Question

perimeter of rectangle whose length = 40, diagonal = 41

Answer 1

Answer:

length of rectangle=40

Diagonal of rectangle=41

Using the formulas

P=2(l+w)

$d = \sqrt{ {w}^{2} + {l}^{2} }$

Solving forP

$p = 2l + 2 \sqrt{ {d}^{2} - {l}^{2} }$

$2 \times 40 + 2 \sqrt{ {41}^{2} - {40}^{2} }$

$= 98$

Answer 2

$\huge{\mathcal{\blue{Answer}}}$

Perimeter:

$\large{\blue{2(L+b)}}$

First the breath of the rectangle is

$(d) {}^{2} = l {}^{2} + b {}^{2} \\ (41) {}^{2} = (40) {}^{2} + b {}^{2} \\ 1681 = 1600 + b {}^{2} \\ 1681 - 1600 = b {}^{2} \\ 81 = b {}^{2} \\ \sqrt{81} = b$

$\huge{\mathcal{\blue{Breath=9}}}$

Now permeter ✔

$\large{2(40cm+9cm}$

$\large{2(49cm}$

$\huge{\mathcal{\green{\boxed{98cm}}}}$