NCERT Solutions Maths Class 10 Exercise 11.1

NCERT Solutions Maths Class 10 Exercise 11.1

Maths Class 10 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.

Solution: To construct: A line segment of length 7.6 cm and to divide the given line segment in the ratio 5 : 8. Also, to measure the two parts.

Steps of construction:

(a) Draw a line segment AB measuring 7.6 cm.

(b) Draw any ray AX, making an acute angle with AB.

(c) Locate 13 (= 5 + 8) points A1, A2, A3, A4, A5, A6, A7, A8, A9, A10, A11, A12 and A13  on AX  such that AA1 = A1A2 = A2A3 = A3A4 = A4A5 = A5A6 = A6A7 = A7A8 = A8A9 = A9A10 = A10A11 = A11A12 = A12A13

(d) Join BA13.

(e) Through the point A5, draw a line parallel to A13B intersecting AB at the point C.

Then, AC : CB = 5 : 8

On measuring, AC = 2.9 cm and CB = 4.7 cm.

Justification:

A5C A13B                                 [By construction]

AA5/A5A13 = AC/CB ….. (i)      [By Basic Proportionality Theorem]

But AA5/A5A13 = 5/8 ….. (ii)     [By construction]

From equations (i) and (ii), we get AC/CB = 5/8

Therefore, AC : CB = 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.

Solution: To construct: A triangle ABC whose sides are 4 cm, 5 cm and 6 cm, and then a triangle similar to the triangle ABC whose sides are 2/3 of the corresponding sides of the triangle ABC.

Steps of construction:

(a) Construct a triangle ABC of sides 4 cm, 5 cm and 6 cm.

(b) Draw any ray BX, making an acute angle with BC on the side opposite to the vertex A.

(c) Locate 3 points B1, B2 and B3 on BX such that BB1 = B1B2 = B2B3.

(d) Join B3C and draw a line segment through the point B2, parallel to B3C intersecting BC at the point C’.

(e) Draw a line segment through C’ parallel to the line segment CA to intersect BA at A’.

Then, A’BC’ is the required triangle.

Justification:

B3C B2C’                                      [By construction]

BB2/B2B3 = BC’/C’C   ….. (i)       [By Basic Proportionality Theorem]

But BB2/B2B3 = 2/1   ….. (ii)        [By construction]

From equations (i) and (ii), BC’/C’C = 2/1

C’C/BC’ = 1/2

C’C/BC’ + 1 = ½ + 1

⇒ (C’C + BC’)/BC’ = 3/2

⇒ BC/BC’ = 3/2

BC’/BC = 2/3     ……… (iii)

CA C’A’                                     [By construction]

BC’A’ ~ BCA                          [AAA similarity criterion]

A’B/AB = A’C’/AC = BC’/BC = 2/3       [From equation (iii)]

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.

Solution: To construct: A triangle ABC of sides 5 cm, 6 cm and 7 cm, and then a triangle similar to triangle ABC whose sides are 7/5 of the corresponding sides of the triangle ABC.

Steps of construction:

(a) Construct a triangle ABC of sides 5 cm, 6 cm and 7 cm.

(b) Draw any ray BX, making an acute angle with BC on the side opposite to the vertex A.

(c) Locate 7 points B1, B2, B3, B4, B5, B6 and B7 on BX such that BB1 = B1B2 = B2B3 = B3B4 = B4B5 = B5B6 = B6B7.

(d) Join B5C and draw a line segment through the point B7 parallel to B5C intersecting BC produced at the point C’.

(e) Draw a line segment through C’ parallel to the line segment CA to intersect BA produced at the point A’.

Then, A’BC’ is the required triangle.

Justification:

C’A’ CA                                      [By construction]

∆ABC ~ A’BC’                           [By AAA similarity criterion]

A’B/AB = A’C’/AC = BC’/BC      [By Basic Proportionality Theorem]

B7C’ B5C                                    [By construction]

∆BB7C’ ~ ∆BB5C                         [By AAA similarity criterion]

But BB5/BB7 = 5/7                       [By construction]

Therefore, BC/BC’ = 5/7

⇒ BC’/BC = 7/5

So, 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 1½ times the corresponding sides of the isosceles triangle.

Solution: To construct: An isosceles triangle ABC whose base is 8 cm and altitude 4 cm and then a triangle similar to it whose sides are 1½ (or 3/2) times the corresponding sides of the first triangle.

Steps of construction:

(a) Draw BC = 8 cm

(b) Construct perpendicular bisector of BC. Let it meet BC at D.

(c) Mark a point A on the perpendicular bisector such that AD = 4 cm.

(d) Join AB and AC. Thus, ABC is the required isosceles triangle.

(e) Draw any ray BX, making an acute angle with BC on the side opposite to the vertex A.

(f) Locate 3 points B1, B2 and B3 on BX such that BB1 = B1B2 = B2B3.

(g) Join B2C and draw a line segment through the point B3 parallel to B2C intersecting BC produced at the point C’.

(h) Draw a line segment through C’ parallel to the line segment CA to intersect BA produced at A’.

Then, A’BC’ is the required triangle.

Justification:

C’A’ CA                                 [By construction]

ABC ~ A’BC’                      [By AAA similarity criterion]

⇒ A’B/AB = A’C’/AC = BC’/BC               [By Basic Proportionality Theorem]

B3C’ B2C                               [By construction]

∆BB3C’ ~ ∆BB2C                     [By AAA similarity criterion]

But BB3/BB2 = 3/2                   [By construction]

Therefore, BC’/BC = 3/2

Hence, A’B/AB = A’C’/AC = BC’/BC = 3/2

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 triangle ABC.

Solution: To construct: A triangle ABC with side BC = 6 cm, AB = 5 cm and ABC = 60° and then a triangle similar to it whose sides are 3/4 of the corresponding sides of the triangle ABC.

Steps of construction:

(a) Construct a triangle ABC with side BC = 6 cm, AB = 5 cm and ABC = 60°.

(b) Draw any ray BX, making an acute angle with BC on the side opposite to the vertex A.

(c) Locate 4 points B1, B2, B3 and B4 on BX such that BB1 = B1B2 = B2B3 = B3B4.

(d) Join B4C and draw a line segment through the point B3, parallel to B4C intersecting BC at the point C’.

(e) Draw a line segment through C’ parallel to the line segment CA to intersect BA at A’.

Then, A’BC’ is the required triangle.

Justification:

B4C B3C’                                           [By construction]

Therefore, BB3/BB4 = BC’/BC           [By Basic Proportionality Theorem]

But BB3/BB4 = 3/4                              [By construction]

Therefore, BC’/BC = 3/4     ……… (i)

CA C’A’                                            [By construction]

Therefore, BC’A’ ~ BCA               [By AAA similarity criterion]

Hence, A’B/AB = A’C’/AC = BC’/BC = 3/4       [From equation (i)]

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 ABC.

Solution: To construct: A triangle ABC with side BC = 7 cm, B = 45° and A = 105° and then a triangle similar to it whose sides are 4/3 times the corresponding sides of the triangle ABC.

Steps of construction:

(a) Construct a triangle ABC with side BC = 7 cm, B = 45° and A = 105°.

(b) Draw any ray BX, making an acute angle with BC on the side opposite to the vertex A.

(c) Locate 4 points B1, B2, B3 and B4 on BX such that BB1 = B1B2 = B2B3 = B3B4.

(d) Join B3C and draw a line segment through the point B4 parallel to B3C intersecting BC produced at the point C’.

(e) Draw a line segment through C’ parallel to the line segment CA to intersect BA produced at A’.

Then, A’BC’ is the required triangle.

Justification:

B4C’ B3C                                   [By construction]

∆BB4C’ ~ ∆BB3C                         [By AAA similarity criterion]

Therefore, BB4/BB3 = BC’/BC    [By Basic Proportionality Theorem]

But BB4/BB3 = 4/3                       [By construction]

Therefore, BC’/BC = 4/3      ……… (i)

CA C’A’                                     [By construction]

BC’A’ ~ BCA                          [By AAA similarity criterion]

Therefore, A’B/AB = A’C’/AC = BC’/BC = 4/3          [From equation (i)]

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.

Solution: To construct: A right triangle in which sides (other than hypotenuse) are of lengths 4 cm and 3 cm, and then a triangle similar to it whose sides are 5/3 times the corresponding sides of the first triangle.

Steps of construction:

(a) Construct a right triangle in which sides (other than hypotenuse) are of lengths 4 cm and 3 cm.

(b) Draw any ray BX, making an acute angle with BC on the side opposite to the vertex A.

(c) Locate 5 points B1, B2, B3, B4 and B5 on BX such that BB1 = B1B2 = B2B3 = B3B4 = B4B5.

(d) Join B3C and draw a line segment through the point B5, parallel to B3C intersecting BC produced at the point C’.

(e) Draw a line segment through C’ parallel to the line segment CA to intersect BA produced at A’.

Then, A’BC’ is the required triangle.

Justification:

B5C’ B3C                                     [By construction]

Therefore, ∆BB5C’ ~ ∆BB3C         [By AAA similarity criterion]

BB5/BB3 = BC’/BC                       [By Basic Proportionality Theorem]

But BB5/BB3 = 5/3                        [By construction]

Therefore, BC’/BC = 5/3       ……… (i)

CA C’A’                                       [By construction]

BC’A’ ~ BCA                            [By AAA similarity criterion]

Therefore, A’B/AB = A’C’/AC = BC’/BC = 5/3          [From equation (i)]

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