Question

find the perimeter of a rhombus the length of whose diagonals are 16 cm and 30 cm
Aman VII J
Give correct answer​

Answer 1

Step-by-step explanation:

According to the Question

It is given that,

• Length of Diagonal ,d = 16cm
• Length of another diagonal ,d' = 30cm

we have to calculate the perimeter of rhombus .

As we know that

• Perimeter of Rhombus = 4 × Side

Firstly we calculate the side of rhombus

Side of Rhombus = ½ √d²+d'²

On substituting the value we get

↠ Side of Rhombus = ½ √16² + 30²

↠ Side of Rhombus = ½ √256+900

↠ Side of Rhombus = ½ √1156

↠ Side of Rhombus = ½ × 34

↠ Side of Rhombus = 17 cm

No, calculating its Perimeter .

• Perimeter of Rhombus = 4 × Side

On substituting the value we get

↠ Perimeter of Rhombus = 4 × 17

↠ Perimeter of Rhombus = 68cm

• Hence, the perimeter of the rhombus is 68 cm .

Additional Information

$\begin{gathered}\boxed{\begin {array}{cc}\\ \quad \Large\underline{\bf Formulas\:of\:Areas:-}\\ \\ \star\sf Square=(side)^2\\ \\ \star\sf Rectangle=Length\times Breadth \\\\ \star\sf Triangle=\dfrac{1}{2}\times Breadth\times Height \\\\ \star \sf Scalene\triangle=\sqrt {s (s-a)(s-b)(s-c)}\\ \\ \star \sf Rhombus =\dfrac {1}{2}\times d_1\times d_2 \\\\ \star\sf Rhombus =\:\dfrac {1}{2}d\sqrt {4a^2-d^2}\\ \\ \star\sf Parallelogram =Breadth\times Height\\\\ \star\sf Trapezium =\dfrac {1}{2}(a+b)\times Height \\ \\ \star\sf Equilateral\:Triangle=\dfrac {\sqrt{3}}{4}(side)^2\end {array}}\end{gathered}$

Answer 2

GIVEN :-

the length of whose diagonals are 16 cm and 30 cm

TO FIND:-

find the perimeter of a rhombus = ?

SOLUTION :-

We know that diagonals of rhombus are perpendicular bisector of each other. 16/2 = 8

30/2 = 15

8² +15² = C²

64 +225 = c²

289 = c² ⇒ C = 17

Perimeter of a rhombus ⇒ 4x side

4 × 17 = 68 cm

Hence, the answer is 68 cm