## CBSE Class 10 Maths Notes Chapter 11 Constructions

**Determining a Point Dividing a given Line Segment, Internally in the given Ratio M : N**

Let AB be the given line segment of length x cm. We are required to determine a point P dividing it internally in the ratio m : n.

**Steps of Construction:**

- Draw a line segment AB = x cm.
- Make an acute ∠BAX at the end A of AB.
- Use a compass of any radius and mark off arcs. Take (m + n) points A
_{1}, A_{2}, … A_{m}, A_{m+1}, …, A_{m+n}along AX such that AA_{1}= A_{1}A_{2}= … = A_{m+n-1}, A_{m+n} - Join A
_{m+n}B. - Passing through A
_{m}, draw a line A_{m}P || A_{m+n}B to intersect AB at P. The point P so obtained is the A required point which divides AB internally in the ratio m : n.

**Construction of a Tangent at a Point on a Circle to the Circle when its Centre is Known**

**Steps of Construction:**

- Draw a circle with centre O of the given radius.

- Take a given point P on the circle.
- Join OP.

- Construct ∠OPT = 90°.

- Produce TP to T’ to get TPT’ as the required tangent.

**Construction of a Tangent at a Point on a Circle to the Circle when its Centre is not Known**

If the centre of the circle is not known, then we first find the centre of the circle by drawing two non-parallel chords of the circle. The point of intersection of perpendicular bisectors of these chords gives the centre of the circle. Then we can proceed as above.

**Construction of a Tangents from an External Point to a Circle when its Centre is Known**

**Steps of Construction:**

- Draw a circle with centre O.
- Join the centre O to the given external point P.
- Draw a right bisector of OP to intersect OP at Q.
- Taking Q as the centre and OQ = PQ as radius, draw a circle to intersect the given circle at T and T’.
- Join PT and PT’ to get the required tangents as PT and PT’.

**Construction of a Tangents from an External Point to a Circle when its Centre is not Known**

If the centre of the circle is not known, then we first find the centre of the circle by drawing two non-parallel chords of a circle. The point of intersection of perpendicular bisectors of the chords gives the centre of the circle. Then we can proceed as above.

**Construction of a Triangle Similar to a given Triangle as per given Scale Factor \(\frac { m }{ n }\) , m < n.**

Let ΔABC be the given triangle. To construct a ΔA’B’C’ such that each of its sides is \(\frac { m }{ n }\) (m < n) of the corresponding sides of ΔABC.

**Steps of Construction:**

- Construct a triangle ABC by using the given data.
- Make an acute angle ∠BAX, below the base AB.
- Along AX, mark n points A
_{1}, A_{2}…, A_{n}, such that AA_{1}= A_{1}A_{2}= … = A_{m-1}A_{m}= … A_{n-1}A_{n}. - Join A
_{n}B. - From A
_{m}, draw A_{m}B’ parallel to AnB, meeting AB at B’. - From B’, draw B’C’ parallel to BC, meeting AC at C’.

Triangle AB’C’ is the required triangle, each of whose sides is \(\frac { m }{ n }\) (m < n) of the corresponding sides of ΔABC.

**Construction of a Triangle Similar to a given Triangle as per given Scale Factor \(\frac { m }{ n }\) , m > n.**

Let ΔABC be the given triangle and we want to construct a ΔAB’C’, such that each of its sides is \(\frac { m }{ n }\) (m > n) of the corresponding side of ΔABC.

**Steps of Construction:**

- Construct a ΔABC by using the given data.
- Make an acute angle ∠BAX, below the base AB. Extend AB to AY and AC to AZ.
- Along AX, mark m points A
_{1}, A_{2}…, A_{n}, ..A_{m}, such that AA_{1}= A_{1}A_{2}= A_{2}A_{3}= … = A_{n-1}A_{n}= … = A_{m-1}A_{m} - Join A
_{n}B. - From A
_{m}, draw A_{m}B’ parallel to A_{n}B, meeting AY produced at B’. - From B’, draw B’C’ parallel to BC, meeting AZ produced at C’.
- Triangle AB’C’ is the required triangle, each of whose sides is (\(\frac { m }{ n }\)) (m > n) of the corresponding sides of ΔABC.