΔABC by AAA similarity condition

From the figure,

AB’/AB = AA2/AA5 = 2/3

AB’ =2/3 AB

AC’ = 2/3 AC
3. Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are
7/5 of the corresponding sides of the first triangle.

Answer

Steps of Construction:
Step I: A triangle ABC with sides 5 cm, 6 cm and 7 cm is drawn.
http://4.bp.blogspot.com/-bB_LlTwhfQk/VYJU2W7FDzI/AAAAAAAAA20/nCzaLNlQzPg/s1600/ch11-maths-class10-construction-3.PNG Step II: A ray BX making an acute angle with BC is drawn                  opposite to vertex A.
Step III: 7 points as B1 B2 B3 B4 B5 B6 and B7 are marked                  on BX.
Step IV; Point B5 is joined with C to draw B5C.
Step V: B7C’ is drawn parallel to B5C and C’A’ is parallel to              CA.
Thus A’BC’ is the required triangle.

Justification
ΔAB’C’ ~  ΔABC by AAA similarity condition
∴ AB/A’B = AC/A’C’ = BC/BC’
and BC/BC’ = BB5/BB7 = 5/7
∴A’B/AB = A’C’/AC = = BC’/BC = 7/5

4. Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are 1.5 times the corresponding sides of the isosceles triangle.

Answer

Steps of Construction:
http://1.bp.blogspot.com/-X0x2rW-rAZo/VYJkzcHdCLI/AAAAAAAAA3M/ZZM0iPL-0Dw/s1600/ch11-maths-class10-construction-4.PNG Step I: BC = 5 cm is drawn.
Step II: Perpendicular bisector of BC is drawn and it intersect BC at O.
Step III: At a distance of 4 cm, a point A is marked on perpendicular                   bisector of BC.
Step IV: AB and AC is joined to form ΔABC.
Step V: A ray BX is drawn making an acute angle with BC opposite to              vertex A.
Step VI: 3 points B1 B2 and B3 is marked BX.
Step VII: B2 is joined with C to form B2C.
http://1.bp.blogspot.com/-49HU1OILkjk/VYJmn-BqKPI/AAAAAAAAA3k/0rBX_xIAXfE/s1600/ch11-maths-class10-construction-4.PNG Step VIII: B3C’ is drawn parallel to B2C and C’A’ is drawn parallel to                   CA.
Thus, A’BC’ is the required triangle formed.

ALSO READ:  NCERT Solutions for Class 6th Geography Chapter 8 : India - Climate, Vegetation and Wildlife

Justification:

ΔAB’C’ ~  ΔABC by AA similarity condition.
∴ AB/AB’ = BC/B’C’ = AC/AC’
also,
AB/AB’ = AA2/AA3 = 2/3
⇒ AB’ = 3/2 AB, B’C’ = 3/2 BC and AC’ = 3/2 AC

5. Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct a triangle whose sides are 3/4 of the corresponding sides of the triangle ABC.

Answer
Steps of Construction:

http://2.bp.blogspot.com/-TAbOxPme2UM/VYJr1o4b5EI/AAAAAAAAA4E/3gijka9nwxQ/s1600/ch11-maths-class10-construction-5.PNG

Step I: BC = 6 cm is drawn.
Step II: At point B, AB = 5 cm is drawn making an
∠ABC = 60° with BC.
Step III: AC is joined to form ΔABC.
Step IV: A ray BX is drawn making an acute angle with BC                     opposite to vertex A.
Step V: 4 points B1 B2 B3 and B4 at equal distance is                              marked on BX.
Step VII: B3 is joined with C’ to form B3C’.
Step VIII: C’A’ is drawn parallel CA.
Thus, A’BC’ is the required triangle.

Justification:

∠A = 60° (Common)
∠C = ∠C’
ΔAB’C’ ~  ΔABC by AA similarity condition.
∴ AB/AB’ = BC/B’C’ = AC/AC’
also,
AB/AB’ = AA3/AA4 = 4/3
⇒ AB’ = 3/4 AB, B’C’ = 3/4 BC and AC’ = 3/4 AC

6. Draw a triangle ABC with side BC = 7 cm, ∠B = 45°, ∠A = 105°. Then, construct a triangle whose sides are 4/3 times the corresponding sides of Δ ABC.

Answer

Sum of all side of triangle = 180°
∴ ∠A  + ∠B  + ∠C  = 180°
∠C = 180° – 150° = 30°
Steps of Construction:
Step I: BC = 7 cm is drawn.
Step II: At B, a ray is drawn making an angle of  45° with BC.
http://3.bp.blogspot.com/-X7u9VsjSaAQ/VYK6HTrMBtI/AAAAAAAAA4g/ovXRMJigE8A/s320/ch11-maths-class10-construction-6.PNG Step III: At C, a ray making an angle of 30° with                BC is drawn intersecting the previous                  ray at A. Thus, ∠A = 105°.
Step IV: A ray BX is drawn making an acute                     angle with BC opposite to vertex A.
Step V: 4 points B1 B2 B3 and B4 at equal                       distance is marked on BX.
Step VI: B3C is joined and B4C’ is made parallel              to B3C.
Step VII: C’A’ is made parallel CA.
Thus, A’BC’ is the required triangle.

Justification:

∠B = 45° (Common)
∠C = ∠C’
ΔAB’C’ ~  ΔABC by AA similarity condition.
∴ BC/BC’ = AB/A’B’ = AC/A’C’
also,
BC/BC’ = BB3/BB4 = 34
⇒ AB = 4/3 AB’, BC = 4/3 BC’ and AC = 4/3 A’C’

7. 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 5/3 times the corresponding sides of the given triangle.

Answer

http://4.bp.blogspot.com/-uEthMYeWO7o/VYL8kNp4q1I/AAAAAAAAA40/rSE7zttiOWE/s1600/ch11-maths-class10-construction-7.PNG

Steps of Construction:
Step I: BC = 3 cm is drawn.
Step II: At B, A ray making an angle of 90° with BC is drawn.
Step III: With B as centre and radius equal to 4 cm, an arc is                 made on previous ray intersecting it at point A.
Step IV: AC is joined to form ΔABC.
Step V: A ray BX is drawn making an acute angle with BC                  opposite to vertex A.
Step VI: 5 points B1 B2 B3 B4 and B5 at equal distance is                           marked on BX.
Step VII: B3C is joined B5C’ is made parallel to B3C.
Step VIII: A’C’ is joined together.
Thus, ΔA’BC’ is the required triangle.

Justification:
As in the previous question 6.

Page No. 221

Exercise 11.2

In each of the following, give also the justification of the construction:

1.  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.

ALSO READ:  NCERT Solutions for Class 10th Hindi Chapter 4 : आत्मकथ्य (Course A)

Answer

http://3.bp.blogspot.com/-3o8vjhUQjTU/VYMEdmHnZoI/AAAAAAAAA5E/a1NaKRHphK0/s1600/ch11-maths-class10-construction-1.PNG

Steps of Construction:
Step I: With O as a centre and radius equal to 6 cm, a circle is                drawn.
Step II: A point P at a distance of 10 cm from the centre O is                   taken. OP is joined.
Step III: Perpendicular bisector OP is drawn and let it                                intersected at M.
Step IV: With M as a centre and OM as a radius, a circle is                        drawn intersecting previous circle at Q and R.
Step V: PQ and PR are joined.
Thus, PQ and PR are the tangents to the circle.
On measuring the length, tangents are equal to 8 cm.
PQ = PR = 8cm.

Justification:
OQ is joined.
∠PQO = 90° (Angle in the semi circle)
∴ OQ ⊥ PQ
Therefor, OQ is the radius of the circle then PQ has to be a tangent of the circle.

Similarly, PR is a tangent of the circle.

2. 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.

Answer

Steps of Construction:

Step I: With O as a centre and radius equal to 4 cm, a circle is                 drawn.

Step II: With O as a centre and radius equal to 6 cm, a                             concentric circle is drawn.

Step III: P be any point on the circle of radius 6 cm and OP is                   joined.

Step IV: Perpendicular bisector of OP is drawn which cuts it at                 M

Step V: With M as a centre and OM as a radius, a circle is                        drawn which intersect the the circle of radius 4 cm at Q              and R

Step VI: PQ and PR are joined.

Thus, PQ and PR are the two tangents.

Measurement:

OQ = 4 cm (Radius of the circle)

PQ = 6 cm ( Radius of the circle)

∠PQO = 90° (Angle in the semi circle)

Applying Pythagoras theorem in ΔPQO,

PQ2 + QO2 = PO2
⇒ PQ2 + 42= 62

⇒ PQ2 + 16 = 36
⇒ PQ2 = 36 – 16
⇒ PQ2 = 20
⇒ PQ = 2√5

Justification:

∠PQO = 90° (Angle in the semi circle)

∴ OQ ⊥ PQ
Therefor, OQ is the radius of the circle then PQ has to be a tangent of the circle.

Similarly, PR is a tangent of the circle.

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

Answer

http://3.bp.blogspot.com/-xFJ6h_aZAps/VYOoVY5GYQI/AAAAAAAAA5o/_3GKFGKSt3A/s1600/ch11-maths-class10-construction-3.PNG

Steps of Construction:
Step I: With O as a centre and radius equal to 3 cm, a                   circle is drawn.
Step II: The diameter of the circle is extended both sides                and an arc is made to cut it at 7 cm.
Step III: Perpendicular bisector of OP and OQ is drawn                 and x and y be its mid-point.
Step IV: With O as a centre and Ox be its radius, a circle                is drawn which intersected the previous circle at                M and N.
Step V: Step IV is repeated with O as centre and Oy as radius and it intersected the circle at R and T.
Step VI: PM and PN are joined also QR and QT are joined.
Thus,  PM and PN are tangents to the circle from P and QR and QT are tangents to the circle from point Q.

Justification:

∠PMO = 90° (Angle in the semi circle)

∴ OM ⊥ PM
Therefor, OM is the radius of the circle then PM has to be a tangent of the circle.

Similarly, PN, QR and QT are tangents of the circle.

4. Draw a pair of tangents to a circle of radius 5 cm which are inclined to each other at an angle of 60°.

Answer

http://4.bp.blogspot.com/-cNdpxyWluwo/VYOr9XUFYmI/AAAAAAAAA58/vuV4BH6FzRs/s1600/ch11-maths-class10-construction-4.PNG

We know that radius of the circle is perpendicular to the tangents.
Sum of all the 4 angles of quadrilateral = 360°
∴ Angle between the radius (∠O)  = 360° – (90° + 90° + 60°) = 120°
Steps of Construction:
Step I: A point Q is taken on the circumference of the circle and OQ              is joined. OQ is radius of the circle.
Step II: Draw another radius OR making an angle equal to 120° with               the previous one.
Step III: A point P is taken outside the circle. QP and PR are joined                  which is perpendicular OQ and OR.
Thus, QP and PR are the required tangents inclined to each other at an angle of 60°.

ALSO READ:  NCERT Solutions for Class 10th Science Chapter 4 : Carbon and its Compounds

Justification:

Sum of all angles in the quadrilateral PQOR = 360°
∠QOR + ∠ORP + ∠OQR + ∠RPQ = 360°
⇒ 120° + 90° + 90° + ∠RPQ = 360°
⇒∠RPQ = 360° – 300°
⇒∠RPQ = 60°
Hence, QP and PR are tangents inclined to each other at an angle of 60°.

5. 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.

Answer

http://1.bp.blogspot.com/-qv7atbtlv6w/VYPar7sjpHI/AAAAAAAAA6M/s4VGJgXkLO8/s1600/ch11-maths-class10-construction-6.PNG

Steps of Construction:
Step I: A line segment AB of 8 cm is drawn.
Step II: With A as centre and radius equal to 4                cm, a circle is drawn which cut the                    line at point O.
Step III: With B as a centre and radius equal to                3 cm, a circle is drawn.
Step IV: With O as a centre and OA as a                         radius, a circle is drawn which                           intersect the previous two circles at P,               Q and R, S.
Step V: AP, AQ, BR and BS are joined.
Thus, AP, AQ, BR and BS are the required tangents.

Justification:
∠BPA = 90° (Angle in the semi circle)

∴ AP ⊥ PB
Therefor, BP is the radius of the circle then AP has to be a tangent of the circle.

Similarly,  AQ, BR and BS are tangents of the circle.

6. 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.

Answer

http://4.bp.blogspot.com/-8G48fYnStIg/VYPmYmv7cAI/AAAAAAAAA6c/AJBN4QOrd4o/s1600/ch11-maths-class10-construction-6.PNG

Steps of Construction:
Step I: A ΔABC is drawn.
Step II:Perpendicular to AC is drawn to point B                which intersected it at D.

Step III: With O as a centre and OC as a radius, a               circle is drawn. The circle through B, C,                D is drawn.
Step IV: OA is joined and a circle is drawn with                 diameter OA which intersected the                         previous circle at B and E.
Step V: AE is joined.
Thus, AB and AE are the required tangents to the circle from A.

Justification:

∠OEA = 90° (Angle in the semi circle)

∴ OE ⊥ AE
Therefor, OE is the radius of the circle then AE has to be a tangent of the circle.

Similarly,  AB is tangent of the circle.

aa.jpg

Do You Liked Our Contents? If Yes! Then Please Spare Us Some Time By Commenting Below. Or To Get Daily Minute by Minute Updates On Facebook Twitter and Google+ From Us (Indiashines.in) Please Like Us On Facebook , Follow Us On Twitter and Follow Us On Google+ . If You also Want To Ask Us/Experts Any Questions Then Please Join Our Forum Here and Be Our Exclusive Member.


GO BACK TO CLASS X MATHS ALL CHAPTERS SOLUTION


Indiashines.in
IMPORTANT : All the contents of this website is for educational and informational purpose only. The data is collected from various websites, links, pages spread all over internet and owner self-made.We take utmost care to keep the information on this webpage as complete and accurate as possible but we do not claim our site to be error free. And in no way Indiashines.In will be liable for any losses & damages arises from its contents/text. Indiashines.In in good faith, makes an effort to become a helpful platform for students, jobseekers etc. This website is not endorsed by any Govt. bodies. Before using any information of ours please READ OUR DISCLAIMER.

Leave a Reply

Top