Question

12. Find the perimeter of a rectangle whose one side
measures 20 m and the diagonal is 29 m.
(h​

Answer 1

Answer:

perimeter of rectangle=82m2

Step-by-step explanation:

one side of rectangle=20m

perimeter of rectangle=2(l+b)

diagonal of rectangle=29

By Pythagoras theorem

AC2=AB2+BC2

29square=20square+X2

881-400=X2

X2=441

x=21

perimeter of rectangle=2(l+b)

=2(20+21)

=2(41)

=82m2

Answer 2

$\huge\mathbb{\underline{SOLUTION:-}}$

$\bold{Given:-}\begin{cases}\sf{One\:Side\:of\: rectangle=20m}\\ \sf{Diagonal\:of\: rectangle=29m}\end{cases}$

$\huge\sf{\red{To\:Find}}$

• We have to find the Perimeter of rectangle

$\boxed{\sf{\pink{Perimeter\:of\: rectangle=2(length+breadth)}}}$

$\bf\underline{\underline{Step\:by\:step\: explanation:-}}$

• First find the 2nd side of rectangle by the Pythagoras theorm

$\boxed{\sf {Diagonal {}^{2}=base{}^{2}+perpendicular {}^{2}}}$

• Now the side of rectangle

• let the side be x

$\sf (29){}^{2}=(20){}^{2}+x{}^{2}\\ \\ \tt\implies 841=400+x{}^{2}\\ \\ \tt\implies 841-400=x{}^{2}\\ \\ \tt\implies 441=x{}^{2}\\ \\ \tt\implies \sqrt{441}=x\\ \\ \sf \implies 21=x\\ \\ \sf\implies or\: x=21m$

• The 2nd side of rectangle or breadth of rectangle= 21m
• Now the perimeter of rectangle

$\bold{we\:have:-}\begin{cases}\sf{length\:of\: rectangle=20m}\\ \sf{Breadth\:of\:rectangle=21m}\end{cases}$

$\sf\longrightarrow Perimeter=2(21m+20m)\\ \\ \sf\longrightarrow Perimeter=2\times 41m\\ \\ \sf\implies Perimeter=82m$

$\large\star{\boxed{\sf{\red{Perimeter\:of\: rectangle=82m}}}}$

#BAL

#answerwithquality