yes
A diagonal is a line between two vertices that is not an edge. As two vertices are connected by an edge, the number of diagonals from a vertex is the number of vertices (or sides or angles) less 3; as each diagonal is counted twice, once for the vertex at each end of it: number_of_diagonals = ½ × number_of_vertices × (number_of_vertices - 3) For a nonagon, it has 9 vertices → number_of_diagonals = ½ × 9 × (9 -3) = ½ × 9 × 6 = 27.
Given: The area of the rhombus is 120 square feet The diagonal of the rhombus is 16 feet think of the rhombus being two identical triangles, connected at their base which is 16 feet long. Each of them would then have an area of 60 feet. Now, in a triangle, area = (base * height) / 2 the area is already given as 60, and the base as 16 we can say then: 60 = (16 * h) / 2 ∴60 = 8h ∴h = 7.5 Now, that 7.5 is half the length of the rhombus (as it's the height of one of our triangles, which each are half our rhombus). So we know that that the other diagonal on the rhombus is twice that. In other words, the answer is 15.
The only pyramid with a square base that has equilateral faces, is one where the diagonal of the base is exactly twice as long as the pyramid is high.
thats a bit of a mouthful...
A series of transformations on quadrilateral S resulted in quadrilateral T. The angles of quadrilateral S and T are congruent but the sides of quadrilateral T are twice as long as quadrilateral S. Which transformation on quadrilateral S must be included to result in quadrilateral T * sorry thats the full question!
The shortest path between two points is a straight line. This is a mathematical fact, which can be proven in another question.The diagonal of a quadrilateral is a straight line between two opposing (non-adjacent) vertices. The perimeter of a quadrilateral will include two separate paths between the same vertices. The difference is that these two paths are each composed of two linked line segments, so each of these paths will be longer than the diagonal.Therefore, the length of the perimeter of a quadrilateral will be greater than twice the length of either diagonal.
yes
octagon-8 sides
You can use pythagorean theorem twice to find the diagonal of a cube
Diagonal = sqrt (twice the square of a side) eg: square of side 8 units, d = sqrt(2 x 64) = sqrt 128 = 11.3137
A diagonal is a line between two vertices that is not an edge. As two vertices are connected by an edge, the number of diagonals from a vertex is the number of vertices (or sides or angles) less 3; as each diagonal is counted twice, once for the vertex at each end of it: number_of_diagonals = ½ × number_of_vertices × (number_of_vertices - 3) For a nonagon, it has 9 vertices → number_of_diagonals = ½ × 9 × (9 -3) = ½ × 9 × 6 = 27.
no thats the obvious answer
If the circle is inscribed in the square, the side length of the square is the same as the diameter of the circle which is twice its radius: → area_square = (2 × 5 in)² = 10² sq in = 100 sq in If the circle circumscribes the square, the diagonal of the square is the same as the diameter of the circle; Using Pythagoras the length of the side of the square can be calculated: → diagonal = 2 × 5 in = 10 in → side² + side² = diagonal² → 2 × side² = diagonal² → side² = diagonal² / 2 → side = diagonal / √2 → side = 10 in / √2 → area _square = (10 in / √2)² = 100 sq in / 2 = 50 sq in.
No sorry thats a fault with the game there sorting it in the next patch
Given: The area of the rhombus is 120 square feet The diagonal of the rhombus is 16 feet think of the rhombus being two identical triangles, connected at their base which is 16 feet long. Each of them would then have an area of 60 feet. Now, in a triangle, area = (base * height) / 2 the area is already given as 60, and the base as 16 we can say then: 60 = (16 * h) / 2 ∴60 = 8h ∴h = 7.5 Now, that 7.5 is half the length of the rhombus (as it's the height of one of our triangles, which each are half our rhombus). So we know that that the other diagonal on the rhombus is twice that. In other words, the answer is 15.
4 times