Question

The perimeter of a rhombus is 240m and of its diagonal is 96m. Find the area of the rhombus and its other diagonal

Answer 1

Perimeter of rhombus = 240

4 × side = 240

$side = \frac{240}{4}$
side = 60 m




××××××××××××

We know, diagonals of rhombus bisects each other at 90°

By Pythagoras theorem,

(96/2)² + (half of other diagonal)² =60²

(48)² + (Half of other diagonal)² = 60²

half of other diagonal = √(60²-48²)

Half of other diagonal = √(3600-2304)

Other diagonal = 2 × 36 = 72 m




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$area \: of \: rhombus \: = \frac{1}{2} \times product \: of \: diagonals \\ \\ here \\ \\ area \: = \frac{1}{2} \times 72 \times 96 \\ \\ = > 36 \times 96 \\ \\ = 3456 {m}^{2}$



I hope this will help you


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Answer 2

Heya !!!





Let ABCD is a rhombus.



In which ,



AB = AD = BC = CD





And,




AC and BD are its two diagonals.




Given that ,





Perimeter of rhombus = 240 m



4 × side = 240




Side = 240/4




Side = 60 m







Let AC = 96 m




We know that,



Diagonals of rhombus bisect each other other at right angle.




OA = OC = 1/2 × AC


OA = OC = 1/2 × 96 = 48





In right angled triangle OAB





(AB)² = (OA)² + (OB)²




(OB)² = ( AB)² - (OA)²




(OB)² = ( 60)² - (48)²






(OB)² = 3600 - 2304




OB² = 1296




OB = root 1296 = 36 m





Diagonal BD = 2 × OB



=> 2 × 36





=> 72 m





Therefore,




Area of rhombus ABCD = 1/2 × ( Product of diagonals )



=> 1/2 × ( AC × BD)



=> 1/2 × ( 96 × 72)





=> 6912/2





=> 3456 m².





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