The lengths of the diagonals of a rhombus are | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let the rhombus be $ABCD$ with diagonals $d_1 = 10$ and $d_2 = 24$.In a rhombus,the diagonals bisect each other at right angles $(90^{\circ})$.Let

Vedclass · Accepted Answer

$13$

Vedclass · Answer

(B) Let the rhombus be $ABCD$ with diagonals $d_1 = 10$ and $d_2 = 24$.In a rhombus,the diagonals bisect each other at right angles $(90^{\circ})$.Let the diagonals intersect at point $O$. Then,the segments of the diagonals are $d_1/2 = 5$ and $d_2/2 = 12$.These segments form a right-angled triangle with the side of the rhombus as the hypotenuse.Using the Pythagorean theorem: $side^2 = (d_1/2)^2 + (d_2/2)^2$.$side^2 = 5^2 + 12^2 = 25 + 144 = 169$.$side = \sqrt{169} = 13$.Therefore,the length of each side of the rhombus is $13$.

The lengths of the diagonals of a rhombus are $10$ and $24$. Then,the length of each side of the rhombus is $\ldots \ldots \ldots \ldots . .$

Similar Questions

In the figure,$DE \parallel AC$ and $DF \parallel AE$. Prove that $\frac{BF}{FE} = \frac{BE}{EC}$.

In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BM}$ is an altitude to the hypotenuse $\overline{AC}$. Then,$BM$ is the geometric mean of..............

In $\Delta PQR$,$m\angle Q = 90^{\circ}$ and $\overline{PM}$ is a median. Prove that $PR^{2} = PM^{2} + 3RM^{2}$.

In $\Delta ABC$,$m \angle B = 90^{\circ}$. If $AB = 20$ and $AC = 29$,find the perimeter of $\Delta ABC$.

In $\Delta ABC$,$m \angle B = 90^{\circ}$. If $AB = 4$ and $BC = 7.5$,find $AC$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module