0000:00:00:00:00Start5Number 5ExplanationA rectangular prism (cuboid) ABCD.EFGHABCD.EFGHABCD.EFGH has a length that is twice its width. Its height is equal to its width. The length of DFDFDF is 999 cm. The area of plane ADFGADFGADFG is...2752 cm2\frac{27\sqrt{5}}{2}\text{ cm}^22275 cm22762 cm2\frac{27\sqrt{6}}{2}\text{ cm}^22276 cm22732 cm2\frac{27\sqrt{3}}{2}\text{ cm}^22273 cm22722 cm2\frac{27\sqrt{2}}{2}\text{ cm}^22272 cm227 cm227\text{ cm}^227 cm2ExplanationProblem Visualization Plane ADFGADFGADFGDiagonal plane ADFGADFGADFG in a prism with p=2lp=2lp=2l and t=lt=lt=l.Toggle GridToggle Play Determining the Width Use the space diagonal DFDFDF to find lll with d2=p2+l2+t2d^2 = p^2 + l^2 + t^2d2=p2+l2+t2. DF2=(2l)2+l2+l2DF^2 = (2l)^2 + l^2 + l^2DF2=(2l)2+l2+l281=4l2+l2+l281 = 4l^2 + l^2 + l^281=4l2+l2+l281=6l281 = 6l^281=6l2 l2=816=272l^2 = \frac{81}{6} = \frac{27}{2}l2=681=227l=272=362l = \sqrt{\frac{27}{2}} = \frac{3\sqrt{6}}{2}l=227=236 Calculating Area of Plane ADFGADFGADFG Plane ADFGADFGADFG is a rectangle with sides AD=lAD = lAD=l and the front-face diagonal AFAFAF. Compute AFAFAF from the right triangle on the front face: AF2=AB2+BF2AF^2 = AB^2 + BF^2AF2=AB2+BF2AF2=(2l)2+l2=5l2AF^2 = (2l)^2 + l^2 = 5l^2AF2=(2l)2+l2=5l2AF=l5AF = l\sqrt{5}AF=l5 So the area is: L=AD×AFL = AD \times AFL=AD×AFL=l×l5=l25L = l \times l\sqrt{5} = l^2\sqrt{5}L=l×l5=l25L=2725L = \frac{27}{2}\sqrt{5}L=2275 Thus, the area of plane ADFGADFGADFG is 2752\frac{27\sqrt{5}}{2}2275 cm².
5Number 5ExplanationA rectangular prism (cuboid) ABCD.EFGHABCD.EFGHABCD.EFGH has a length that is twice its width. Its height is equal to its width. The length of DFDFDF is 999 cm. The area of plane ADFGADFGADFG is...2752 cm2\frac{27\sqrt{5}}{2}\text{ cm}^22275 cm22762 cm2\frac{27\sqrt{6}}{2}\text{ cm}^22276 cm22732 cm2\frac{27\sqrt{3}}{2}\text{ cm}^22273 cm22722 cm2\frac{27\sqrt{2}}{2}\text{ cm}^22272 cm227 cm227\text{ cm}^227 cm2ExplanationProblem Visualization Plane ADFGADFGADFGDiagonal plane ADFGADFGADFG in a prism with p=2lp=2lp=2l and t=lt=lt=l.Toggle GridToggle Play Determining the Width Use the space diagonal DFDFDF to find lll with d2=p2+l2+t2d^2 = p^2 + l^2 + t^2d2=p2+l2+t2. DF2=(2l)2+l2+l2DF^2 = (2l)^2 + l^2 + l^2DF2=(2l)2+l2+l281=4l2+l2+l281 = 4l^2 + l^2 + l^281=4l2+l2+l281=6l281 = 6l^281=6l2 l2=816=272l^2 = \frac{81}{6} = \frac{27}{2}l2=681=227l=272=362l = \sqrt{\frac{27}{2}} = \frac{3\sqrt{6}}{2}l=227=236 Calculating Area of Plane ADFGADFGADFG Plane ADFGADFGADFG is a rectangle with sides AD=lAD = lAD=l and the front-face diagonal AFAFAF. Compute AFAFAF from the right triangle on the front face: AF2=AB2+BF2AF^2 = AB^2 + BF^2AF2=AB2+BF2AF2=(2l)2+l2=5l2AF^2 = (2l)^2 + l^2 = 5l^2AF2=(2l)2+l2=5l2AF=l5AF = l\sqrt{5}AF=l5 So the area is: L=AD×AFL = AD \times AFL=AD×AFL=l×l5=l25L = l \times l\sqrt{5} = l^2\sqrt{5}L=l×l5=l25L=2725L = \frac{27}{2}\sqrt{5}L=2275 Thus, the area of plane ADFGADFGADFG is 2752\frac{27\sqrt{5}}{2}2275 cm².
