The lengths of the diagonals of a rhombus are 24cm and 32cm, then the length of the altitude of the rhombus is
(a) 12cm
(b) 12.8cm
(c) 19 cm
(d) 19.2cm
(d) 19.2cm
Explanation:
Let, ABCD be the given rhombus
Where,
AC = 32cm
Bd = 24 cm
First, let's finding the side of rhombus
We Know that,
Diagonals of rhombus are perpendicular bisector of each other
∴ AC 丄 BD
OB = `"BD"/2`
= `"24"/2` = 12 cm
OA = `"AC"/2`
= `"32"/2` = 16 cm
By Pythagoras Theorem,
`"AB"^{2}` = `("OA")^{2}` + `("OB")^{2}`
`"AB"^{2}` = `("16")^{2}` + `("12")^{2}`
`"AB"^{2}` = 256 + 144
`"AB"^{2}` = 400
`"AB"^{2}` = `("20")^{2}`
Now, To find altitude,
we use the help of Area
Area using Diagonals
Area of Rhombus = `frac{1}{2}` x Diagonal 1 x Diagonal 2
= `frac{1}{2}` x 24 x 32
= 384 `"cm"^{2}`
Area using Base and Height
Area of Rhombus = Base x Height
= 12 x 32 = 20 x Height
= 12 x 32 x `frac{1}{20}` = Height
= 19.2 = Height
0 Comments