Constructions

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Questions:

Generic 2

5 x 2 = 10

Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then construct another triangle whose sides are \(3\over 4\) times the corresponding sides of the isosceles triangle.

Construct a \(\triangle ABC\) in which BC = 9 cm, and \(AB=6cm.\) Then construct another triangle whose sides are \(2\over 3\) of the corresponding sides of \(\triangle ABC\).

Construct a triangle ABC in which \(AB=5\ cm, BC=6\ cm\) and \(AC=7cm.\) Construct another triangle similar to \(\triangle ABC\) such that its sides are \(3\over 5\) of the corresponding sides of \(\triangle ABC\).

Draw a triangle ABC with sides BC = 6 cm, AB = 5 cm and (Then construct a triangle whose sides are \({3\over 2}\) of the corresponding sides of the triangle ABC.

Construct a triangle with sides 5 cm, 6 cm, and 7cm and then another triangle whose sides are \(7\over 5\) of the corresponding sides of the first triangle

1 Marks

7 x 1 = 7

Give three sides such that construction of a triangle is possible.

In figure, \(\angle ADE\) is constructed similar to, \(\triangle ABC\) write down the scale factor.

To divide a line segment LM in the ratio a : b, where a and b are positive integer, draw a ray LX so that angle \(\angle\)MLX is an acute angle, then find the minimum number of such points marked at equal distances on the ray LX.

To divide a line segment PQ in the ratio 2 : 7 first a ray PX is drawn so that angle \(\angle\)QPX is an acute angle then find the minimum number of such points marked at equal distances on the ray PX.

10.

Given a triangle with side AB = 8 cm. To get a line segment AB^' = \(\frac{3}{4}\) of AB, find the ratio in which line segment AB is divided.

11.

To construct a triangle similar to a given \(\Delta ABC\) with its sides \(\frac{3}{7}\) of the corresponding sides of \(\Delta ABC\), first draw a ray BX such that \(\angle CBX\) is an acute angle and X lies on the opposite side of A with respect to BC. Then locate points B₁, B_2, B₃,......... on BX at equal distance and then which points are joined in the next step.

12.

To draw a pair of tangents to a circle which are inclined to each other at an angle of 60° , it is required to draw tangents at end points of those two radii of the circle, then find the angle between them.

Generic 3

2 x 3 = 6

13.

Construct a right triangle in which the sides, (other than the hypotenuse) are of length 6 cm and 8 cm . Then construct another triangle, whose sides are \(\frac { 3 }{ 5 } \) times the corresponding sides of the given triangle.

14.

Draw a triangle with sides 5 cm, 6 cm and 7 cm. Then draw another triangle whose sides are \(\frac { 4 }{ 5 } \) of the corresponding sides of first triangle.

Fill up / 1 Marks

3 x 1 = 3

15.

A pair of tangents drawn through a point outside the circle are always ............................. in length.

16.

The sum of all the angles of a triangle is________________

17.

The ratio of the sides of the triangle to be constructed with the corresponding sides of the given triangle is known as their ____________

True or False

5 x 1 = 5

18.

A tangent line meets the circle only in one point

(a)

False

(b)

True

19.

In all geometrical constructions only two geometrical instruments viz. graduated ruler and compasses are required.

(a)

False

(b)

True

20.

Two tangents can be drawn to a circle through a point outside the circle.

(a)

False

(b)

True

21.

In order to divide a line segment internally in the ratio m : n, both m and n are real numbers.

(a)

True

(b)

False

22.

At least three parts are sufficient for construction of a triangle.

(a)

False

(b)

True

Generic 4

3 x 4 = 12

23.

Draw a right-angled triangle, in which the sides (other than the hypotenuse) are lengths 8 cm and 6 cm. Then, construct another triangle, whose sides are \(\frac { 3 }{ 4 } \) times of the corresponding sides of given triangle. Justify your construction.

24.

Draw a \(\triangle\)ABC with BC = 7 cm, \(\angle B=45°\) and \(\angle C=60°\). Then, construct another triangle, whose sides are \(\frac { 3 }{ 5 } \) times of the corresponding sides of \(\Delta ABC\) and justify your construction.

25.

Construct a pair of tangents PQ and PR to a circle of radius 4 cm from a point P outside the circle 8 cm away from the centre. Measure PQ and PR.

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Answers Key:

Generic 2

5 x 2 = 10

Steps of Construction:
(i) A line segment BC = 9 cm is drawn.
(ii) ∠ABC = 60^o is constructed at B.
(iii) An arc of 6 cm radius to be drawn with B as centre, cutting BA at A.
(iv) A and C are joined. Then triangle ABC is constructed.
(v) An acute angle CBX is drawn below BC.
(vi) Points B₁,B₂,B₃ are taken on BX, such that BB₁=B₁B₂=B₂B₃.
(vii) B₃ and C are joined.
(viii) B₂C' is drawn parallel to B₃C, meeting BC at C'
(ix) C'A' is drawn parallel to CA, meeting BA at A'.
(x) A' BC' is the required triangle similar to ΔABC whose sides are \(\frac { 2 }{ 3 } \)of the corresponding sides of ΔABC.

Steps of Construction:
(i) A line segment BC = 9 cm is drawn.
(ii) ㄥABC = 60^o is constructed at B.
(iii) An arc of 6 cm radius to be drawn with B as centre, cutting BA at A.
(iv) A and C are joined. Then triangle ABC is constructed.
(v) An acute angle CBX is drawn below BC.
(vi) Points B₁,B₂,B₃ are taken on BX, such that BB₁= B₁B₂= B₂B₃.
(vii) B₃ and C are joined.
(viii) B₂C' is drawn parallel to B₃C, meeting BC at C'
(ix) C'A' is drawn parallel to CA, meeting BA at A'.
(x) A' BC' is the required triangle similar to ΔABC whose sides are \(\frac { 2 }{ 3 } \) of the corresponding sides of Δ ABC.

Steps of Construction:
(i) A line segment BC = 6 cm is drawn.
(ii) An arc is drawn from B of radius 5 cm.
(iii) An arc is drawn from C of radius 7 cm, cutting the first arc at A. AB and AC are joined to get ΔABC.
(iv) An acute angle CBX is drawn below BC.
(v) On BX, points B₁,B₂,B₃,B₄,B₅ are taken such that BB1=B₁B₂=B₂B₃=B₃B₄=B₄B_5.
(vi) B₅ and C are joined.
(vii) B₃C' is drawn parallel to B₅C meeting BC at C'.
(viii) C'A' is drawn parallel to CA, meeting BA at A'.
(ix) Then ΔA'BC' is the required triangle similar to ΔABC, where sides are \(\frac { 3 }{ 5 } \) corresponding sides of ΔABC.

Steps of Construction:
1. Draw a line segment BC = 6 cm and at point B draw a ㄥABC = 60^o.
2. Cut AB 5 cm. Join AC. We obtain ABC is triangle.
3. Draw a ray BX making an acute angle with BC on the side opposite to the vertex A.
4. Locate 4 points A₁,A₂,A₃ and A₄ on the ray BX so that BA₁=A₁A₂=A₂A₃=A₃A₄.
5. Join A₄ to C.
6. At A₃ draw A₃C' || A₄C. Where C' is a point on the line segment BC.
7. At C' draw C'A' || CA, where A' is a point on the line segment BA.

Δ A'BC' is the required triangle.
Justification:
In Δ A'BC' and ΔABC A'C' || AC
∴ By BPT \(\frac { A'B }{ AB } =\frac { BC' }{ BC } \) ...(i)
From (i) and (ii),
\(\frac { A'B' }{ AB } =\frac { 3 }{ 4 } \Rightarrow A'B=\frac { 3 }{ 4 } AB\)
In ΔBA₃C' and ΔBA₄C
\(\frac { BC' }{ BC } =\frac { { BA }_{ 3 } }{ { BA }_{ 4 } } =\frac { 3 }{ 4 } \) ...(ii)
∴ Sides of new triangle formed are \(\frac { 3 }{ 4 } \) times the corresponding sides of first triangle.

Steps of Construction:
1. Draw a ΔABC with AB = 5 cm, BC = 7 cm, and AC = 6 cm.
2. Draw an acute angle CBX below BC at point B.
3. Mark the ray BX as B₁,B₂,B₃,B₄,B₅,B₆ and B₇ such that BB₁ = B₁,B₂ = B₂,B₃= B₃B₄= B₄B₅= B₅B₆= B₆B₇.
4. Join B₅ to C.
5. Draw B₇C' Parallel B₅C, where C' is the point on extended line BC.

6. Draw A'C' || AC.
Where A' is a point on BA extended to A'. A'BC' is the required triangle.
Justification:
In ΔABC and ΔA'BC'
AC || A'C'
∴ By BPT \(\frac { AB }{ A'B } =\frac { BC }{ BC' } \) ...(i)
Equating (i) and (ii)
\(\frac { AB }{ A'B } =\frac { { BB }_{ 5 } }{ { BB }_{ 7 } } \Rightarrow \frac { AB }{ A'B } =\frac { 5 }{ 7 } \Rightarrow A'B=\frac { 7 }{ 5 } AB\)
In ΔBB₅C and ΔBB₇C'
B₅C || B₇C'
∴ By BPT \(\frac { BC }{ BC' } =\frac { { BB }_{ 5 } }{ { BB }_{ 7 } } =\frac { 5 }{ 7 } \) ...(ii)
∴ Sides of new triangle formed is \(\frac { 7 }{ 5 } \) times the corresponding sides of first triangle.

1 Marks

7 x 1 = 7

()
The sum of two sides of a triangle must be greater than third side.
Let the sides are 2.5 cm, 4.5 cm and 6.5 cm

()

Scale factor = \(\frac { 3 }{ 4 } \)

()
a + b

()
9

10.

()
3 : 1

11.

()
B₇ to C

12.

()
Angle between the radii = 180° - 60° = 120°

Generic 3

2 x 3 = 6

13.

Given: A right triangle with sides of length 6 cm and 8 cm making a right angle.

Required: Triangle whose sides are \(\frac { 3 }{ 5 } \) times the corresponding sides of the given triangle.
Steps of Construction :
1. Construct a right triangle ABC right-angled at B with sides BC = 8 cm and AB = 6 cm.
2. Through B, construct an acute angle ∠CBX, (<90^o).
3. Mark five points B₁, B₂, B₃, B₄ and B₅, such that BB₁ = B₁B₂ = B₂B₃ = B₃B₄ = B₄B₅.
4. Join B₅C.
5. Through B₃, draw B₃C'||B₅C, intersecting BC in C'.
6. Through C', draw C'A'||CA, intersecting AB in A'.
Hence, ΔA'BC' is the required triangle.

14.

Given : A ΔABC, in which AB = 5 cm, BC = 6 cm and CA = 7 cm.

Required: ΔA'BC' ∼ ΔABC with \(\frac { 4 }{ 5 } \) (reduced) scale-factor.
Steps of Construction :
1. Construct ΔABC, such that AB=5cm, BC = 6 cm and CA = 7 cm.
2. Through B, construct an acute ㄥCBX, 90^o).
3. Mark five points on BX such that BB₁ = B₁B₂ = B₂B₃ - B₃B₄ = B₄B₅.
4. Join B₅C.
5. Through B₄, draw B₄C'||B₅C, intersecting BC in C'.
6. Through C', draw C'A'||CA, intersecting BA in A'.
Hence, ΔA'BC' is the required triangle.

Fill up / 1 Marks

3 x 1 = 3

15.

()
equal

16.

()
180^o

17.

()
Scale factor.

True or False

5 x 1 = 5

18.

(a)
False

19.

(a)
False

20.

(a)
False

21.

(b)
False

22.

(a)
False

Generic 4

3 x 4 = 12

23.

Given A right-angled triangle with sides of lengths 8 cm and 6 cm making a right angle. Required Triangle whose sides are \(\frac { 3 }{ 4 } \) times of the corresponding sides of given triangle.
Steps of Construction
1. Construct a right angled \(\angle ABC\) right angle at B with sides BC = 8 cm and AB = 6 cm
2. Through B, construct an acute LCBX on the side \(\angle CBX\) opposite to the vertex A.

3. Mark four points B₁, B₂ , B₃, and B₄ on BX such that BB₁ = B₁B₂ = B₂B₃,= B₃B₄
4. Join B₄C
5. Through B₃, draw B₃C' II B₄C intersecting BC .at C'.
6. Through C', draw C'A' II CA intersecting AB at A'. Hence, \(\Delta A'BC'\) is the required triangle.

24.

Given A \(\Delta ABC\) in which BC = 7 cm, \(\angle B=45°\) and \(\angle C=60°\) Required Draw \(\Delta A'BC'-\Delta ABC\)with scale factor \(\frac { 3 }{ 5 } <1\)
Steps of Construction
1. Draw a line segment BC = 7 cm.
2. At B, construct an \(\angle CBY=45°\)
3. At C, construct an \(\angle BCA=60°\) intersecting BY at A.

4. Join AB and CA. Then, \(\Delta ABC\) is the required triangle.
5. Through B, construct an acute angle \(\angle CBX\) on the side opposite to the vertex A.
6. Mark five points B₁, B₂, B₃, B₄ and B₅ on BX such that BB₁ = B₁B₂ = B₂B₃ =B₃B₄ = B₄B₅
7. Join B₅C

25.

Given, a circle of radius 4 cm whose centre is O and a point P, 8 cm away from its centre.
Steps of Construction
1. Draw a circle with O as centre and radius is equal to 4cm.
2. Take a point P such that QP = 8 cm and bisect it. Let M be the mid-point of QP.
3. Taking M as centre and MO as radius, draw a dotted circle. Let this circle cuts the given circle at Qand R.
4. Join PQ and PR. Thus, PQ and PR are the required tangents. By measurement (using scale), PQ = PR = 7 cm