1. A rectangular plank $$\sqrt{10}$$ metre wide, is placed symmetrically along the diagonal of a square of side 10 metres as shown in the figure. The area of the plank is:





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  • By: anil on 05 May 2019 03.05 pm
    In the given diagram AB=$$sqrt{10}$$ m Given that PQRS is a square and the plank is placed symmetrically $$ riangle$$BPA and $$ riangle$$AQC will be isosceles right triangles.           So PA=PB=$$frac{sqrt{10}}{sqrt{2}}$$=$$sqrt{5}$$ m           PQ= PA+AQ             AQ= PQ-PA=10-$$sqrt{5}$$ m We know that AQ=QC ($$ riangle$$AQC is isosceles right triangle)      So AC=$$sqrt{2}$$AQ=$$sqrt{2}$$*(10-$$sqrt{5}$$) m Now we can calculate area of plank       Area of ABCD= AB*AC= $$sqrt{10}$$*$$sqrt{2}$$(10-$$sqrt{5}$$)=10($$sqrt{20}$$-1) sq. mt

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