Question 4 Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are 1.5 times the corresponding sides of the isosceles triangle.
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Solution
Steps of Construction: Step I: BC = 8 cm is drawn. Step II: Perpendicular bisector of BC is drawn and it intersects BC at O.
Step III: At a distance of 4 cm, a point A is marked on the perpendicular bisector of BC.
Step IV: AB and AC are joined to form ΔABC. Step V: A ray BX is drawn making an acute angle with BC opposite to vertex A. Step VI: 3 points B1,B2andB3 are marked on BX. Step VII:B2 is joined with C to form B2C. Step VIII:B3C′ is drawn parallel to B2C and C'A' is drawn parallel to CA. Thus, A'BC' is the required triangle formed. Justification: ΔAB′C′ ~ ΔABC by AA similarity condition. ∴ABAB′=BCB′C′=ACAC′ Also, ABAB′=AA2AA3=23 ⇒AB′=32AB,B′C′=32 BCandAC′=32AC