Question

The length and breadth of a rectangle are (5x+3) cm and (3x-2) cm. If the area of rectangle is 230 sq. cm, find the values of x.

Answer 1

$\large\underline{\sf{Given- }}$

Length of a rectangle, L = 5x + 3 cm

Breadth of a rectangle, B = 3x - 2 cm

Area of rectangle, A = 230 sq. cm.

We know,

$\rm \: Area_{(rectangle)} = Length \times Breadth$

So, on substituting the values, we get

$\rm \: 230 = (5x + 3)(3x - 2)$

$\rm \: 15 {x}^{2} - 10x + 9x - 6 = 230$

$\rm \: 15 {x}^{2} - x - 6 = 230$

$\rm \: 15 {x}^{2} - x - 6 - 230 = 0$

$\rm \: 15 {x}^{2} - x - 236 = 0$

So, on splitting the middle terms, we get

$\rm \: 15 {x}^{2} - 60x + 59x - 236 = 0$

$\rm \: 15x(x - 4) + 59(x - 4) = 0$

$\rm \: (x - 4)(15x + 59) = 0$

$\rm\implies \:x = 4 \: \: or \: \: x = - \dfrac{59}{15} \: \{rejected \: as \: x > 0 \}$

So,

$\rm\implies \:x = 4 \:$

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$\begin{gathered}\begin{gathered}\boxed{\begin {array}{cc}\\ \dag\quad \Large\underline{\bf Formulas\:of\:Areas:-}\\ \\ \star\sf Square=(side)^2\\ \\ \star\sf Rectangle=Length\times Breadth \\\\ \star\sf Triangle=\dfrac{1}{2}\times Base\times Height \\\\ \star \sf Scalene\triangle=\sqrt {s (s-a)(s-b)(s-c)}\\ \\ \star \sf Rhombus =\dfrac {1}{2}\times d_1\times d_2 \\\\ \star\sf Rhombus =\:\dfrac {1}{2}d\sqrt {4a^2-d^2}\\ \\ \star\sf Parallelogram =Base\times Height\\\\ \star\sf Trapezium =\dfrac {1}{2}(a+b)\times Height \\ \\ \star\sf Equilateral\:Triangle=\dfrac {\sqrt{3}}{4}(side)^2\end {array}}\end{gathered}\end{gathered}$

Answer 2

Given :-

The length and breadth of a rectangle are (5x+3) cm and (3x-2) cm. If the area of rectangle is 230 sq. cm

To Find :-

x

Solution :-

We know that

Area = lb

230 = (5x + 3)(3x - 2)

230 = (5x × 3x) + (3 × 3x) - (2 × 5x) - (2 × 3)

230 = 15x² + 9x - 10x - 6

230 = 15x² - x - 6

230 + 6 = 15x² - x

236 = 15x² - x

15x² - x - 236 = 0

15x² - (-59x + 60x) - 236

15x² + 59x - 60x - 236 = 0

x(15x + 59) - 4(15x + 59) = 0

(x - 4)(15x + 59) = 0

x = 4 & -59/15

As length can't be negative

x = 4