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)]