Question

IF THE AREA OF THE RHOMBUS IS 24 CM ^ AND ONE OF ITS DIAGONAL BE 4CM FIND THE PERIMETER OF THE RHOMBUS. PLS GOVE STEP BY STEP ANSWER TO UNDERSTAND I DONT JUST WRITE THE ANSWER LIKE THT PLS WRITE IN SUCH A WAY THT I UNDERSTAND I AM NEW TO THIS CHAPTER THS WHY PLSS.....

Answer 1

Answer:

area of rhombus = 24 cm square

one of its diagonal = 4 cm

let its other diagonal be X

so , 1/2 * 4 * X = 24cm square

x = 12 cm

we know that the diagonals of a rhombus bisect each other at eight angle .

now from one of the right triangle within it we will calculate the side of the rhombus .

the one of the side we get from above process = ( 36 + 4 ) ^ 1/ 2 = 40 ^ 1/2 = 2√ 10

so perimeter of the given rhombus = 4 * 2√10

= 8√ 10

Step-by-step explanation:

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Answer 2

Answer:

Step-by-step explanation:

Area of rhombus =product of diagonals /2

24=product of diagonals /2

48/4=II Diagonal =12

By Pythagoras theorem

6²+2²=side²

36+4=side²

√40=side

2√10=side

Perimeter =4*2√10

Perimeter =8√10