Let the diagonal of the triangle have length 17*d. * Consider the *k*-th triangle from the left as shown with vertex E on the diagonal. Let F be the point on AB so that EF is perpendicular to AB. Then the height of the triangle is the length of EF which is, by similar triangles, AE Ã— BC/AC = 40*k*/17
Hence, area of *k-*th triangle is (1/2)(70/21)(40*k*/17)
Hence, the sum of the area of the 17 triangles is (1/2)(70/21)(40/17)(1 + 2 + 3+ ... +17) = 600
Hence, the required area is 600 squared units.
Hence, option 4. |