Construct a triangle similar to a given triangle ABC, with sides measuring 5cm, 6cm and 8cm, such that its sides are of the corresponding sides of 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 times the corresponding sides of the given triangle.
Asked by Topperlearning User | 4th Jun, 2014, 01:23: PM
Steps of construction:
1. Draw line segment BC of length 5 cm. With B and C as centres and radii 6 cm and 8 cm respective draw two arcs intersecting each other at A. Join AB and AC. Now, triangle ABC is the triangle with sides 5 cm, 6 cm and 8 cm.
2. Draw an acute angle XBC on the side opposite to the vertex A.
3. On BX, mark six points B1, B2, B3, B4, B5, B6 and B7 such that:
BB1 = B1B2 = B2B3 = B3B4 = B4B5 = B5B6
4. Join B5 to C.
5. From B6, draw B6C' parallel to B5C, intersecting BC produced in C'.
6. From C' draw C'A' parallel to CA, meeting AB produced in A'.
Triangle A'BC' is the required triangle.
Step of Constructions:
1. Draw BC = 4 cm.
2. At B, draw a ray BY making angle 90o with BC i.e. YBC = 90o.
3. From BP, cut BA = 3 cm.
4. Join A and C to get ABC.
5. Through vertex B, draw ray BX making an acute angle with BC on the side opposite to vertex A.
6. On BX cut 5 equal line segments BB1 = B1B2 = B2B3 = B3B4 = B4B5.
7. Join B5 to C.
8. Through B3, draw a line parallel to B5C to meet BC at point C'.
9. Through C', draw a line parallel to side CA to meet BA at A'.
A'BC' is the required triangle.
Answered by | 4th Jun, 2014, 03:23: PM
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