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MathematicsWAEC SSCE2008

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Earlier discussion

1 older comment kept for context.

Oluwaseun Adeyemi

For a rhombus, all the sides are equal and diagonals bisect each other at 90o. Hence, the triangles formed are congruent: (under RHS).


Thus: \(\bar{PS}^2\) = 52 + 122


\(\bar{PS} = \sqrt{25 + 144}\)


= \(\sqrt{169}\)


\(\bar{PS}\) = 13cm


perimeter = 4 x length of a side


= 4 x 13cm


= 52cm