Question

if one diagonal of a rhombus is equal to one of its side, them the diagonals are in the ratio??

Answer 1

hey mate

here is your answer

ratio is root 3:2 or 2:root 3

hope it helped!

Answer 2

Hey mate!!!

$<b >Here is your answer ;</b >$

Given,
ABCD is a rhombus, in which,
Diagonals AC and BD intersect each other at O.

Also it is stated that ;

$<b >Diagonal AC = side of the Rhombus ABCD.</b >$

Let side AB be taken as $a$.

Now,
A rhombus is a special parallelogram, in which all the sides are equal.
Also, the diagonals bisect each other.

So by the above fact we can say that;

$\frac{AC} {2} = OA = OC$

That is $OA = \frac{a}{2}$

Similarly, $\frac{BD} {2} = OB = OD$ _eq1

In triangle AOB,
Angle. AOB = $90^o$

Then,

By the pythagorean theorem,

[tex ] OB^2[/tex] = $AB^2$ - $OA^2 [/tex ] = = > > [tex ] OB<SUP >2</SUP >$ =$a^2$ + $(\frac{a}{2})^2 [/tex ] = = > > [tex ] OB^2$ =$a^2 - \frac{a^2}{4} [/tex ] = = > > [tex ] OB^2 = \frac {4a^2 - a^2}{4} [/tex ] = = > > [tex ] OB^2 = \frac {3a^2}{4}$

= = > > $OB = √(\frac{3a^2}{4}$

= = > > $OB = \frac{√3}{2}a$ _eq2.

Then,

$BD = 2OB$ [from eq1]

$BD = 2 * \frac{√3}{2}a$

$BD = √3a$

Now,
Ratio of AC and BD is,

a : √3a

= 1 : √3

$<b >I hope it helps you up and you understand my answer. And I am sorry if any thing is wrong in my answer. BEST OF LUCK for your exams. If any questions, you may ask in the comments below... </b >$

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