1. if the area of rhombus is 48 cm^2 . and one diagonal is 6cm. find its altitude. i want answer.
2. NAME A POLYHEDRON THAT HAVE 4 VERTICES 6 FACES NAD 8 EDGES
Here's (1)
Area of a rhombus = (1/2) d1*d2 ....where d1, d2 are the diagonals......so let d1 = 6 and we have
48 =(1/2) 6 *d2
48 = 3 * d2 divide both sides by 3
16 = d2
And since the diagonals bisect each other at right angles, they form two legs of a right triangle.....so...the length of the base = √[8^2 + 3^2] = √[64 + 9] = √73
And the area can also be expressed as a*b where a is the altitude and b is the base....so we have
48 = a * b
48 = a * √73
altitude = 48 / √73 = about 5.618 cm
Here's (1)
Area of a rhombus = (1/2) d1*d2 ....where d1, d2 are the diagonals......so let d1 = 6 and we have
48 =(1/2) 6 *d2
48 = 3 * d2 divide both sides by 3
16 = d2
And since the diagonals bisect each other at right angles, they form two legs of a right triangle.....so...the length of the base = √[8^2 + 3^2] = √[64 + 9] = √73
And the area can also be expressed as a*b where a is the altitude and b is the base....so we have
48 = a * b
48 = a * √73
altitude = 48 / √73 = about 5.618 cm