Question

the diagonals of a rohmbus are 16 and 30 cm . find its perimeter​

Answer 1

Answer:

Given: Diagonals AC=30cm and DB=16cm.

Since the diagonals of the rhombus bisect at right angle to each other.

Therefore, OD=2DB=216=8cm

and OC=2AC=230=15cm

Now, In right angle triangle DOC,

(DC)2=(OD)2+(CO)2

⇒(DC)2=(8)2+(15)2

⇒(DC)2=64+225=289

⇒DC=289=17cm

Perimeter of the rhombus=4× side

=4×17=68cm

Thus, the perimeter of rhombus is 68 cm.

Answer 2

Answer:

68 cm

Step-by-step explanation:

• Diagonals of rhombus bisect each others at right angle. Then we got, base and height of right triangle 8 cm and 15 cm

Now,

$side \: of \: rhombus \: (hypotnuse \: of \: triangle \: ) = \sqrt{ {8}^{2} + {15}^{2} }$

• Therefore, side of rhombus =
• $\sqrt{289}$
• Side = 17 cm
• Now, perimeter of rhombus = 4 (side)
• Therefore, perimeter = 4 (17) = 68 cm
• I hope it will help you. Thank you