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In the figure shown above, lengths AB, BD, and CD are all 3√2 units. Angles A and C are both 45°. What is the perimeter of ABCD? What is the area of ABCD? see attachment for the figure
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Look at 2 triangles, notice they are isosceles triangles w/ 2 equal sides this means corresponding angles are equal -> y and z are both 45 degrees From this we can say the third angle must be 90 degrees Now we have 2 45-45-90 right triangles where BC and AD are hypotenuse BC = AD = (3sqrt(2))*sqrt(2) = 6 perimeter just add up all outside sides area find height by drawing straight line down one of the triangles forming a new right triangle of lengths 3:h:3sqrt(2) solve for h and multiply by 6 to get area
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