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

kes1968 Mar 13, 2016

#1**+10 **

Here's (1)

Area of a rhombus = (1/2) d_{1}*d_{2} ....where d_{1}, d_{2} are the diagonals......so let d_{1} = 6 and we have

48 =(1/2) 6 *d_{2 }

48 = 3 * d_{2} divide both sides by 3

16 = d_{2}

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

CPhill Mar 13, 2016

#1**+10 **

Best Answer

Here's (1)

Area of a rhombus = (1/2) d_{1}*d_{2} ....where d_{1}, d_{2} are the diagonals......so let d_{1} = 6 and we have

48 =(1/2) 6 *d_{2 }

48 = 3 * d_{2} divide both sides by 3

16 = d_{2}

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

CPhill Mar 13, 2016