In an isosceles trapezoid the length of a diagonal is 25 cm and the length of an altitude is 15 cm. Find the area of the trapezoid. Lots of points

Accepted Solution

Answer: The area of the trapezoid is [tex]525\ cm^{2}[/tex]Step-by-step explanation:we know thatThe area of a isosceles trapezoid is equal to the area of two isosceles right triangles plus the area of a rectanglestep 1Find the area of the  isosceles right triangleRemember thatIn a isosceles right triangle the height is equal to the base of the trianglewe have[tex]h=15\ cm[/tex]so[tex]b=15\ cm[/tex]The area is equal to[tex]A=\frac{1}{2}(b)(h)[/tex]substitute the values[tex]A=\frac{1}{2}(15)(15)=112.5\ cm^{2}[/tex]step 2Find the area of the rectangleThe area of the rectangle is equal to[tex]A=LW[/tex]we have[tex]W=15\ cm[/tex] -----> is the height of the trapezoid[tex]d=25\ cm[/tex]  -----> the diagonal of the rectangleApplying the Pythagoras Theorem[tex]25^{2}=L^{2}+15^{2}\\L^{2}=25^{2}-15^{2} \\ L^{2} =400\\L=20\ cm[/tex]The area of the rectangle is[tex]A=(20)(15)=300\ cm^{2}[/tex]step 3Find the area of the trapezoid[tex]A=2(112.5\ cm^{2})+300\ cm^{2}=525\ cm^{2}[/tex]