Question

1. Find the perimeter of rhombus, the lengths of whose diagonals are 16 cm and 30 cm.
15cm
Bem
to the​

Answer 1

To find :

• we need to find the perimeter of rhombus .

Solution :

• Length of Diagonals of rhombus = 16cm and 30cm.

As we know that,

Perimeter of rhombus = 2 √(d1)² + (d2)²

• Length of Diagonal 1 (d1) = 16cm
• Length of Diagonal 2 (d2) = 30cm

››➔ p = 2√16² + 30²

››➔ p = 2√256 + 900

››➔ p = 2√1156

››➔ p = 2 × 34

››➔ p = 68 cm

Hence,

• Perimeter of rhombus is 68 cm

━━━━━━━━━━━━━━━━━━━━━━━━━

Some Formulas : --

›› Area of rhombus = (d1 × d2)/2

›› Perimeter of rhombus = 4a or 2√(d1)² + (d2)²

Where,

• d1 = Length of Diagonal 1
• d2 = Length of Diagonal 2
• a = side of rhombus.

━━━━━━━━━━━━━━━━━━━━━━━━━

Answer 2

$\huge\mathcal{\underbrace{QUESTION:-}}$

✍️ Find the perimeter of Rhombus, if the diagonals of the rhombus measure 16cm and 30cm .

$\huge\mathcal{\underbrace{SOLUTION:-}}$

✍️ See the attachment picture .

✍️ According to the question,

• Diagonal AC = 30cm

• Diagonal BD = 16cm

✍️ Since, the diagonals of the rhombus bisects at right angle to each other .

✍️ Therefore,

✔️ OB = $\rm{\dfrac{BD}{2}\:=\:\dfrac{16}{2}\:=\:8\:cm\:}$

✔️ OA = $\rm{\dfrac{AC}{2}\:=\:\dfrac{30}{2}\:=\:15\:cm\:}$

✍️ Now, in ∆AOB ,

$\checkmark\:\rm{\purple{(AB)^2\:=\:(OA)^2\:+\:(OB)^2\:}}$

$\rm{\implies\:(AB)^2\:=\:15^2\:+\:8^2\:}$

$\rm{\implies\:(AB)^2\:=\:225\:+\:64\:}$

$\rm{\implies\:AB\:=\:\sqrt{289}\:}$

$\rm{\implies\:AB\:=\:±\:17\:}$

[NOTE :- Length can never be negative]

$\rm{\red{\implies\:AB\:=\:17\:cm\:}}$

✍️ Now,

✔️ Perimeter of the rhombus = 4 × (side)

• Side of rhombus = 17 cm

$\rm{\implies\:Perimeter\:of\:rhombus\:=\:4\times{17}\:}$

$\bigstar\:\rm{\blue{\boxed{\implies\:Perimeter\:of\:Rhombus\:=\:68\:cm\:}}}$

$\rm{\green{\therefore}}$ Perimeter of the rhombus is ‘68cm’ .