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Construct a triangle whose sides are 3.6 cm , 3.0 cm and 4.8 cm. Bisect the smallest angle and measure each part.

Answer»

Solution :To construct a triangle ABCin which `AB = 3.6cm , AC = 3.0cm and BC= 4.8cm` , use the following STEPS.
(i) Draw a line segment BC of length `4.8cm`.
(ii) From B, point A is at a distance of `3.6cm` . So having B as centre, draw an arc of radius `3.6cm`.

(III) From C, point A is at a distance of 3cm. So, having C as centre, draw an arc of radius 3cm which intersect previous arc at A.
(iv) Join AB and AC. Thus, `DeltaABC` is the required triangle. Here, angle B is smallest , as AC is the smallest side . To DIRECT angle B, we use the following steps.
(i) Taking B as centre, we draw an are intersecting AB and BC at D and E, respectively.
(ii) Taking D and E as centres we draw ARCS intersecting at P.
(iii) Joining BP, we obtain angle bisector of `angleB`.
(iv) Here, `angleABC = 39^(@)`
Thus, `angleABD = angleDBC = (1)/(2)xx139^(@)= 19.5^(@)`


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