Diagonal of a rectangle is 25 cm and length is 20 cm find the area of

1.	Diagonal of a rectangle is 25 cm and length is 20 cm find the area of rectangle
Answer» $\huge\mathbb{\underline{SOLUTION:-}}$ $\sf\underline{\purple{\:\:\:\:Given\:\:\:\:}}$ $\sf\bullet diagonal \:of\: Rectangle=25cm\\ \\ \sf\bullet <klux>LENGTH</klux>\:of\: Rectangle=20cm$ Now by Pythagoras $\boxed{\sf{Hypotenuse {}^{2}=base{}^{2}+perpendicular {}^{2}}}$ $\sf\implies (25){}^{2}=(20){}^{2}+b{}^{2}\\ \\ \sf\implies 625=400+b{}^{2}\\ \\ \sf\implies 625-400=b{}^{2}\\ \\ \sf\implies 225=b{}^{2}\\ \\ \sf\implies \sqrt{225}=b\\ \\ \sf\implies b=15cm$ Now the AREA of rectangle whose Length =20cm Breadth=15cm $\boxed{\sf{Area\:of\: rectangle=length\times breadth}}$ $\sf\implies Area\:of\: rectangle=15\times 20\\ \\ \sf\implies Area=300cm{}^{2}$ $\boxed{\sf{\red{Area\:of\: rectangle=300cm{}^{2}}}}$

Diagonal of a rectangle is 25 cm and length is 20 cm find the area of rectangle

Answer»

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

$\sf\underline{\purple{\:\:\:\:Given\:\:\:\:}}$

$\sf\bullet diagonal \:of\: Rectangle=25cm\\ \\ \sf\bullet <klux>LENGTH</klux>\:of\: Rectangle=20cm$

Now by Pythagoras

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

$\sf\implies (25){}^{2}=(20){}^{2}+b{}^{2}\\ \\ \sf\implies 625=400+b{}^{2}\\ \\ \sf\implies 625-400=b{}^{2}\\ \\ \sf\implies 225=b{}^{2}\\ \\ \sf\implies \sqrt{225}=b\\ \\ \sf\implies b=15cm$