if the area of a rhombus is 48cm^2 , and one of its diagonal is 5 cm . find its altitude

Asked by ramaysah12 | 12th Mar, 2016, 06:38: PM

Expert Answer:

Given,
Area of Rhombus = 48 cm2
one diagonal = 5 cm 
 
Area of a rhombus = fraction numerator d 1 space cross times space d 2 over denominator 2 end fraction = altitude × side 
 
48 = fraction numerator 5 space cross times space d 2 over denominator 2 end fraction
fraction numerator 48 space cross times space 2 over denominator 5 end fraction = d2
length of diagonal 2 = 19.2 cm 
 
We know, diagonals of a rhombus are perpendicular and bisect each other.
So, according to Pythagoras theorem, 
side2 = open parentheses fraction numerator d 1 over denominator 2 end fraction close parentheses squared plus space open parentheses fraction numerator d 2 over denominator 2 end fraction close parentheses squared
open parentheses 5 over 2 close parentheses squared plus space open parentheses fraction numerator 19.2 over denominator 2 end fraction close parentheses squared
= (2.5)2 + (9.6)2
= 6.25 + 92.16
= 98.41
side = square root of 98.41 end root space equals space 9.92
So, each side of the rhombus = 9.92 cm 
 
altitude = area ÷ side 
= 48 ÷ 9.92
= 4.83 cm 
Hence, the height of the rhombus is 4.83 cm 

Answered by Keravi Thanani | 5th Dec, 2017, 02:27: PM