ABCD is a rhombus with each side being equal to 8 cm. If BD = 10 em, AC = $$2\sqrt{x}$$cm, what is the value of $$1+\sqrt{x}$$(by approximation)
AO=Â $$\sqrt{x}$$, BO = 5 cm, AB = 8 cm
By pythagoras theorem,$$AB^2=BC^2+AC^2$$
$$64=25+x$$
$$x=39$$
$$1+\sqrt{x}=1+\sqrt{39}$$ = 7(approx)
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