**Q.1 Construct a triangle ABC in which BC = 7 cm, and AB + AC = 13 cm **

**Sol. ****Steps of Construction :**

1.Draw a ray BX and cut off a line segment BC = 7 cm

2. Construct

3. From BY, cut off BD = 13 cm

4. Join CD.

5. Draw the perpendicular bisect of CD, intersecting BA at A.

6. Join AC.

The triangle ABC thus obtained is the required triangle.

**Q.2 Construct a triangle ABC in which BC = 8 cm, and AB – AC = 3.5 cm **

**Sol. **

**Steps of Construction :**

1. Draw a ray BX and cut off a line segment BC = 8 cm from it.

2. Construct

3. Cut off a line segment BD = 3.5 cm from BY.

4. Join CD.

5. Draw perpendicular bisector of CD intersecting BY at a point A.

6. Join AC

Then ABC is the required triangle.

**Q.3 Construct a triangle PQR in which QR = 6 cm and PR – PQ = 2cm**

**Sol.**

**Steps of Construction :**

1. Draw a ray QX and cut off a line segment QR = 6 cm from it.

2. Construct a ray QY making an angle of 60º with QR and produce YQ to form a line YQY'

3. Cut off a line segment QS = 2cm from QY'.

4. Join RS.

5. Draw perpendicular bisector of RS intersecting QY at a point P.

6. Join PR.

Then PQR is the required triangle.

**Q.4 Construct a triangle XYZ in which , and XY + YZ + ZX = 11 cm. **

**Sol. **

**Steps of Construction :**

1. Draw a line segment PQ = 11 cm

2. At P, draw a ray PL such that

3. At Q, draw ray QM such that intersecting PL at X.

4. Draw perpendicular bisectors of XP and XQ intersecting PQ in Y and Z respectively.

Then XYZ is the required triangle.

**Note : -** For clarity in figure, method of drawing angles of 15º and 45º have not been shown. Students should draw these angles with the help of ruler and compass only by the method as shown earlier.

**Q.5 Construct a right triangle whose base is 12 cm and sum of its hypotenuse and other side is 18 cm. **

**Sol. ****Steps of Construction :**

1. Draw a ray BX and cut off a line segment BC = 12 cm

2. Construct

3. From by cut off a line segment BD = 18 cm .

4. Join CD.

5. Draw the perpendicular bisector of CD intersecting BD at A.

6. Join AC

Then ABC is the required triangle.

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Why should we draw perpendicular

You should mention how the lenght has been adjused in q1 AB+AC=13

You shuld mention where is 13

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