Question

find the base of a parallelogram whose perimeter is [4x^(2)+10x-50]/[(x-3)(x+5)]. and one side is 5/(x-3​

Answer 1

Gɪᴠᴇɴ :-

• Perimeter of parallelogram = [4x²+10x-50]/[(x-3)(x+5)].
• One sides of parallelogram = 5/(x - 3) .

Tᴏ Fɪɴᴅ :-

• Base or Other Side of parallelogram .

Fᴏʀᴍᴜʟᴀ ᴜsᴇᴅ :-

• Perimeter of Parallelogram = 2( Base + Other Adjacent Side) .

Sᴏʟᴜᴛɪᴏɴ :-

Comparing The Perimeter now, we get,

➻ [4x²+10x-50]/[(x-3)(x+5)] = 2[ 5/(x - 3) + Base ]

Taking 2 Common From LHS Numerator,

➻2[(2x² + 5x - 25)/(x -3)(x+5)] = 2[ 5/(x - 3) + Base ]

Cancelling 2 from Both Numerator, and taking LCM in RHS part,

➻ [(2x² + 5x - 25)/(x -3)(x+5)] = [ (5 + Base*(x-3))/(x - 3) ]

Cancel (x - 3) From Both Denominator,

➻ [(2x² + 5x - 25)/(x+5)] = (5 + Base*(x-3))

Cross - Multiply,

➻ (2x² + 5x - 25) = (x + 5)((x-3)*Base + 5)

Splitting The Middle Term of LHS part,

➻(2x² + 10x - 5x - 25) = (x + 5)((x-3)*Base + 5)

➻ 2x(x + 5) - 5(x + 5) = (x + 5)((x-3)*Base + 5)

➻ (x + 5)(2x - 5) = (x + 5)((x-3)*Base + 5)

Cancel (x + 5) From Both sides

➻ (2x - 5) = ((x-3)*Base + 5)

➻ (x-3)*Base = (2x - 5) - 5

➻ Base = [(2x - 10)/(x-3)] (Ans.)

Hence, Base of The Given Parallelogram is [(2x - 10)/(x-3)].

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★★Extra Brainly Knowledge★★

✯✯ Some Properties of a Parallelogram ✯✯

→ Opposite sides are parallel by definition.

→ Opposite sides are congruent.

→ Opposite angles are congruent.

→ Consecutive angles are supplementary.

→ The diagonals bisect each other.

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Answer 2

Answer:

⋆ DIAGRAM :

$\setlength{\unitlength}{1.5cm}\begin{picture}(8,2)\thicklines\put(8.6,3){\large{A}}\put(7.7,0.9){\large{B}}\put(9.5,0.7){\sf{\large{Base}}}\put(11.1,0.9){\large{C}}\put(8,1){\line(1,0){3}}\put(11,1){\line(1,2){1}}\put(9,3){\line(3,0){3}}\put(7.7,2){\large{ \sf{}^{5}\!/{}_{(x-3)} }}\put(8,1){\line(1,2){1}}\put(12.1,3){\large{D}}\end{picture}$

$\rule{150}{1}$

$\underline{\bigstar\:\textbf{According to the Question :}}$

$:\implies\sf Perimeter_{(Parallelogram)}=2(Sum\:of\:Adjacent\:Sides)\\\\\\:\implies\sf \dfrac{4x^2+10x-50}{(x-3)(x+5)}=2(AB+BC)\\\\\\:\implies\sf \dfrac{2(2x^2+5x-25)}{(x-3)(x+5)}=2\bigg(\dfrac{5}{x-3}+Base\bigg)\\\\\\:\implies\sf \dfrac{2x^2+5x-25}{(x-3)(x+5)}=\dfrac{5}{x-3}+Base\\\\\\:\implies\sf \dfrac{2x^2+10x-5x-25}{(x-3)(x+5)}=\dfrac{5}{x-3}+ Base\\\\\\:\implies\sf \dfrac{2x(x+5)-5(x+5)}{(x-3)(x+5)}=\dfrac{5}{x-3}+ Base\\\\\\:\implies\sf \dfrac{(2x-5)(x+5)}{(x-3)(x+5)}=\dfrac{5}{x-3}+ Base\\\\\\:\implies\sf \dfrac{2x-5}{x-3}=\dfrac{5}{x-3}+ Base\\\\\\:\implies\sf \dfrac{2x-5}{x-3}-\dfrac{5}{x-3}=Base\\\\\\:\implies\sf \dfrac{2x-5-5}{x-3}=Base\\\\\\:\implies\underline{\boxed{\textsf{ \textbf{Base = \dfrac{\text{2x - 10}}{\text{x - 3}} }}}}$