Question

the diagonals of a rhombus measure 16 cm and 30 cm, then it perimeter is
answer is 68 but how​

Answer 1

Step-by-step explanation:

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Answer 2

SOLUTION

Let ABCD be the given rhombus.

We know that the diagonals of a rhombus are perpendicular bisector of each other.

Let the point of intersection of diagonals AC & BD be M.

Given, AC= 30cm & BD= 16cm

Now,

AM= AC/2= 30/2= 15cm

& DM= BD/2= 16/2= 8cm

Now, In right ∆AMD,

By applying Pythagoras theorem,

${</u></strong><strong><u>AD</u></strong><strong><u>}^{2} = {</u></strong><strong><u>AM</u></strong><strong><u>}^{2} + {</u></strong><strong><u>MD</u></strong><strong><u>}^{2} \\ = > {</u></strong><strong><u>AD</u></strong><strong><u>}^{2} = {15}^{2} + {8}^{2} \\ = > {</u></strong><strong><u>AD</u></strong><strong><u>}^{2} = 225 + 64 \\ = > {</u></strong><strong><u>AD</u></strong><strong><u>}^{2} = 289 \\ = > </u></strong><strong><u>AD</u></strong><strong><u> = \sqrt{289} = 17cm$

Again, all the sides of a rhombus are equal.

Therefore,

AB=BC=CD=AD= 17cm

Now the Perimeter of rhombus:

Sum of all sides (4×side)

=) 4× 17cm

=) 68cm

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