Question 1118632: The base of an isosceles triangle is 2a cm in length and its perpendicular height is h cm. On each of the three sides of the triangle, a square is constructed. Show that the area of the complete figure is 6a^2 +ah+2h^2 cm^2.
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! The length of the sides of the triangle are 2a for the base, and for the equal sides, one uses the Pythagorean Theorem, which would have leg a, for half the base, height h, and hypotenuse sqrt(a^2+h^2).
The square on one of the sides has area a^2+h^2, because the area of a square is the hypotenuse squared given this figure.
Both squares have area 2a^2+2h^2
the square on the base has area 4a^2.
The area of the triangle, part of the figure, is (1/2)(2a)*h=ah
The sum of these three is 6a^2+ah+2h^2 cm^2
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