Question

The length of one diagonal of a rhombus is less than 4 cm .the area of the rhombus is 30sqcm. Find the length of rhombus

Answer 1

Answer:

Step-by-step explanation:

Let the length of second diagonal of a rhombus be ‘x’ cm.

∴ the length of first diagonal of a rhombus = x – 4

Area of rhombus = ½ ×  Product of length of diagonals

Area of rhombus = ½ (x) (x – 4)

According to given condition,

½ (x) (x – 4) = 30

∴ x ( x – 4) = 60

∴ x2 – 4x = 60

∴ x2 – 4x – 60 = 0

∴ x2 – 10x + 6x – 60 = 0

∴ x(x – 10) + 6(x – 10) = 0

∴ (x – 10) (x + 6 ) = 0

∴ x – 10 = 0  or x + 6 = 0

∴ x = 10  or x = - 6

∵ The length of diagonal of the rhombus cannot be negative.

∴ x = 10

∴ x – 4 = 10 – 4 = 6

∴ The length of first diagonal of a rhombus is 6 cm and second  diagonal is 10 cm.

Answer 2

$\mathfrak{Answer:}$

= 10 cm and 6 cm.

$\mathfrak{Step-by-Step\:Explanation:}$

$\underline{\bold{Given\:in\:the\:Question:}}$

• Area of the rhombus = 30 cm².
• The difference of diagonals = 4 cm.

$\tt{To\quad Find:}$

• Length of both diagonals.

$\bold{Solution:}$

Let the first diagonal be x cm.

∴ Second one =( x - 4 ) cm.

$\mathfrak{According\:to\:question}:\\\\\\\boxed{\bold{Area\:of\:Rhombus=\dfrac{1}{2}\times d_1\times d_2}}\\\\\\\implies\tt{\dfrac{1}{2}\times x\times(x-4)=30}\\\\\\\implies\tt{x^2-4x=60}\\\\\\\implies\tt{x^2-4x-60=0}\\\\\\\implies\tt{x^2+6x-10x-60=0}\\\\\\\implies\tt{x(x+6)-10(x+6)=0}\\\\\\\implies\tt{(x+6)(x-10)=0}\\\\\\\therefore\tt{\quad x=-6\:or\:10.}\\\\\\\underline{\bold{Possible\:value\:of\:x=10.}}$

$\boxed{\boxed{\bold{First\:diagonal=10\:cm\quad\&\quad Second\:diagonal=6\:cm.}}}$