Question

find the diagonal of a rectangle whose length is 16cm and areas is 192sq.cm
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Answer 1

➤ Given :-

Length of the rectangle = 16 cm

Area of the rectangle = 192² cm

➤ To Find :-

Diagonal of the rectangle

➤ Solution :-

We know that,

Area of rectangle = Length × Breadth

$\sf 192 = 16 × b$

$\sf b = 192/16$

$\sf b = 12 \: \: cm$

Using Pythagoras theorem, .i.e.,

$\boxed{\bf AB² = BC² + AC²}$

According to the question

BC = 12cm

AC = 16cm

$:\implies\sf AB² = 12² + 16²$

$: \implies \sf AB = \sqrt{{12}^{2} + {16}^{2}}$

$: \implies \sf AB = \sqrt{144 + 256}$

$: \implies \sf AB = \sqrt{400} = 20 \: \: cm$

$\Huge\therefore$The diagonal of rectangle is $20$ cm

Answer 2

$\huge{\mathbb{\red{ANSWER:-}}}$

Given :-

$\sf{For \: a \: rectangle:-}$

$\sf{length(L) = 16 \: cm}$

$\sf{Area = 192 \: cm^2}$

To Find out :-

$\sf{diagonal \: of \: this \: rectangle = ? }$

Using Formula :-

$\sf{Area \: of \: rectangle = length\times breadth}$

$\sf{Diagonal =\sqrt{L^2 + b^2}}$

Solution :-

$\sf{Area \: of \: rectangle = 192 \: cm^2}$

$\sf{Length\times breadth = 192}$

$\sf{16\times breadth = 192}$

$\sf{breadth(b)=\dfrac{192}{16}}$

$\sf{breadth(b)=12 \: cm}$

$\sf{Now}$

$\sf{Diagonal(d)=\sqrt{L^2 + b^2}}$

$\sf{d =\sqrt{16^2 + 12^2}}$

$\sf{d =\sqrt{256 + 144}}$

$\sf{d =\sqrt{400}}$

$\sf{d = 20 \: cm}$

Result :-

$\sf{Diagonal \: of \: this \: rectangle=20 \: cm}$