NCERT Solutions for Class 10 Maths Chapter 11 - Facts that Matter and Constructions (Exercise 11.1)

We know that: 

• To divide a line segment in a given ratio m:n, we divide this segment into (m + n) equal parts. Then we take m parts on one side and n on the other.
• The idea of dividing a line segment in any ratio is used in the construction of a triangle similar to a given triangle, whose sides are in a given ratio with the corresponding sides of the given triangle.
• The scale factor means the ratio of the sides of the triangle to be constructed with the corresponding sides of the given triangle.

Tangents to a Circle

Remember:

• If a point lies inside a circle, then there cannot be a tangent to the circle through this point.
• If a point lies on the circle, then there is only one tangent to the circle at this point and it is perpendicular to the radius through that point.
• If the point lies outside the circle, there will be two tangents to the circle from this point.

Note:

• For drawing a tangent at a point of a circle, simply draw the radius through this point and draw a line perpendicular to this radius through this point.
• The two tangents to a circle from an external point are equal.

Construction of Tangents to a Circle from a Point outside it.

Steps of construction:

• Let the centre of the circle be O and P be a point outside the circle.
• Join O and P.
• Bisect OP and let M be the midpoint of OP.
• Taking M as centre and MP or MO as radius, draw a circle intersecting the given circle at points A and B.
• Join PA and PB.

Thus, PA and PB are the required two tangents.

Note:
In case, the centre of the circle is not known, then to locate its centre, we take any two non-parallel chords and then find the point of intersection of their perpendicular bisectors.

Page No. 219 - 220

Exercise 11.1

In each of the following, give the justification of the construction also:
Q1. Draw a line segment of length 7.6 cm and divide it in the ratio 5:8.
Measure the two parts.
Ans: Steps of construction:

• Draw a line segment AB = 7.6 cm.
• Draw a ray AX making an acute angle with AB.
• Mark 13 (8 + 5) equal points on AX, and mark them as X₁, X₂, X₃, ........, X₁₃.
• Join ‘point X₁₃’ and B.
• From ‘point X₅’, draw X₅C ║ X₁₃B, which meets AB at C.

Thus, C divides AB in the ratio 5:8.
On measuring the two parts, we get: AC = 4.7 cm and BC = 2.9 cm.

Justification: 
In Δ ABX₁₃ and Δ ACX₅, we have

∴
⇒ AC : CB = 5 : 8.

Q2. 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.
Ans: Steps of construction:

• Draw a ∆ABC such that BC = 6 cm, AC = 5 cm and AB = 4 cm.
• Draw a ray BX making an acute angle ∠CBX.
• Mark three points X₁, X₂, X₃ on BX such that BX₁ = X₁X₂ = X₂X₃.
• Join X₃C.
• Draw a line through X₂ such that it is parallel to X₃C and meets BC at C′.
• Draw a line through C′ parallel to CA to intersect BA at A′.

Thus, A′BC′ is the required triangle.
Justification: 
By construction, we have:

⇒
But

⇒
⇒
Adding, 1 to both sides, we get

⇒
⇒  
Now, in ΔBC′A′ and ΔBCA
we have CA ║ C′A′
∴ Using AA similarity, we have:

⇒       [each equal to 2/3]

Q3. 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.
Ans: Steps of construction:

• Construct a Δ ABC such that AB = 5 cm, BC = 7 cm and AC = 6 cm.
• Draw a ray BX such that ∠CBX is an acute angle.
• Mark 7 points of X₁, X₂, X₃, X₄, X₅, X₆ and X₇ on BX such that BX₁ = X₁X₂ = X₂X₃ = X₃X₄ = X₄X₅ = X₅X₆ = X₆X₇
• Join X₅ to C.
• Draw a line through X₇ intersecting BC (produced) at C′ such that X₅C ║ X₇C′.
• Draw a line through C′ parallel to CA to intersect BA (produced) at A′. Thus, ΔA′BC′ is the required triangle.
  Justification: 
  By construction, we have
  C′A′ ║ CA
  ∴ Using AA similarity, ΔABC ~ ΔA′BC′
  
  Also X₇C′ ║ X₅C      [By construction
  ∴
  ∴
  ∴
  

Q4. Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are times the corresponding sides of the isosceles triangle.
Ans: Steps of construction:

• Draw BC = 8 cm
• Draw the perpendicular bisector of BC which intersects BC at D.
• Mark a point A on the above perpendicular such that DA = 4 cm.
• Join AB and AC.
  Thus, ΔABC is the required isosceles triangle.
• Now, draw a ray BX such that ∠CBX is an acute angle.
• On BX, mark three points X₁, X₂ and X₃ such that: BX₁ = X₁X₂ = X₂X₃
• Join X₂ to C.
• Draw a line through X₃ parallel to X₂ C and intersecting BC (extended) to C′.
• Draw a line through C′ parallel to CA intersecting BA (extended) at A′, thus, ΔA′BC′ is the required triangle.
  Justification:
  We have C′A′ ║ CA   [By construction]
  ∴ Using AA similarity, Δ ABC ~ ΔA′BC′
  ⇒
  Since, X₃C′ ║ X₂C       [By construction]
  ⇒
  ⇒     [By BPT]
  But
  ⇒
  Thus,

Q5. 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 the triangle ABC.
Ans: Steps of construction:

• Construct a ΔABC such that BC = 6 cm, AB = 5 cm and ∠ABC = 60°.
• Draw a ray  such that ∠CBX is an acute angle.
• Mark four points X₁, X₂, X₃ and X₄ on BX such that BX₁ = X₁X₂ = X₂X₃ = X₃X₄
• Join X₄C and draw X₃C′ ║ X₄C such that C′ is on BC.
• Also, draw another line through C′ and parallel to CA to intersect BA at A′.

Thus, ΔA′BC′ is the required triangle.
Justification:
By construction, we have:
X₄C ║ X₃C′
     [By BPT]
But      [By construction]
⇒      ...(1)
Now, we also have
     [By construction]
    [using AA similarity]
⇒      [From (1)]

Q6. 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 Δ ABC.
Ans: Steps of construction:

• Construct a Δ ABC such that BC = 7 cm, ∠B = 45° and ∠A = 105°.
• Draw a ray BX making an acute angle ∠CBX with BC.
• On BX, mark four points X₁, X₂, X₃ and X₄ such that BX₁ = X₁X₂ = X₂X₃ = X₃X₄.
• Join X₃ to C.
• Draw X₄C′ ║ X₃C such that C′ lies on BC (extended).
• Draw a line through C′ parallel to CA intersecting the extended line segment BA at A′.
  

Thus, ΔA′BC′ is the required triangle.
Justification:
By construction, we have:
C′A′ ║ CA
∴ ΔABC ~ ΔA′BC′     [AA similarity]
⇒       ...(1)
Also, by construction,
X₄C′ ║ X₃C
Δ BX₄C′ ~ Δ BX₃C
⇒
But  

⇒      ..(2)
From (1) and (2), we have:

Q7. 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.
Ans: Steps of construction: 

• Construct the right triangle ABC such that ∠B = 90°, BC = 4 cm and BA = 3 cm.
• Draw a ray BX such that an acute angle ∠CBX is formed.
• Mark 5 points X₁, X₂, X₃, X₄ and X₅ on BX such that BX₁ = X₁X₂ = X₂X₃ = X₃X₄ = X₄X₅.
• Join X₃ to C.
• Draw a line through X₅ parallel to X₃C, intersecting the extended line segment BC at C′.
• Draw another line through C′ parallel to CA intersecting the extended line segment BA at A′.

Thus, ΔA′BC′ is the required triangle.
Justification: 
By construction, we have:
C′A′ ║ CA
∴ ΔABC ~ Δ A′BC′     [By AA similarity ]
⇒       ...(1)
Also, X₅C′ ║ X₃C    [By construction]
∴ Δ BX₅C′ ~ Δ BX₃C
⇒
But      ...(2)
From (1) and (2) we get