Question

In a rhombus if diagonals are 30 and 40 cm find perimeter

Answer 1

mark me as a brainliest

Answer 2

❏ Used ForMuLaS:-

•For a Rhombus of:-

1) Diagonals bisects each other at right angle.

2) All sides are equal in length.

3) The opposite internal angles are equal.

4) Sum of the two consequtive internal angles is 180°.

5) The diagonals are not equal in length.

•Perimeter:-

$\sf\longrightarrow\boxed{\red{ Perimeter=4\times side}}$

•Area:-

$\sf\longrightarrow\boxed{\red{ Area=d_1\times d_2}}$

where, $d_1\: and\:d_2\:are \: diagonals$

❏ Question:-

Q) In a rhombus if diagonals are 30 and 40 cm find perimeter.

➪ We know that , the diagonals of a Rhombus bisect each other at right angles.

Let, the Rhombus is ABCD . ( see in the fig.)

∴sides, AB=BC=CD=DA

➾The Diagonals are AC and BD.

•Given, AC=40 cm.

BD=30 cm.

Let, the Diagonals AC and BD bisect each other at the point O.

$\therefore AO=OC=\frac{AC}{2}=\frac{\cancel{40}}{\cancel2}\:cm=20\:cm$

AND

$\therefore BO=OD=\frac{BD}{2}=\frac{\cancel{30}}{\cancel2}\:cm=15\:cm$

Now, For the ∆AOB,

$\sf\longrightarrow\angle AOB=90\degree$

∴ ∆AOB is a right angled triangle.

•Now, from Pythagorean Theorem

$\sf\longrightarrow AB{}^{2}=AO{}^{2}+BO{}^{2}$

$\sf\longrightarrow AB{}^{2}=20{}^{2}+15{}^{2}$

$\sf\longrightarrow AB=\sqrt{400+225}$

$\sf\longrightarrow AB=\sqrt{625}$

$\sf\longrightarrow\boxed{ \large{\red{AB=25}}}$

$\therefore Side \:of \:the \:Rhombus= 25\: cm$

$\therefore Perimeter= 4\times25\: cm$

$\therefore \boxed{\large{\red{Perimeter= 100\: cm}}}$

━━━━━━━━━━━━━━━━━━━━━━━

$\underline{ \huge\mathfrak{hope \: this \: helps \: you}}$

#answerwithquality & #BAL