Question

The area of a rectangle is 21 cm² and its perimeter is 20 m. The length and breadth of the rectangle are to be found.

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Answer 1

Question

The area of a rectangle is 21 cm² and its perimeter is 20 cm. The length and breadth of the rectangle are to be found.

More Information

Surface area of cuboid = 2(lb+lh+bh)

Surface area of cube = 6 (length of an edge)²

Area of four walls = 2(l+b)h

Lateral surface area of cuboid = 2(l+b)h

Lateral surface area of cube = 4 (length of an edge)²

Answer 2

Answer:

$\huge\mathfrak\red{\bold{\underline{☯︎{Lεศɢนε \: σʄ \: Mıŋɖร}☯︎ }}}$

Given :

• The area of rectangle is 21 cm² .
• Its perimeter is 20 cm.

To find :

• Length and breadth of the rectangle.

$\huge\underbrace\mathfrak\color{lime}{ †\: Solution:–}$

Consider ,

• Length of rectangle = x cm
• Breadth of rectangle = y cm

Formula Used :-

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

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

According to the 1st condition :-

• Area of the rectangle is 21 cm.

→xy=21..............(i)

According to the 2nd condition :-

• Perimeter of rectangle is 20 cm.

→2(x+y)=20

→x+y=10

→x=10−y.................(ii)

• Now take eq(i) and put x=10-y from eq(ii) .

xy=21

→(10−y)y=21

→10y−y²=21

→−y² +10y−21=0

→y²−10y+21=0

→y ²−(7+3)y+21=0

→y²−7y−3y+21=0

→y(y−7)−3(y−7)=0

→(y−7)(y−3)=0

Either,

• y - 7 = 0

→ y = 7

Or,

• y - 3 = 0

→ y = 3

• Now put y = 3 in eq (ii).

x = 10-y

→ x = 10-3

→ x = 7

Therefore ,

$\huge\pink{ \underline{ \boxed{ \sf{Length = 7 cm}}}}$

$\huge\pink{ \underline{ \boxed{ \sf{Breadth=3cm}}}}$

$\huge {\red {\ddot {\smile}}} {\huge {\pink {\ddot \smile}}} {\huge {\purple{\ddot {\smile}}}} {\huge {\orange{\ddot {\smile}}}}$

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