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Answer :

Step of construction

1. Draw a line segment BC with the measure of 8 cm.

2. Now draw the perpendicular bisector of the line segment BC and intersect at the point D

3. Take the point D as centre and draw an arc with the radius of 4 cm which intersect the perpendicular bisector at the point A

4. Now join the lines AB and AC and the triangle is the required triangle.

5. Draw a ray BX which makes an acute angle with the line BC on the side opposite to the vertex A.

6. Locate the 3 points P, Q and R on the ray BX such that BP= PQ = QR

7. Join the points B2C and draw a line from B3 which is parallel to the line B2C where it intersects the extended line segment BC at point C’.

8. Now, draw a line from C’ the extended line segment AC at A’ which is parallel to the line AC and it intersects to make a triangle.

9. Therefore, \(\triangle \) A’BC’ is the required triangle.

- Draw a line segment of length 7.6 cm and divide it in the ratio 5:8. Measure the two parts.
- Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are \(\frac{2}{3} \) of the corresponding sides of the first trianngle.
- Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are \(\frac{7}{5} \) of the corresponding sides of the first triangle.
- Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are \(1\frac { 1 }{ 2 }\) times the corresponding sides of the isosceles triangle.
- Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and \(\angle \)ABC=60° ;. Then construct a triangle whose sides are \(\frac{3}4\) of the corresponding sides of the triangle ABC.
- Draw a triangle ABC with side BC = 7 cm, \(\angle \) B=45°, \(\angle \) A=105°;. Then construct a triangle whose sides are \(\frac{4}3 \) times the corresponding sides of \(\triangle \) ABC.
- 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 \(\frac{5}3\)times the corresponding sides of the given triangle.
- Q.1In each of the following, give the justification of the construction also: 1. Draw a line segment of length 7.6 cm and divide it in the ratio 5 : 8. Measure the two parts.
- Q.2 Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are \(\frac{2}{3} \) of the corresponding sides of the first triangle.
- Q.3 Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are \(\frac{7}{5} \) of the corresponding sides of the first triangle
- Q.5 Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and \(\angle \)ABC = 60°. Then construct a triangle whose sides are \(\frac{3}{4} \) of the corresponding sides of the triangle ABC.
- Q.6 Draw a triangle ABC with side BC = 7 cm, \(\angle \) B = 45°, \(\angle \) A = 105°. Then, construct a triangle whose sides are \(\frac{4}{3} \)times the corresponding sides of \(\triangle \)ABC.
- Q.7 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 \(\frac{5}{3} \) times the corresponding sides of the given triangle.

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