#### Exercise 17.3

Q1. Draw \(\triangle ABC\)

__Steps of construction:__

- Draw a line segment AB of length 3 cm.
- Draw \(\angle XBA =70^{\circ}\)
. - Cut an arc on BX at a distance of 5 cm at C.
- Join AC to get the required triangle.

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Q2. Draw \(\triangle ABC\)

__Steps of construction: __

- Draw a line segment AC of length 6 cm.
- Draw \(\angle XAC =70^{\circ}\)
. - Cut an arc on AX at a distance of 4 cm at B.
- Join BC to get the desired triangle.
- We see that BC = 6 cm.

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Q3. Draw an isosceles triangle in which each of the equal sides is of length 3 cm and the angle between them is \(45^{\circ}\)

__Steps of construction:__

Draw a line segment PQ of length 3 cm.

Draw \(\angle QPX =45^{\circ}\)

Cut an arc on PX at a distance of 3 cm at R.

Join QR to get the required triangle.

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Q4. Draw \(\triangle ABC\)

__Steps of construction:__

- Draw a line segment AC of length 3 cm.
- Draw \(\angle XAC =120^{\circ}\)
. - Cut an arc on AX at a distance of 3 cm at B.
- Join BC to get the required triangle.

By measuring, we get \(\angle B\)

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Q5. Draw \(\triangle ABC\)

__Steps of construction:__

- Draw a line segment BC of length 4 cm.
- At C, draw \(\angle BCY=90^{\circ}\)
. - Cut an arc on CY at a distance of 4 cm at A.
- Join AB. ABC is the required triangle.

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Q6. Draw a triangle ABC in which BC= 4 cm, AB = 3 cm and \(\angle B=45^{\circ}\)

__Steps of construction: __

- Draw a line segment AB of length 3 cm.
- Draw an angle of \(45^{\circ}\)
and cut an arc at this angle at a radius of 4 cm at C. - Join AC to get the required triangle.
- With A as centre, draw intersecting arcs at M and N.
- With centre M and radius more than half of MN, cut an arc on the opposite side of A.
- With N as centre and radius the same as in the previous step, cut an arc intersecting the previous arc at E.
- Join AE, it meets BC at D, then AE is the required perpendicular.

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Q7. Draw a triangle ABC with AB = 3 cm, BC = 4 cm and \(\angle B=60^{\circ}\)

__Steps of construction: __

Draw a line segment BC = 4 cm.

Draw \(\angle CBX=60^{\circ}\)

Draw an arc on BX at a radius of 3 cm cutting BX at A.

Join AC to get the required triangle.

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__Angle bisector for angle A: __

- With A as centre, cut arcs of the same radius cutting AB and AC at P and Q, respectively.
- From P and Q cut arcs of same radius intersecting at R.
- Join AR to get the angle bisector of angle A.

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__Angle bisector for angle C: __

- With A as centre, cut arcs of the same radius cutting CB and CA at M and N, respectively.
- From M and N, cut arcs of the same radius intersecting at T
- Join CT to get the angle bisector of angle C.

Mark the point of intersection of CT and AR as 0.

Angle \(\angle COA=120^{\circ}\)