SEBA Class 10 Mathematics Chapter 11 Constructions Solutions, SEBA Class 10 Maths Textbook Notes in English Medium, SEBA Class 10 Mathematics Chapter 11 Constructions Notes in English to each chapter is provided in the list so that you can easily browse throughout different chapter Assam Board SEBA Class 10 Mathematics Chapter 11 Constructions Notes and select needs one.
SEBA Class 10 Mathematics Chapter 11 Constructions
Also, you can read the SCERT book online in these sections Solutions by Expert Teachers as per SCERT (CBSE) Book guidelines. SEBA Class 10 Mathematics Chapter 11 Constructions Question Answer. These solutions are part of SCERT All Subject Solutions. Here we have given SEBA Class 10 Mathematics Chapter 11 Constructions Solutions for All Subject, You can practice these here.
Constructions
Chapter – 11
| Exercise 11.1 |
1. Draw a line segment of length 7.6 cm and divide it in the ratio 5 : 8. Measure the two parts of construction.
Ans:
(i) Take a line segment AB =7.6 cm.
(ii) Draw any ray AX, making an acute angle ∠BAX.
(iii) Locate 5 + 8 = 13 (given ra- tio 5: 8) points A1, A2, A3, A4….A12A13 on ray AX such that A1,A2 = A2,A3,= A3A4, = A11A12 = A12A13
(iv) Join BA3
(v) Through points A5, draw a line A5C II A13 B (by making angle equal to ∠AB13B) at A5 intersecting AB at ‘C’. The AC : CB = 5 : 8.
2. Construct a triangle of sides 4cm, 5cm and 6cm and then a triangle similar to it whose sides are ⅔ of the corresponding sides of the first triangle.
Ans: 1. Construct a triangle ABC with given measurements AB = 5 cm, AC = 4 cm and BC = 6 cm.
3. Construct a triangle with sides 5cm, 6 cm and 7cm and then another triangle whose sides are 7/5 of the corresponding sides of the first triangle.
Ans:
(i) Construct a triangle ABC in which AB = 7cm BC = 6 cm and AC = 5cm.
(ii) Make any acute angle ∠BAX below the base AB.
(iii) Located seven points A1, A2, A3, A4 ,A5, A6, A7, on the ray AX such that AA1 = A1A2 = A2A3 = A3A4 = A4A5 = A5A6 = A6A7
(iv) Join BA5
(v) Through A7, draw a line parallel A5B. Let it meets AB at Bl’ on being produced such that ABl = 7/5 AB.
(vi) Through Bl, draw a line parallel to BC which meets AC at Cl on being produced.
∆ABlCl is the required triangle.
4. Construct an isosceles triangle whose base is 8 cm and altitude 4cm and then another triangle whose sides are 1 ½ times the corresponding sides of the isosceles triangle.
Ans: Steps of constrictions:
(i) Take base AB = 8cm.
(ii) Draw the perpendicular bisector of AB. Let it intersect AB at ‘M’.
(iii) With M as centre and radius = 4cm, draw an arc which intersects the perpendicular bisector at ‘C’.
(iv) Join CA and CB.
(v) ∆ABC is an isosceles with CA = CB.
(vi) Make any acute ∠BAX below the side BC.
(vii) Locate three (greater of 2 and 3 in 1 ½ or 3/2) A1, A2, A3, an ‘AX’ an such that AA1 = A1A2=A2A3.
(viii) Join A₂(2nd point smaller of 2 and 3 in 3/2) and B.
(ix) Through A3, draw a line parallel to A2B meet AB is Bl on being produced.
(x) Through Bl, draw a line parallel to BC which meets AC in Cl on being produced.
∆АВlСl is the required triangle whose sides are 1 ½ times the corresponding sides of ∆AВС.
5. Draw a triangle ABC with side BC = 6 cm, AB = 5cm and ∠ABC = 60°. Then construct a triangle whose sides are ¾ of the corresponding sides of the triangle ABC.
Ans:
6.Draw a triangle ABC with side BC = 7cm, ∠B = 45°, ∠A = 105°. Then construct a triangle whose sides are 4/3 times the corresponding sides ∆ABC.
Ans:
7. Draw a right triangle in which the sides (other than hypotenuse) are of length 4 cm and 3 cm. Then construct another triangle whose sides are 5/3 times the corresponding sides of the given triangle.
Ans: Steps of construction:
| Exercise 11.2 |
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.
Ans: Steps of constructions:
5. Then PT and PT are two required tangents. Length of PT and PTl is 8.2cm (approx).
2. Construct a tangent to a circle of radius 4 cm from a point on the concentric circle of radius 6cm and measure its length. Also verify the measurement by actual calculation.
Ans: Steps of constructions:
3. Draw a circle of radius 3 cm. Take two points P and Q on one of its extended diameters each at a distance of 7 cm from its centre. Draw tangents to the circle from these two points P and Q.
4. Draw a pair of tangents to a circle of radius 5cm which are inclined to each other an angle of 60°.
Ans: Steps of constructions:
5. Draw a line segment AB of length 8cm. Taking A as centre, draw a circle
of radius 4cm and taking B as centre, draw another circle of radius of 3cm. Construct tangents to each circle from the centre of other 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 though B, C, D is drawn. Construct the tangents from A to this circle.
Ans: Steps of construction:
1. Construct a right angled triangle ABC according to given conditions and measurements.
2. Draw BD⟂AC.
3. Take midpoint of side BC take it as ‘M’.
4. Take ‘M’ as centre and BC as diameter, draw a circle through B, C, D during property, angle in semicircle is 90°. (∠BDC = 90°) Take this circle as I.
5. Now join ‘A’ and ‘M’.
6. Draw perpendicular bisector of ‘N’. Now with ‘N’ as centre and ‘NA’
or ‘NM’ as radius, draw a circle (II) which intersects the circle (I) at ‘B’ and ‘P’.
7. Join AP.
8. AP and AB are the required tangents.
7. 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.
Ans: Steps of construction:
1. Draw a circle using bangle (I).
2. Take any two chords AB and CD (non parallel) on circle.
3. Draw the perpendicular bisectors of chords AB and CD. The perpendicular bisectors intersect each other at ‘O’.
∴ OA = OB and OC = OD
∴ OA = OB = OC = OD (radii of circle)
∴ ‘O’ is the centre of circle.
4. Take any point ‘P’ outside the circle.
5. Join OP.
6. Draw the perpendicular bisector of OP. Let ‘M’ the mid point of OP.
7. With M as centre and radius ‘MP’ or ‘MO’ draw a circle which intersects the circle (I) at T and TI.
8. Join PT and PTI, which is required pair of tangents.
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