Question

Each side of a rhombus is 15 cm. If the length of one of its diagonals is 18 cm, then what is the length of the other diagonal?

Answer 1

Answer:

That's the answer to the question.

Mark as the brainliest answer.

Thank you

Answer 2

$\huge{\mathbb{\red{ANSWER:-}}}$

In a rhombus ABCD :-

Each side is equal.

$AB=BC=CD=DA=15 \: cm$

$diagonal(AC)=18 \: cm$

$second \: diagonal \: is \: BD.$

both diagonals perpendicular to each other & intersect on O.

$So , \: OA = OC =\frac{AC}{2}=9 \: cm$

$Also , \: OB = OD.$

$Here , \: \angle{AOB}=90$

In right angled triangle AOB:-

$From \: pythogoras \: theorem-$

$AB^2 = OA^2 + OB^2$

$(OB)^2 = (AB)^2 - (OA)^2$

$OB =\sqrt{15^2 - 9^2}$

$\: \: =\sqrt{225 - 81}$

$\: \: =\sqrt{144}$

$\: \: = 12 \: cm$

$So , \: OB = OD = 12 \: cm$

$Second \: diagonal (BD)-$

$= OB + OD$

$= 2(OB)$

$= 2 \times 12$

$BD = 24 \: cm$

Second diagonal of rhombus ABCD

is 24 cm.

other related Formulas :-

$For \: rhombus \: ABCD-$

$Area \: of \: rhombus=\dfrac{1}{2} \times AC\times BD$

$perimeter \: of \: rhombus=4a$

$where -$

$AC \: and \: BD \: are \: diagonals.$

$a = each \: side$