The length of a diagonal of a square is $5 \sqrt{2}$. Then,the length of its sides is............

  • A
    $5 \sqrt{2}$
  • B
    $10 \sqrt{2}$
  • C
    $10$
  • D
    $5$

Explore More

Similar Questions

$\square ABCD$ is a trapezium in which $\overline{ AD } \| \overline{ BC }, \overline{ AC } \cap \overline{ BD }=\{ P \} .$ If $PD =9, PA =5$ and $PB =7.2,$ then $AC =\ldots \ldots \ldots \ldots$

Difficult
View Solution

For the correspondence $ABC \leftrightarrow XYZ$ being a similarity, which of the following correctly matches the information in Part $I$ and Part $II$?
Part $I$ Part $II$
$1.$ In $\Delta ABC$ and $\Delta XYZ, \frac{AB}{XY} = \frac{BC}{YZ}$ and $\angle B \cong \angle Y$ $a. SSS$ condition
$2.$ In $\Delta ABC$ and $\Delta XYZ, \angle A \cong \angle X, \angle B \cong \angle Y$ and $\angle C \cong \angle Z$ $b. SAS$ condition
$3.$ In $\Delta ABC$ and $\Delta XYZ, \frac{AB}{XY} = \frac{BC}{YZ} = \frac{CA}{ZX}$ $c. AAA$ condition
- $d. \text{None of the conditions apply}$

In $\Delta XYZ$,$m\angle Y = 90^{\circ}$ and $\overline{YM}$ is an altitude to the hypotenuse $\overline{XZ}$. If $YM = 12$ and $XM = 8$,find $XZ$.

In a triangle $PQR$,$N$ is a point on $PR$ such that $QN \perp PR$. If $PN \cdot NR = QN^2$,prove that $\angle PQR = 90^{\circ}$.

Difficult
View Solution

Legs (sides other than the hypotenuse) of a right triangle are of lengths $16 \,cm$ and $8 \,cm$. Find the length of the side of the largest square that can be inscribed in the triangle. (in $cm$)

Difficult
View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo