**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 A_{1}, A_{2},
A_{3}, A_{4}, A_{5}, A_{6}, A_{7}, A_{8},
A_{9}, A_{10}, A_{11}, A_{12 }and A_{13 }on AX such that AA_{1} = A_{1}A_{2}
= A_{2}A_{3} = A_{3}A_{4} = A_{4}A_{5}
= A_{5}A_{6} = A_{6}A_{7} = A_{7}A_{8}
= A_{8}A_{9} = A_{9}A_{10} = A_{10}A_{11}
= A_{11}A_{12} = A_{12}A_{13}

**(d) **Join BA_{13}.

**(e) **Through the point A_{5}, draw a line parallel to A_{13}B intersecting
AB at the point C.

Then,
AC : CB = 5 : 8

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

**Justification**:

A_{5}C
∥ A_{13}B [By construction]

AA_{5}/A_{5}A_{13}
= AC/CB ….. (i) [By Basic Proportionality Theorem]

But AA_{5}/A_{5}A_{13
}= 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 B_{1}, B_{2
}and B_{3 }on BX such that
BB_{1 }= B_{1}B_{2 }= B_{2}B_{3}.

**(d) **Join B_{3}C and draw a line segment through the point B_{2}, parallel to B_{3}C 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**:

B_{3}C
∥ B_{2}C’ [By
construction]

BB_{2}/B_{2}B_{3}
= BC’/C’C ….. (i) [By Basic Proportionality Theorem]

But
BB_{2}/B_{2}B_{3} = 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 B_{1}, B_{2},
B_{3}, B_{4}, B_{5}, B_{6} and B_{7} on
BX such that BB_{1} = B_{1}B_{2} = B_{2}B_{3}
= B_{3}B_{4} = B_{4}B_{5} = B_{5}B_{6}
= B_{6}B_{7}.

**(d) **Join B_{5}C and draw a
line segment through the point B_{7}
parallel to B_{5}C 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]

B_{7}C’
∥ B_{5}C [By construction]

∆BB_{7}C’
~ ∆BB_{5}C [By AAA similarity
criterion]

But
BB_{5}/BB_{7} = 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 B_{1}, B_{2 }and
B_{3} on BX such that BB_{1} = B_{1}B_{2}
= B_{2}B_{3}.

**(g) **Join B_{2}C and draw a
line segment through the point B_{3} parallel to B_{2}C 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]

B_{3}C’
∥ B_{2}C [By construction]

∆BB_{3}C’
~ ∆BB_{2}C [By AAA similarity
criterion]

But
BB_{3}/BB_{2} = 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 B_{1, }B_{2},
B_{3} and B_{4 }on BX
such that BB_{1} = B_{1}B_{2} = B_{2}B_{3} = B_{3}B_{4}.

**(d) **Join B_{4}C and draw a
line segment through the point B_{3}, parallel to B_{4}C 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**:

B_{4}C
∥ B_{3}C’ [By construction]

Therefore, BB_{3}/BB_{4} = BC’/BC [By Basic Proportionality Theorem]

But
BB_{3}/BB_{4} = 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 B_{1, }B_{2},
B_{3} and B_{4} on
BX such that BB_{1} = B_{1}B_{2} = B_{2}B_{3 }= B_{3}B_{4}.

**(d) **Join B_{3}C and draw a
line segment through the point B_{4} parallel to B_{3}C 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**:

B_{4}C’
∥ B_{3}C [By
construction]

∆BB_{4}C’
~ ∆BB_{3}C [By AAA similarity criterion]

Therefore, BB_{4}/BB_{3} = BC’/BC [By Basic Proportionality Theorem]

But
BB_{4}/BB_{3} = 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 B_{1}, B_{2},
B_{3, }B_{4} and B_{5} on BX such that BB_{1 }=
B_{1}B_{2 }= B_{2}B_{3} = B_{3}B_{4 }= B_{4}B_{5}.

**(d) **Join B_{3}C and draw a
line segment through the point B_{5},
parallel to B_{3}C 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**:

B_{5}C’
∥ B_{3}C [By
construction]

Therefore,
∆BB_{5}C’ ~ ∆BB_{3}C [By AAA similarity criterion]

BB_{5}/BB_{3}
= BC’/BC [By Basic
Proportionality Theorem]

But
BB_{5}/BB_{3} = 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)]