Given base BC = 8 cm
∠ B = 45°
And AB - AC = 3.5 cms.
Steps of construction:
i. Draw a base line BC of 8 cms.
ii. Construct ∠ B = 45°.
a. With B as centre and with any radius, draw another arc cutting the line BC.
b. With D as centre and with the same radius, draw an arc cutting the first arc (drawn in step b) at point E.
c. With E as centre and with the same radius, draw another arc cutting the first arc (drawn in step b) at point F.
d. With E and F as centers, and with a radius more than half the length of EF, draw two arcs intersecting at point G.
e. Join points B and G. The angle formed by GBC is 90°. i.e.
∠ GBC = 90°.
f. Now the point H will be the point of intersection of the ray BG and the first arc (from step b). With points H & D as centers, with any radius more than half the length of HD, draw two arcs such that they meet at point I. By joining points I and B, we get the ray BI which forms 45° with ray BD.
As, AB – AC = 3.5
AB > AC
So the ray BI will be above BC
iii. With B as centre draw an arc with length 3.5 cms ( = AB - AC given), such that it intersects ray BI at J.
iv. Join CJ and we draw a perpendicular bisector for CJ.
a. By drawing arcs on both sides of the line CJ, with C and J as centers and with same lengths. These arcs intersect at K and L on either side of line CJ.
v. The perpendicular bisector for CJ will intersect the ray BJ at point A. Join AC.
Thus the formed triangle ABC is the required triangle.
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