Question

find the area of the Rhombus for which one side has length 10 cm and the diagonal is difference by 4 cm is​

Answer 1

AC = 16cm, BD = 12 cm

Step-by-step explanation:

Given,Side = 10 cm

Difference of diagonal = 4 cm

⇒If AC = x,BD=x+4

⇒AO=$\frac{x}{2}$

In a rhombus ABCD,if AC and BD are the diagonals which intersect at O.

Diagonals of a rhombus intersect each other at 90°.

In ΔAOB

∠AOB = 90°

By applying Pythagoras Theorem,

$(AB)^2 = (AO)^2+(BO)^2$

$(10)^2= (\frac{x}{2})^2+(\frac{x+4}{2})^2$

$100=\frac{x^2}{4}+(\frac{x+4}{2})^2$

$100=\frac{x^2}{4}+\frac{x^2+16+8x}{4}$

$100= 2\frac{x^2}{4}+4+\frac{8x}{4}$

$96=\frac{x^2}{2}+2x$

By Pythagorean triplets

(6,8,10)

Hence,AO=8 cm

AC = $2\times 8$

     =16 cm

BO = 6cm

BD=12cm

Since difference of AC and BD = 16 cm-12cm

                                                    = 4 cm.