Construct an isosceles triangle whose base is 8cm and altitude is 4cm and then draw another triangle whose sides are 3/2 yes the corresponding sides of the triangle.
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Step of Construction : --------
- Draw a line segment BC = 8 cm.
- Draw a perpendicular bisector AD (4 cm) of BC.
- Joining AB and AC we get isosceles ΔABC.
- Construct an acute ∠CBX downwards.
- Along BX mark off 3 equal points B1, B2, B3 such that BB1 = B1B2 = B2B3.
- Join C to B2 and draw a line through B3 parallel to B2C intersecting the extended line segment BC at C’.
- Again draw a parallel line C’ A’ to AC cutting BP at A’.
- ΔA ‘BC’ is the required triangle.
C’A’ || CA [By construction]
ΔABC ~ ΔA 'BC' (By AA similarity)
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