Question

Find the diagonal of rectangle whose length is 16cm and area is 192cm²​

Answer 1

So, In order to attempt this question....

Figure it out in rough way, similar to my attachment ... (here I've named rectangle as ABCD)

As, we can now observe

What are we given?

Given here are the following things -

• Length of rectangle (here, ABCD) = 16 cm
• Area of rectangle = 192 cm²

Now, here comes another question,

What we're gonna have to find? (o o?_

• (Length of) breadth of given (ABCD for this question)
• Length of diagonal of Rectangle

& here is the actual solution (^-^;)

Solution:

So, if we wish to follow correct order for solving this question, we have to start with calculating, length of Breadth

$\blue{\because Area of rectangle is = Length \times \: \: Breadth}\\\implies 192cm^{2} = 16 cm \times BC (= breadth, here)\\\implies BC \:\: = \mathrm{\frac{192cm^{2}}{16cm}}\\ \implies BC \: \: = \: \: 12cm$

Moving to next step,

As we know, sides of rectangle are perpendicular to each other,

$\blue{\mathrm{So, \: according \: to \: Pythagoras' \: Theorem}} - \\ Length^{2} \: \: + \: \: Breadth^{2} \: \: = \: \: Diagonal^{2} \\\implies (16cm)^2 \: + \: (12cm)^2 = \: \: BD^{2}\\ \implies BD^2 = \: \: 256cm^2 + 144cm^2 \\\implies BD^2 = \: \: 400cm^2 \\\implies \mathrm {\sqrt{BD^{2}} = \sqrt{400cm^2}}\\\implies BD = 20cm (= Diagonal)$

Check, if I made it clear o((*^▽^*))o

Happy Learning! :D