# Notes of Ch 11 Construction| Class 10th Math

## Revision Notes for Ch 11 Construction Class 10th Mathematics

**Division of a line segment in a given ratio**

**Q. Draw Line segment PQ=9cm and divide it in the ratio 2:5. Justify your construction.**

**Answer**

__Steps of construction:__

(i) Draw Line Segment PQ=9cm

(ii) Draw a Ray PX, making an acute angle with PQ.

(iii) Mark 7 points A

_{1}, A

_{2}, A

_{3}…A

_{7}along PX such that PA

_{1}= A

_{3}A

_{2}= A

_{2}A

_{3}= A

_{3}A

_{4}= A

_{4}A

_{5}= A

_{5}A

_{6}= A

_{6}A

_{7}

(iv) Join QA

_{7}

(v) Through the point A

_{2}, draw a line parallel to A

_{7}Q by making an angle equal to ∠PA

_{7}Q at A

_{2}, intersecting PQ at point R. PR:RQ = 2:5

__Justification:__

We have A

_{2}R || A

_{7}Q

**Construction of a triangle similar to a given triangle as per the given scale factor when Scale Factor is less than 1**

**Q. Draw a ΔABC with sides BC = 8 cm, AC = 7 cm, and ÐB = 70°. Then, construct a similar triangle whose sides are (3/5)th of the corresponding sides of the**

**Δ**

**ABC.**

**Answer**

__Steps of construction:__

(i) Draw BC = 8 cm

(ii) At B, draw ∠XBC = 70°

(iii) With C as centre and radius 7 cm, draw an arc intersecting BX at A.

(iv) Join AB, and DABC is thus obtained.

(v) Draw a ray , making an acute angle with BC.

(vi) Mark 5 points, B

_{1}, B

_{2}, B

_{3}, B

_{4}, B

_{5}, along BY such that

(vii) BB

_{1}= B

_{1}B

_{2}= B

_{2}B

_{3}= B

_{3}B

_{4}= B

_{4}B

_{5}

(viii) Join CB

_{5}

(ix) Through the point B

_{3}, draw a line parallel to B

_{5}C by making an angle equal to ∠BB

_{5}C, intersecting BC at C´.

(x) Through the point C´, draw a line parallel to AC, intersecting BA at A´. Thus, ΔA´BC´ is Required Triangle.

__Justification__

Using BPT

**Construction of a triangle similar to a given triangle as per the given scale factor when Scale Factor is more than 1**

**Q. Construct an isosceles triangle with base 5 cm and equal sides of 6 cm. Then, construct another triangle whose sides are of the corresponding sides of the (4/3)th of first triangle .**

**Answer**

__Steps of construction:__

(i) Draw BC = 5 cm

(ii) With B and C as the centre and radius 6 cm, draw arcs on the same side of BC, intersecting at A.

(iii) Join AB and AC to get the required ΔABC.

(iv) Draw a ray , making an acute angle with BC on the side opposite to the vertex A.

(v) Mark 4 points B

_{1}, B

_{2}, B

_{3}, B

_{4}, along BX such that BB

_{1}= B

_{1}B

_{2}= B

_{2}B

_{3}= B

_{3}B

_{4}

(vi) Join B

_{3}C. Draw a line through B

_{4}parallel to B

_{3}C, making an angle equal to ∠BB

_{3}C intersecting the extended line segment BC at C´.

(vii) Through point C´, draw a line parallel to CA, intersecting extended BA at A´.

(viii) The resulting ΔA´BC´ is the required triangle.

**Construction of tangents to a circle**

**Q. Draw a circle of radius 3 cm. From a point 5 cm away from its centre, construct a pair**

**Answer**

__Steps of construction:__

1. Draw a circle with centre O and radius 3 cm. Take a point P such that OP = 5 cm, and then join OP.

2. Draw the perpendicular bisector of OP. Let M be the mid point of OP.

3. With M as the centre and OM as the radius, draw a circle. Let it intersect the previously drawn circle at A and B.

4. Joint PA and PB. Therefore, PA and PB are the required tangents. It can be observed that PA=PB=4cm.