Question

calculate the area of a rectangle whose area is 25x²-35x+12​

Answer 1

Correct Question :

Give possible expression for the length and the breadth of rectangle, whose area is 25x² - 35x + 12

Solution :

Area = 25x² - 35x + 12

We'll solve this by splitting the middle term :

Factorisation of polynomial ax² + bx + c by splitting middle term is as :

Let its factor be (px + q) and (rx + s). Then,

★ ax² + bx + c = (px + q) (rx + s) = pr x² + (ps + qr) x + ps

After comparing the coefficient of x², we'll get a = pr

then comparing the coefficient of x, we'll get b = ps + qr

and comparing the coefficient of constant term, we'll get c = qs.

It shows that b is the sum of two numbers ps and qr, whose product is (ps)(qr) = (pr)(qs) = ac

Therefore, to Factorise ax² + bx + c, we'll have to write b as the sum of two number whose product is ac.

Now,

→ 25x² - 35x + 12

→ 25x² - (20x + 15x) + 12

→ 25x² - 20x - 15x + 12

→ 5x(5x - 4) - 3(5x - 4)

→ (5x - 3) (5x - 4)

Possible dimensions are :

$\sf \: \bullet \: \: Length = 5x - 3$

$\sf \: \bullet \: \: Breadth = 5x - 4$

Answer 2

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Question:-

calculate the area of a rectangle whose area is 25x²-35x+12

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Given

Area of Rectangle = 25x² – 35x + 12

$\large \blue{ \boxed{ \red{ \bf To \:find}}}$

• Perimeter of the Rectangle.

$\large \blue{ \boxed{ \red{ \bf Solution:-}}}$

$\blacksquare \: \scriptsize\boxed {\bold{Area \: of \: Rectangle = Length \times Breadth}}$

So, By Factoring 25x² – 35x + 12, the Length and Breadth can be obtained.

$\large \leadsto \tt{25 {x}^{2} - 35x + 12 = 0}$

$\large \leadsto \tt{25 {x}^{2} - 15x - 20x + 12 = 0}$

$\large \leadsto \tt{5x(5x - 3) - 4(5x - 3) = 0}$

$\pink{\large \leadsto \tt{(5x - 3)(5x - 4) = 0}}$

So, the Length and Breadth are (5x – 3)(5x – 4).

$\blacksquare \: \scriptsize\boxed {\bold{Perimeter = 2(Length + Breadth)}}$

$\large \leadsto \tt{Perimeter= 2(5x - 3 + 5x - 4)}$

$\orange{\large \leadsto \boxed{\tt{Perimeter= 2(10x - 7)}}}$

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