Assignments for Class 10 Mathematics Construction have been developed for Standard 10 students based on the latest syllabus and textbooks applicable in CBSE, NCERT and KVS schools. Parents and students can download the full collection of class assignments for class 10 Mathematics Construction from our website as we have provided all topic wise assignments free in PDF format which can be downloaded easily. Students are recommended to do these assignments daily by taking printouts and going through the questions and answers for Grade 10 Mathematics Construction. You should try to do these test assignments on a daily basis so that you are able to understand the concepts and details of each chapter in your Mathematics Construction book and get good marks in class 10 exams.

**Construction**

To divide a line segment in a given ratio.(lets say m:n).

Let AB line has to divide in 3:2 ratio. Here m=3 and n=2.

- Draw a ray AX making acute angle with AB.
- Locate m + n point on AX such that for all points.
- Join BA
_{m + n}. - Draw a line A3C from A3 to AB which is parallel to BX.
- AC and CB are the desired ratio (here 3:2) in which line has to be divided.

**To construct the pair of tangents from an external point to a circle.**

If we have to draw a tangent from point A to a given circle of center O, then.

- Draw a line AN.
- Bisect this line AN. Midpoint on line is M.
- With AN as diameter, draw a circle with center M.
- Join A with the points B and C where this circle cuts the given circle.
- AB and AC are the required two tangents.

**To construct a triangle similar to a given triangle as per a given scale factor.**Scale factor less than one. (Triangle will be smaller than the given triangle).

Scale factor greater than one (Triangle will be greater than the given triangle).

- Draw a ray BX making an acute angle with BC on the side opposite to the vertex A.
- When scale factor is less than one : locate denominator it is greater number of fraction. For example in scale we 3/4, have to locate 4).
- When scale factor is greater then one : locate numerator it is greater number of fraction. For example in scale 5/4, have to locate 5.
- Point B
_{1},B_{2},B_{3},B_{4}Bon BX so that BB_{1 }=B_{1}B_{2}=B_{2}B_{3}=B_{3}B_{4=}B_{4}B_{5 }. - When scale factor is less than one : Join B
_{4}C and draw a line through B_{3}(the 3^{rd }point, 3 being smaller number in fraction ¾) Parallel B_{4}C to intersect BC at C’. - Draw a line through C’ parallel to the line CA to intersect BA at A’ then,Δ A’B C’ is the required triangle.
- When scale factor is greater than one : Join B
_{4}C (the 4th point, 4 being smaller number in fraction 5/4). - Draw a line through B
_{5 }(the 5^{th}point, 5^{th}being greater number in fraction 5/4) parallel to the line B_{4}C to intersect BC externally at C” ., Δ A”B C” Then, is the required triangle.

**Question. Construct a triangle similar to a given triangle with sides 7 cm, 9 cm and 10 cm and whose sides are 5/7 th of the corresponding sides of the given triangle. ****Solution.** Steps of Constructions :

1. With the given measurements construct the triangle ABC in which AB = 7 cm, BC = 9 cm and AC = 10 cm

2. Draw a ray AP, making any suitable angle with AB and on opposite side of vertex C.

3. Starting from A, cut off seven equal line-segments AX_{1}, X_{1}X_{2}, X_{2}X_{3}, X_{3}X_{4}, X_{4}X_{5}, X_{5}X_{6} and X_{6}X_{7} on AP.

4. Join BX7 and draw a line X5B’ parallel to X_{7}B which meets AB at B’.

5. Through B’, draw B’C’ || BC which meets AC at point C’.

The ΔAB’C’, so obtained, is similar to the given ΔABC and each side of ΔAB’C’ is 5/7 times the corresponding side of ΔABC.

**Question. Draw a pair of tangents to a circle of radius 3 cm which are inclined to each other at an angle of 60°. ****Solution.** Step of Constructions :

1. Draw a circle with centre O and radius = 3 cm

2. Draw any radius OA.

3. Draw another radius OB such that ∠AOB = 180° – 60° = 120°.

4. At point A, draw AM ⊥ OA i.e. draw tangent AM.

5. At B, draw tangent BN to the given circle.

6. The two tangents meet at point P in such a way that ∠APB = 60°.

∴ PA and PB are the required tangents.

**Question. 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.** Step of Constructions :

1. Draw BC = 4 cm

2. At B, draw a ray BP making angle 90° with BC i.e. ∠PBC = 90°

3. From BP, cut BA = 3 cm

4. Join A and C to get the given ΔABC.

5. Through vertex B, draw ray BX making any suitable angle with BC.

6. On BX cut 5 equal line segments BB_{1} = B_{1}B_{2} = B_{2}B_{3} = B_{3}B_{4} = B_{4}B_{5}.

7. Join B3 to C.

8. Through B5, draw a line parallel to B3C to meet BC produced at point C’.

9. Through C’, draw a line parallel to side CA to meet BA produced to A’. A’BC’ is the required triangle.

**Question. Draw a line segment of length 7 cm and divide it in the ratio 2 : 3.****Solution.** Step of constructions :

1. Draw AB = 7 cm

2. Draw ray AX making a suitable acute angle with AB.

3. Cut 2 + 3 = 5 equal segments AA_{1}, A_{1}A_{2}, A_{2}A_{3}, A_{3}A_{4} and A_{4}A_{5} on AX.

4. Join A5 with B.5. Through A_{2}, draw A2P parallel to A5B by making corresponding angles AA_{2}P and AA_{5}B equal.

6. The line through A2 and parallel to A5B will meet the given line segment at point P.

Then P is the required point which divides AB in the ratio 2 : 3 i.e. AP : PB = 2 : 3**Proof :**

Since in a triangle the line drawn parallel to one side of the triangle divides other two sides

proportionally, ∴ In ΔAA_{5}B, A_{2} P || A_{5}B

⇒ AP/PB = AA_{2}/A_{2}A_{5} i.e. AP/PB = 2/3

**Question. Draw a line segment of length 7.8 cm and divide it in the ratio 5 : 8. Measure the two parts.****Solution.** Steps:

(i) Draw a line segment AB = 7.8 cm.

(ii) Draw ray AX making any suitable angle with AB.

(iii) Draw ray BY parallel to ray AX by making alternate angles BAX and ABY equal.

(iv) Since 5 + 8 = 13, from AX cut 13 equal segments i.e., AA_{1}, A_{1}A_{2}, A_{2}A_{3}, …, A12A_{13}.

(v) Also, from BY cut 13 equal segments [of the same size as taken in step (iv)] i.e., BB_{1}, B_{1}B_{2}, B_{2}B_{3}, …, B_{12}B_{13}.

(vi) Join A_{5} with B8.

Line joining A_{5} and B8 cuts line segment AB at point P.

∴ P is the required point such that AP : PB = 5 : 8.

On measuring we find: AP = 3 cm and BP = 4.8 cm.

**Question. In the figure given on the next page, if B _{1}, B_{2}, B_{3},…… and A_{1}, A_{2}, A_{3},….. have been marked at equal distances. In what ratio C divides AB? **

**Solution.** 8 : 5 (1)

**Question. To divide a line segment BC internally in the ratio 3 : 5, we draw a ray BX such that ∠CBX is an acute angle. What will be the minimum number of points to be located at equal distances, on ray BX?****Solution.** 8

**Question. Draw a line segment AB of length 7 cm. Using ruler and compasses, find a point P on AB such that AP : PB = 3 : 5. ****Solution.** Steps of construction:

(i) Draw a line segment AB = 7 cm.

(ii) Draw AX || BY such that ∠A and ∠B are acute angles.

(iii) Divide AX and BY in 3 and 5 parts equally by

compass and mark A_{1}, A_{2}, A_{3}, B_{1}, B_{2}, B_{3}, B_{4} and B_{5} respectively.

(iv) Join A3B5 which intersects AB at P and divides AP :

PB = 3 : 5.

Hence, P is the required point on AB which divides it in 3 : 5.

**Question. Draw a line segment of length 6 cm. Using compasses and ruler, find a point P on it which divides it in the ratio 3 : 4.****Solution.** Steps of construction:

(i) Draw a line segment AB = 6 cm.

(ii) Draw any ray AX making an acute angle XAB with AB.

(iii) Along AX mark 7 (3 + 4) points A_{1}, A_{2}, A_{3}, A_{4},

……, A7 at equal distances such that AA_{1} = A_{1}A_{2}

= ….. = A6A7.

(iv) Join A7B.

(v) From A3, draw A3P parallel to A7B (by making an angle equal to ∠AA_{7}B at A_{3}), to meet AB at point P.

Then AP : PB = 3 : 4.

**Question. Draw a line segment of length 7.6 cm and divide it in the ratio 5 : 8. Measure the two parts. ****Solution.** Steps of construction:

(i) Draw a line segment AB = 7.6 cm.

(ii) Draw a ray AX making an acute angle with AB.

(iii) Mark 13 (8 + 5) equal points on AX, and mark them

as X1, X_{2}, X_{3}, …….., X_{13}.

(iv) Join ‘point X13’ and B.

(v) From ‘point X_{5}’, draw X5C || X_{13}B, which meets AB at C.

Thus, C divides AB in the ratio 5 : 8

On measuring the two parts, we get:

AC = 2.9 cm and BC = 4.7 cm.

**Question. ABC be a right triangle in which AB = 6 cm, BC = 4 cm and ∠B = 90°. BD is the perpendicular from B on AC. The circle through B, C, D is drawn. Construct the tangents from A to this circle. ****Solution.** Steps:

(i) Draw perpendicular bisectors of BC and CD which meet at point O.

O is the centre of the circle through B, C and D.

(ii) Join A and O.

(iii) Draw a circle with AO as diameter which meets the given circle at points P and B.

(iv) Join AP and AB.

AP and AB are the required tangents.

**Question. Draw a circle of radius 6 cm. From a point 10 cm away from its centre, construct the pair of tangents to the circle and measure its length. ****Solution.** Steps:

(i) With point O as centre, draw a circle of radius = 6 cm.

(ii) Mark a point A such that OA = 10 cm.

(iii) Join O and A. Then, draw the perpendicular bisector of OA which cuts OA at point B.

(iv) With B as centre and AB = BO as radius, draw a circle which cuts the circle drawn in step (i) at points P and Q.

(v) Join PA and QA.

PA and QA are the required tangents.

On measuring, we find PA = QA = 8 cm.

**Question. Construct a tangent to a circle of radius 4 cm from a point on the concentric circle of radius 6 cm and measure its length. Also, verify the measurement by actual calculation. ****Solution.** Steps:

(i) Mark a point as O.

(ii) With O as centre, draw two circles with radii 4 cm and 6 cm respectively.

(iii) On the bigger circle, mark a point A.

(iv) Join O and A.

(v) Draw perpendicular bisector of OA to get its mid-point P.

(vi) With P as centre and PA = PO as radius, draw a semicircle which meets smaller circle at point B.

(vii) Join A and B, and produce it to meet outer circle at point C. ABC is the required tangent.

The length of AB = 4.5 cm (approx.)

Verification: Join OB.

Since, angle between radius and tangent is 90°.

⇒ ∠ABO = 90°.

In right ΔABO, AB^{2} + OB^{2} = OA^{2}

⇒ AB^{2} = 62 – 42 = 36 – 16 = 20

and AB = √20 cm = 4.5 cm (approx.)

**Question. Draw a right triangle ABC in which AB = 6 cm, BC = 8 cm and ∠B = 90°. Draw BD perpendicular from B on AC and draw a circle passing through the points B, C and D. Construct tangents from A to this circle. ****Solution.** Steps of construction:

(i) Draw a right angle triangle ABC, right angled at B,

AB = 6 cm and BC = 8 cm.

(ii) Draw BD ⊥ AC.

(iii) Draw a circumcircle of DBDC.

(iv) From point A draw a pair of tangents AB and AP.

Then AB and AP are the required tangents.

**Question. Draw two concentric circles C _{1} and C_{2} of radii 3 cm and 5 cm. Taking a point on outer circle C2, construct the pair of tangents to the other. Measure the length of a tangent and verify it by actual calculation. **

**Solution.**4 cm

**Question. Draw two concentric circles of radii 3 cm and 5 cm. Construct a tangent to smaller circle from a point on the larger circle. Also measure its length. ****Solution.** Steps of construction:

(i) Take point O. Draw two concentric circles of radii 3 cm and 5 cm respectively.

(ii) Locate point P on the circumference of larger circle.

(iii) Join OP and bisect it. Let M be mid-point of OP.

(iv) Taking M as centre and MP as radius, draw an arc intersecting smaller circle at A and B.

(v) Join PA and PB. Thus, PA and PB are required tangents.

**Question. Is it possible to construct a pair of tangents from point P to circle of radius 5 cm situated at a distance of 4.9 cm from the centre?****Solution.** No

**Question. Draw a pair of tangents to a circle of radius 5 cm which are inclined to each other at an angle of 60°. ****Solution.** Steps of construction:

(i) A circle, with centre O and radius 5 cm is drawn.

(ii) As tangents are inclined at 60°.

∴ ∠TOS = 120°.

(iii) Two radius OT and OS, inclined at an angle of 120° are drawn.

(iv) Tangents are drawn to the circle at T and S meeting at P.

Then PT and PS are the required tangents.

**Question. Draw a circle of radius 3 cm. Take two points P and Q on one of its extended diameters each at a distance of 7 cm from its centre. Draw tangents to the circle from these two points P and Q. ****Solution.** Steps of construction:

(i) Join P and O.

(ii) Bisect PO such that M be its mid-point.

(iii) Taking M as centre and MO as radius, draw a circle.

Let it intersects the given circle at A and B.

(iv) Join PA and PB.

Thus, PA and PB are the two required tangents from P.

(v) Now, join O and Q.

(vi) Bisect OQ such that N is its mid-point.

(vii) Taking N as centre and NO as radius, draw a circle.

Let it intersects the given circle at C and D.

(viii) Join QC and QD.

Thus, QC and QD are the required tangents to the given circle.

**Question. Is it possible to construct a pair of tangents from point P lying on circle of radius 4 cm and centre O?****Solution.** No

**Question. 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. ****Solution.** Steps of construction:

(i) AB = 8 cm is taken.

(ii) With centre A, a circle of radius 4 cm is drawn and with centre B, a circle of radius 3 cm is drawn.

(iii) With AB as diameter, a circle is drawn meeting circle with centre A at S and T respectively and circle with centre B at P and Q respectively.

(iv) Then AP and AQ are tangents from A to circle with centre B and BS and BT are tangents from B to circle with centre A.

**Question. Draw a circle of radius 3 cm. From a point 6 cm away from its centre, construct the pair of tangents to the circle and measure their lengths.****Solution.** 5.20 m (approx)

## Assignments for Class 10 Mathematics Construction as per CBSE NCERT pattern

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