Bihar Board Class 10th Maths Chapter 11 Constructions Solution

 

BSEB Bihar Board Class 10th Maths Solutions Chapter 11 Constructions Ex 11.1

Question 1.
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 :

1. Draw a line segment AB = 7.6 cm.

2. Draw a ray AC making any acute angle with AB, as shown in the figure.

3. On ray AC, starting from A, mark 5 + 8 = 13 equal line segments : AA1, A1A2, A2A3, A3A4, A4A5, A5A6, A6A7, A7A8, A8A9, A9A10, A10A11, A11A12 and A12A13.

4. Join A13B.

5. From A5, draw A5P || A13B, meeting AB at P.

6. Thus, P divides AB in the ratio 5 : 8.
On measuring the two parts, we find that AP = 2.9 cm and PB = 4.7 cm (approx.).

Justification :
In ∆ ABA13, PA5 || BA13.
So, APPB = AA5 A5 A13 = 58 [By BPT]
Thus, AP : PB = 5 : 8

Question 2.
Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are 23 of the corresponding sides of it.
Solution:
Steps of Construction :

1. Draw a line segment BC = 6 cm.

2. With B as centre and radius equal to 5 cm, draw an arc.

3. With C as centre and radius equal to 4 cm, draw an arc intersecting the previously drawn arc at A.

4. Join AB and AC. Then ∆ ABC is the required triangle.

5. Below BC, make an acute angle CBX.

6. Along BX, mark off three (bigger of the numerator and denominator) points : B1, B1 and B3 such that BB1 = B1B2 = B2B3

7. Join B3C. [Note that 3 is denominator]

8. From B2 draw B2D || B3C, meeting BC at D. [Note that 2 is numerator]

9. From D, draw DE || CA, meeting BA at E. Then, ∆ EBD is the required triangle whose sides are 23 of the corresponding sides of ∆ ABC.

Justification :
Since DE || CA, therefore
∆ ABC ~ ∆ EBD
and EBAB = DECA = BDBC = 23 [Since BDBC = 23 ,by construction]
Hence, we get the new triangle similar to the given triangle whose sides are equal to 23 of the corresponding sides of ∆ ABC.

Question 3.
Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are 75 of the corresponding sides of the first triangle.
Solution:
Steps of Construction :
1. With the given data, construct ∆ ABC in which BC = 7 cm, CA = 4 cm and AB = 5 cm.

2. Below BC, make an acute angle CB1.

3. Along BX, mark off seven (bigger of the numerator and denominator) points : B1, B2, B3, B4, B5, B6 and B7 such that BB1 = B1B2 = B2B3 = B3B4 = B4B5 = B5B6 = B6B7

4. Join B5C. [Note that 5 is denominator]

5. From B7, draw B7D || B5C, meeting BC produced at D. [Note that 7 is numerator]

6. From D, draw DE || CA, meeting BA produced at E. Then, ∆ EBD is the required triangle whose sides are 75 of the corresponding sides of ∆ ABC.

Justification :
Since DE || CA, therefore
∆ ABC ~ ∆ EBD and EBAB = DECA = BDBC = 75
Hence, we get the new triangle similar to the given triangle whose sides are equal to 75 of the corresponding sides of ∆ ABC.

Question 4.
Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are 112 times the corresponding sides of the isosceles triangle.
Solution:
Steps of Construction :

1. Draw BC = 8 cm.

2. Construct PQ, the perpendicular bisector of line segment BC meeting BC at M.

3. Along MP, cut off MA = 4 cm.

4. Join BA and CA. Then, ∆ ABC so obtained is the required triangle.

5. Make an acute angle CBX below BC.

6. Mark three points B1, B2 and B3 on BX such that BB1 = B1B2 = B2B3

7. Join BA2C.

8. From B3, draw B3D || B2C meeting BC produced at D.

9. From D, draw DE || CA meeting BA produced at E. Then, ∆ EBD is the required triangle.

Justification :
Since DE || CA, therefore ∆ ABC ~ ∆ EBD
and EBAB = DECA = BDBC = 32
Hence, we get the new triangle similar to the given 32 triangle whose sides are 32, i.e., 1 12 times of the corresponding sides of the isosceles ∆ ABC.

Question 5.
Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct a triangle whose sides are 34 of the corresponding sides of the triangle ABC.
Solution:
Steps of Construction :
1. With the given data, construct ∆ ABC in which BC 6 cm, ∠ABC = 60° and AB = 5 cm.

2. Below BC, make an acute angle CB1.

3. Along BX, mark off 4 points : B1, B2, B3 and B4 such that BB1 = B1B2 = B2B3 = B3B4.

4. Join B4C.

5. From B3, draw B3D || B4C to meet BC at D.

6. From D, draw ED || AC, meeting BA at E. Then, ∆ EBD is the required triangle whose sides are 34 th of the corresponding sides of ∆ ABC.

Justification :
Since DE || CA, therefore ∆ ABC ~ ∆ EBD
and EBAB = DECA = BDBC = 34
Hence, we get the new triangle similar to the given triangle whose sides are equal to 34 of the corresponding sides of ∆ ABC.

Question 6.
Draw a triangle ABC with side BC = 7 cm, ∠B = 45% ∠A = 105°. Then construct a triangle whose sides are 43 times the corresponding sides of ∆ ABC.
Solution:
Steps of Construction :
1. With the given data, construct ∆ ABC in which BC = 7 cm, ∠B = 45°, ∠C = 180° – (∠A + ∠B).

i. e., ∠C = 180° – (105° + 45°)
= 180° – 150° = 30°.

2. Below BC, make an acute ∠CBX.

3. Along BX, mark off four points : B1, B2, B3 and B4 such that BB1 = B1B2 = B2B3 = B3B4.

4. Join B3C.

5. From B4, draw B4D || B3C, meeting BC produced at D.

6. From D, draw DE || CA, meeting BA produced at E. Then, ∆ EBD is the required triangle whose sides are 43 times of the corresponding sides of ∆ ABC.

Justification :
Since DE || CA, therefore ∆ ABC ~ ∆ EBD
and EBAB = DECA = BDBC = 43
Hence, we get the new triangle similar to the given triangle whose sides are equal to 43 times of the corresponding I sides of ∆ ABC.

Question 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 53 times the corresponding sides of the given triangle.
Solution:
Steps of Construction :
1. With given data, construct ∆ ABC in which BC = 4 cm, ∠B = 90° and BA = 3 cm.

2. Below BC, make an acute angle CBX.

3. Along BX, mark off four points : B1, B2, B3 and B4 such that BB1 = B1B2 = B2B3 = B3B4.

4. Join B3C.

5. From B5, draw B5D || B3C, meeting BC produced at D.

6. From D, draw DE || CA, meeting BA produced at E.
Then, EBD is the required triangle whose sides are 53 times of the corresponding sides of ∆ ABC.

Justification :
Since DE || CA, therefore ∆ ABC ~ ∆ EBD
and EBAB = DECA = BDBC = 53
Hence, we get the new triangle similar to the given triangle whose sides are equal to 53 times of the corresponding sides of ∆ ABC.

BSEB Bihar Board Class 10th Maths Solutions Chapter 11 Constructions Ex 11.2

Question 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.
Solution:
Steps of Construction :
1. Take a point 0 as centre and draw a circle of radius 6 cm.
2. Mark a point P at a distance of 10 cm from the centre O.

3. Join OP and bisect it. Let M be its mid-point of OP.
4. With M as centre and MP as radius, draw a circle to intersect the previous circle at Q and R.
5. Join PQ and PR. Then, PQ and PR are the required tangents.
On measurement, we find that PQ = PR = 8 cm.

Justification :
On joining OQ, we find that ∠PQO = 90°, as ∠PQO is the angle in a semi-circle.
∴ PQ ⊥ OQ.
Since OQ is the radius of the given circle, so PQ has to be a tangent to the circle. Similarly, PR is also a tangent to the circle.

Question 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.
Solution:
Steps of Construction :
1. Take a point O and draw two concentric circles of radii 4 cm and 6 cm respectively.
2. Mark a point P on the bigger circle.

3. Join OP and bisect it. Let M be its mid-point.
4. With M as centre and MP as radius, draw a circle to intersect the smaller circle at Q and R.
5. Join PQ and PR. Then, PQ and PR are the required tangents.

On measuring, we find that PQ = PR = 4.5 cm (approx.).
Calculation : From A OQP,
OP² = OQ² + PQ²
or 6² = 4² + PQ²
or PQ² = 36 – 16 = 20
So, PQ = 20−−√cm = 4-47 cm (approx.)
Similarly, PR = 4.47 cm (approx.).

Justification :
On joining OQ, we find that ∠PQO = 90°, as ∠PQO is the angle in a semi-circle.
∴ PQ ⊥ OQ.
Since OQ is the radius of the given circle, so PQ has to be a tangent to the circle. Similarly, PR is also a tangent to the circle.

Question 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.
Solution:
Steps of Construction :
1. Take a point O, draw a circle of radius 3 cm with this point as centre.
2. Take two points P and Q on one of its extended diameter such that OP = OQ = 7 cm.
3. Bisect OP and OQ. Let their respective mid-points be M1 and M2.
4. With M1 as centre and M1P as radius, draw a circle to intersect the circle at T1 and T2.

5. Join PT1 and PT2. Then, PT1 and PT2 are the required tangents.
Similarly, the tangents QTS and QT4 can be obtained.

Justification :
On joining OT4, we find ∠PT10 = 90°, as an angle in a semi-circle.
∴ PT1 ⊥ OT1
Since OT1 is a radius of the given circle, so PT1 has to be a tangent to the circle.
Similarly, PT2, QT3 and QT4 are also tangents to the circle.

Question 4.
Draw a pair of tangents to a circle of radius 5 cm which are inclined to each other at an angle of 60°.
Solution:
Steps of Construction :
1. With O as centre and radius = 5 cm, draw a circle.
2. Draw any diameter AOC.

3. Draw a radius OL such that ∠COL = 60° (i.e., the given angle).
4. At L, draw LM ⊥ OL.
5. At A, draw AN ⊥ OA.
6. These two perpendiculars intersect each other at P. Then, PA and PL are the required tangents.

Justification :
Since OA is the radius, so PA has to be a tangent to the circle.
Similarly, PL is the tangent to the circle.
∠APL = 360° – ∠OAP – ∠OLP – ∠AOL
= 360° – 90° – 90° – (180° – 60°)
= 360° – 360° + 60° = 60°
Thus, tangents PA and PL are inclined to each other at an angle of 60°.

Question 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.
Solution:
Steps of Construction :
1. Draw a line segment AB = 8 cm.
2. Taking A as centre, draw a circle of radius 4 cm and taking B as centre, draw a circle of radius 3 cm.

3. Draw perpendicular bisector of the line segment AB. It intersects the line segment AB at O. Clearly, O is the mid-point of AB. With O as centre, draw a circle of radius OA or OB, intersecting circle with centre B at T4 and T4, circle with centre A at T3 and T4.
4. Join AT1( AT2, BT3 and BT4). Then, these are the required tangents.

Justification :
On joining BT1 we find that ∠BT1A = 90°, as ∠BT1A is the angle in a semi-circle.
∴ AT1 ⊥ BT1
Since BT1 is the radius of the given circle, so AT1 has to be a tangent to the circle.
Similarly, AT2, BT3 and BT4 are the tangents.

Question 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.
Solution:
Steps of Construction :
1. With the given data, draw a A ABC, in which AB = 6 cm, BC = 8 cm and ∠B = 90°.
2. Draw BD ⊥ AC.

3. Draw perpendicular bisectors of BC and BD. They meet at a point O.
4. Taking O as centre and OD as radius, draw a circle. This circle passes through B, C and D.
5. Join OA.
6. Draw perpendicular bisector of OA. This perpendicular bisector meets OA at M.
7. Taking M as centre and MA as radius, draw a circle which intersects the circle drawn in step 4 at point B and P.
8. Join AB and AP.
These are the required tangents from A.

Question 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.
Solution:
Steps of Construction :
1. Draw a circle with the help of a bangle.
2. Draw a secant ARS from an external point A. Produce RA to C such that AR = AC.

3. With CS as diameter, draw a semi-circle.
4. At the point A, draw AB ⊥ AS, cutting the semi-circle at B.
5. With A as centre and AB as radius, draw an arc to intersect the given circle, in T and T’. Join AT and AT’. Then, AT and AT’ are the required pair of tangents.

Aliter:

1. Mark any three points P, Q and R on the circle.
2. Draw the perpendicular bisectors of PQ and QR to meet at O, the centre of the circle.
3. Join OA and locate its mid-point M.
4. With M as centre and OM as radius, draw a circle to intersect the previous circle at T1 and T2.
5. Join AT1 and AT2, which are the required tangents.