To construct a right-angled triangle such that its sides are 3/4 times of the corresponding sides of the given triangle:
(a). Draw a line BC = 7 cm as base.
(b). Taking BC as base draw 60° angle as ∠CBA = 60°, where AB = 6 cm.
(c). Connect A to C as hypotenuse.
Now we shall construct another triangle whose sides are 3/4 times the corresponding sides of the triangle.
(d). Set the compass to 1 cm, rest one of the legs of the compass at point B and start making arcs B1 starting from B, B2 from B1, B3 from B2 and B4 from B3, since we have to construct a triangle having side equal to 3/4 times of corresponding sides of the given triangle.
(e). Join B4 to C.
(f). Now, draw a line to BC from B3, parallel to CB4.
(g). Say this point on BC is C’.
(h). Similarly, make an arc from C’ and cut the arc at 60° from its foot, then join C’A’.
(i). We get BA’ automatically.
And hence, we have the required triangle BA’C’.
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