# Ncert Solutions For Class 10 Maths Ex 11.1

## Ncert Solutions For Class 10 Maths Chapter 11 Ex 11.1

1) Draw a line segment of length 7.6 cm and divide it in 5: 8 ration. In addition, find the measure of two parts.

Solution:

Procedure for construction:

1. Draw any ray AX, making an acute angle with AB.
2. Locate 13(= 5 + 8) points A1, A2, A3 ……… A13 on AX so that AA1 = A1A2 Al2 A13.
3. Join BA13.
4. Through the point A5(m = 5), draw a line parallel to BA13 (by making an angle equal to L AA13 B at A5 intersecting AB at C. Then AC: CB = 5 : 8)

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

Solution:

Procedure for construction:

1. Draw a line segment BC with length 5 cm.
2. With B as centre and radius of 4 cm draw an arc.
3. With C as centre and radius of 6 cm draw an arc.
4. Join AB and AC. Then, ∆ABC is the required triangle.
5. Below BC, make an acute angle ∠CBX
6. Along BX, mark up three points B1, B2, B3 such that BB1 = B1 B2 = B2B3
7. Join B3C
8. From B2, draw B2C’llB3c, meeting BC at C’
9. From C’ draw C’ All CA, meeting BA at A’
10. Then ∆A’BC’ is the required triangle, each of whose sides is two-third of the corresponding sides of ∆ABC.

3) Construct a triangle with side lengths 5 cm, 6 cm and 7 cm and then another triangle whose sides are  of the corresponding sides of the first triangle.

Solution:

Procedure for  construction :

1. Draw a line segment BC with length 6 cm.
2. With B as centre and keeping radius as 5 cm, draw an arc.
3. With C as centre and keeping radius as 7 cm, draw another arc, intersecting the previously drawn arc at Point A.
4. Join AB and AC. Then, ∆ABC is the required triangle.
5. Below BC, make an acute angle∠CBX.
6. Along BX, mark up seven points B1, B2, B3….. B7 such that BB1 = B1,B2, B6B7.
7. Join B5 to C (5 being smaller of 5 and 7 in7/5) and draw a line through B7 parallel to B5C, intersecting the extended line segment BC at C’.
8. Draw a line through C’ parallel to CA intersecting the extended line segment BA at A’. Then A’BC’ is the required triangle.

4) Draw a triangle ABC with sides BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct a triangle whose sides are ¾ of the corresponding sides of ∆ABC.

Solution:

Procedure for construction:

(i)            Draw a triangle ABC with BC = 6 cm, AB = 5 cm and ∠ABC = 60°.

(ii)           Draw any ray BX making an acute angle with BC on the side opposite to the vertex X.

(iii)          Locate 4(the greater of 3 and 4 in ¾) points B1, B2, B3, B4 on BX so that BB1 = B1B2 = B2B3 = B3B4.

(iv)         Join B4C and draw a line through B3(the 3rd point, 3 being smaller of 3 and 4 in ¾) parallel to B4C to intersect BC at C’.

(v)          Draw a line through C’ parallel to the line CA to intersect BA at A’. Then ∆A’BC’ is the required triangle.

Justification of construction

∆ABC ~ ∆A’BC’ , Therefore,

ABAB=ACAC=BCBC$\frac{AB}{A’B} = \frac{AC}{A’C’} = \frac{BC}{BC’}$

But, BCBC=BB3BB4=34$\frac{BC}{BC’} = \frac{BB_{3}}{BB_{4}} = \frac{3}{4}$

So, ABAB=ACAC=BCBC=34$\frac{AB}{A’B} = \frac{AC}{A’C’} = \frac{BC}{BC’} = \frac{3}{4}$
5) Draw a triangle ABC with side BC = 7 cm, ∠B = 45°, ∠A = 105°. Then, construct a triangle whose sides are4/3 times the corresponding sides of ∆ABC.

Solution:

Procedure for  construction :

(i)            Draw a triangle ABC with BC = 7cm, ∠B = 45° and ∠A = 105°.

(ii)           Draw any ray BX making an acute angle with BC on the side opposite to the vertex X.

(iii)          Locate 4(the greater of 3 and 4 in 4/3) points B1, B2, B3, B4 on BX so that 3 BB1 = B1 B2 = B2B3 = B3B4.

(iv)         Join NC’ and draw a line through B3(the 3rd point, 3 being smaller of 3 and 4 in 1) parallel to NC’ to intersect BC’ at C. 3

(v)          Draw a line through C’ parallel to the line CA to intersect BA at A’. Then A NBC’ is the required triangle.

6) Construct a triangle of isosceles type, whose base is 8 cm and height 4 cm and then another triangle whose sides are 1.5 times the corresponding sides of the isosceles triangle.

Solution:

Given:An isosceles triangle whose base is 8 cm and height 4 cm. Scale factor: 1  =

Required: To construct a similar triangle to above whose sides are 1.5 times the above triangle.

Procedure for construction:

(i)            Draw a line segment BC = 8 cm.

(ii)           Draw a perpendicular bisector AD of BC.

(iii)          Join AB and AC we get a isosceles ∆ABC.

(iv)         Construct an acute angle∠CBX downwards.

(v)          On BX make 3 equal parts.

(vi)         Join C to B2 and draw a line through B3 parallel to B2C intersecting the extended line segment BC at C’.

(vii)        Again draw a parallel line C’A’ to AC cutting BP at A’.

(viii)       ∆A’BC’ is the required triangle.