Question

The diagonals of a rhombus measure 50cm and 48cm. Find its perimeter

Answer 1

Step-by-step explanation:

$primeter = 2 \sqrt{ {d1}^{2} + {d2}^{2} }$

so

p=2×√50^2+48^2

P=138.62cm

Answer 2

Concept:

The perimeter of a rhombus is $2\sqrt{d_{1} ^{2}+d_{2} ^{2} }$

Given:

Two diagonals are given to us, 50cm and 48cm.

To find:

The perimeter of the rhombus.

Solution:

We know the perimeter of a rhombus is $2\sqrt{d_{1} ^{2}+d_{2} ^{2} }$

Let

$d_{1} = 50cm\\d_{2}= 48cm$

So, now we will put the values of the given diagonals in the formula of the perimeter of the rhombus.

We will get,

$2\sqrt{d_{1^{2} }+d_{2}^{2} } \\2\sqrt{50^{2} +48^{2} } \\2\sqrt{2500+2304} \\2\sqrt{4804} \\2*69.3\\138.6$

∴The required answer to this question is the determining perimeter of the given rhombus which is 138.6cm.

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