Some of the important questions are covered in this question paper from the chapter Constructions.
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Questions:

Generic 2

5 x 2 = 10

Draw a line segment of length 7 cm. Find a point P on it which divides it in the ratio 3 : 5.

Draw a circle of radius 3.2 cm. Draw any diameter of the circle. At the end points of the diameter of the circle, draw tangents. Are they parallel?

Is it possible to construct a triangle whose sides measure 3 cm, 6 cm, 7 cm?

Draw two tangents from the end points of the diameter of a circle of radius 3.5 cm. After these tangents parallel?

Draw a pair of tangents to a circle of radius 5 cm which are inclined to each other at an angle of \(60^o\).

Generic 3

4 x 3 = 12

Draw a line segment AB of length 8 cm. Taking A as centre, draw a circle of radius 4 cm and taking B as centre, draw another circle of radius 3 cm . Construct tangents to each circle from the centre of the other circle

Draw a right angled \(\Delta ABC\) , in which BC = 12 cm, AB = 5 cm and \(\angle B=90°\). Then, construct a triangle similar to it and of scale factor \(\frac { 2 }{ 3 } \). Is the new triangle also a right angled triangle?

Draw two tangents to a circle of radius 3.5 cm from a point P at a distance of 6.2 cm from its centre.

Let PQR be a right triangle in which PQ = 3cm, QR = 4cm and \(< Q=90^o\). QS is the perpendicular from Q on PR. The circle through Q, R, S is drawn. Construct the tangents from P to this circle.

1 Marks

5 x 1 = 5

10.

Construct a \(\Delta ABC\) with sides AB = 4 cm, BC = 5 cm and CA = 7cm. Then, construct a triangle to it whose sides are \(\frac { 5 }{ 7 } \) times of the corresponding sides of the given triangle. First, we draw a ray BX such that \(\angle CBX\) is an acute angle and X lies on the opposite side of A with respect to Be. Then, locate points B₂ , B₂ , B₃, ... on BX at equal distances. What is the next step to join?

11.

Find the ratio in which E divides the line segment PQ.

12.

To divide a line segment LM in the ratio 4 : 3, a ray LX is drawn first such that angle \(\angle\)MLX is an acute angle and then points L₁, L₂, L₃,... are located at equal distances on the ray LX, then point M is joined to which point?

13.

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.

14.

To divide a line segment AB in the ratio 5 : 7, first a ray AX is drawn so that \(\angle BAX\) is an acute angle then find the minimum number of such points marked at equal distances on the ray AX.

Fill up / 1 Marks

5 x 1 = 5

15.

If a line and a circle have two points common then the line is called ___________

16.

In reduced scale-factor, the geometric figure to be constructed is .______________in size.

17.

The sum of any two sides of a triangle is always _____________ than the third side.

18.

The sum of all the angles of a triangle is________________

19.

The tangent line is ___________ to the radius through the point of contact.

True or False

5 x 1 = 5

20.

In a \(\Delta ABC\) , we draw \(\Delta AB'C'\sim \Delta ABC\) with scale factor \(\frac { 13 }{ 15 } \) . Then, perimeter of \(\Delta ABC\) > perimeter of \(\Delta AB'C'\)

(a)

False

(b)

True

21.

A triangle similar to \(\Delta ABC\) is to be drawn. If scale factor is greater than 1, then we get enlarged figure.

(a)

False

(b)

True

22.

In enlarged scale-factor, the geometric figure to be constructed is larger in size.

(a)

False

(b)

True

23.

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

(a)

False

(b)

True

24.

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

(a)

False

(b)

True

Generic 4

3 x 4 = 12

25.

Construct a triangle ABC with BC = 7 cm, B = 60° and AB = 6 cm. Construct another triangle whose sides are 3/4 times the corresponding sides ABC

26.

To a circle of radius 4 cm, draw two tangents which are inclined to each other at an angle of 60°.

27.

Draw a circle of radius 3.5 cm. Take a point T out side the circle at a distance of 7 cm from the centre and construct a pair of tangents from this point T to the circle and justify your construction.

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Answers Key:

Generic 2

5 x 2 = 10

Steps of construction:
1. Draw a line segment AB = 7
1. Draw a line segment AB = 7 cm.
2. Draw any ray AX making an acute angle with AB.
3. Draw the point A ll such that
\(A_{ 1 }A_{ 2 },A_{ 3 }.....A_{ 6 }\)
4.Join BA_g
5.Through the point A31 draw a line parallel to BAs.

Then AP : PB = 3 : 5

Yes

Yes , because sum of any two sides is greater than third side i.e., 3 + 6 > 7, 6 + 7 > 3 and 3 + 7 >6.

Steps of Construction :
1. Draw a circle with centre O and radius 3.5 cm.
2. Draw the diameter POQ.

3. Construct angle of 90^o at the end points P and Q.
4. XPY and LQM are the two tangents at P and Q to the circle with centre O.
XPY || LQM, because ㄥP + ㄥQ = 180^o.

Steps of Construction:
1. Draw a circle of radius 5 cm.
2. As tangents are inclined to each other at an angle of 60^o.
∴ Angle between the radii of circle is 120^o. (Use quadrilateral property)
3. Draw radii OA and OB inclined to each other at an angle 120^o.
4. At points A and B, draw 90^o angles. The arms of these angles intersect at point P.

5. PA and PB are the required tangents.
Justification:
In quadrilateral AOBP
AP and BP are the tangents to the circle.
Join OP.
In right angled ΔOAP
OA 丄 PA
OAPB forms a quadrilateral
∴ ㄥAOB =120^o
ㄥAOP = 60^o
OA = 5cm
∴ \({ tan\quad 60 }^{ o }=\frac { AP }{ OA } =\frac { AP }{ 5 } \Rightarrow \sqrt { 3 } =\frac { AP }{ 5 } \Rightarrow AP=5\sqrt { 3 } cm.\)
Similarly BP = \(5\sqrt { 3 } cm\)
A pair of tangents can be drawn to a circle from an external point outside the circle. These two tangents are equal in lengths.
∴ PA = PB.

Generic 3

4 x 3 = 12

Following steps will be followed for constructing the tangents on the given circles:
1. Draw a line segment AB of 8 cm. Taking A and B as centre draw two circles of 4 cm and 3 cm radius.
2. Bisect the line AB. Let mid-point of AB is C. Taking C as centre draw a circle of radius AC which will intersect the circles at point P, Q, R and S. Join BP, BQ, AS and AR. These are required tangents.

Given, scale factor = \(\frac { 2 }{ 3 } <1\)
Steps of Construction:
1. Draw a line segment Be = 12 cm.
2. From B, draw a line AB = 5 cm, which makes right angle at B.
3. Join AC. Thus, \(\Delta ABC\) is the required right angled triangle.
4. From B, draw an acute \(\angle CBY\) downwards.
5. On ray BY, mark three points B₁ , B₂ and B₃ such that BB₁ = B₁B₂ = B₂B₃
6. Join B₃C,
7. From point B₂, draw B₂N || B₃C intersecting BC at N.
8. From point N, draw NM II CA intersecting BA at M.Thus \(\Delta MBN\) is the required triangle.
Hence, \(\Delta MBN\) is also a right angled triangle, right-angled at B.

Given: A circle of radius 3.5 cm and a point P, 6.2 cm away from its centre.
Required: A pair of tangents.
Steps of Construction :

1. Draw a circle C (O, r) with centre O and radius 3.5 cm.
2. TakeapointP, such that OP = 6.2 cm.
3. Draw AB, the perpendicular bisector of OP and let it intersect OP in M.
4. With M as centre and PM or OM as radius, draw another circle intersecting the given circle in T and T^'.
5. Join PT and PT^'.
6. Thus, PT and PT^' are the required tangents from point P to the circle C (O, r).

Steps of Construction:
(i) Draw perpendicular bisector of QR intersecting at O.
(ii) Draw a circle with O as centre and OR as radius.
(iii) Join OP.
(iv) Draw a circle with OP as diameter intersecting the given circle at Q and T. Join PT.
∴ PQ and PT are required tangents.

1 Marks

5 x 1 = 5

10.

()
Here, scale factor =\(\frac { 5 }{ 7 } \).So, we locate points B₁ , B₂ , B₃ , B₄ , B₅ , B₆ and B₇ on BX at equal distances and in next step, join the last p .nt B₇ to C.

11.

()
3 : 2

12.

()
L₇

13.

()
B₇ to C

14.

()
Minimum number of points = 5 + 7 = 12

Fill up / 1 Marks

5 x 1 = 5

15.

()
secant line

16.

()
smaller

17.

()
greater

18.

()
180^o

19.

()
Perpendicular.

True or False

5 x 1 = 5

20.

(a)
False

21.

(a)
False

22.

(a)
False

23.

(a)
False

24.

(a)
False

Generic 4

3 x 4 = 12

25.

Steps of construction:
Draw a line segment BC = 7 cm.

1. Draw a line segment BC of length 5 cm.
2. At B, draw LMBC = 60° and produced line BM.
3. From point C draw a line making an angle of 30°.
4. Both the lines intersect at A.
5.MBC is the given triangle.
6. Draw a ray BXmaking an acute angle.
7. Locate three points B1, Bz, and B3on line segment BX.
8. Join BC
9. Draw a parallel line through B3to B3C intersecting extended line BCat C.
10. Through C' draw a line parallel to AC intersecting extended line segment BA at A'. A'BC is the required triangle.

26.

Steps of construction:
1. Draw a circle of radius 4 cm with 0 as centre.
2. Take a point A on the circumference of the circle and join OA. Draw perpendicular to OA at point A.
3. Draw a radius OB, making an angle of 1200 with OA.
4. Draw the perpendicular to OB at point B. Let both the perpendiculars intersect at point P.
5. Join OP. PA and PB are required tangents, which make an angle of 600 to each other.

27.

\(2\sqrt { 15 } cm\)