Question

if the length of diagonals of a rhombus is 16and12 cm the side of rhombus is​

Answer 1

${\Huge {\fbox{\bf {\red{Given:-}}}}}$

The length of the first diagonal is 12 cm, whereas the length of the second diagonal is 16 cm.

${\Huge {\fbox{\bf {\green{To\:Find:-}}}}}$

The sides of Rhombus

${\Huge {\fbox{\bf {\green{Solution:-}}}}}$

Now as we know that BD = 16 cm and AC = 12 cm; therefore to find S

We use,

$\bf \left(\dfrac{B D}{2}\right)^{2}+\left(\dfrac{A C}{2}\right)^{2}=S^{2}$

$\bf \left(\frac{16}{2}\right)^{2}+\left(\frac{12}{2}\right)^{2}=S^{2}$

$\bf (8)^{2}+(6)^{2}=S^{2}(8)$

$\bf \sqrt{100}=S$

$\bf S=10cm$

$\therefore {\bold{\underline{Side\:of\:rhombus\:is\:10cm}}}$

Answer 2

Answer:

Diagonals of rhombus are cut each other at 90

Since d ¹ =16and d ² =12

Therefore, by Pythagoras theorem

a ² = (d¹/2)² + (d² /2)²

⇒ a ² =8 ² +6 ²

⇒ a= 100 =10

Step-by-step explanation:

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