The length of a diagonal of a square is 12. T | vedclass

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

Question

(B) Let the side of the square be $a$.We know that the diagonal $d$ of a square is given by the formula $d = a \sqrt{2}$.Given that the diagonal $d =

Vedclass · Accepted Answer

$6 \sqrt{2}$

Vedclass · Answer

(B) Let the side of the square be $a$.We know that the diagonal $d$ of a square is given by the formula $d = a \sqrt{2}$.Given that the diagonal $d = 12$,we have:$12 = a \sqrt{2}$$a = \frac{12}{\sqrt{2}}$To rationalize the denominator,multiply the numerator and denominator by $\sqrt{2}$:$a = \frac{12 	imes \sqrt{2}}{\sqrt{2} 	imes \sqrt{2}}$$a = \frac{12 \sqrt{2}}{2}$$a = 6 \sqrt{2}$Therefore,the length of each side of the square is $6 \sqrt{2}$.

The length of a diagonal of a square is $12$. Then,the length of each side of the square is...........

Similar Questions

In $\Delta ABC$,the bisector of $\angle B$ intersects $\overline{AC}$ at $D$. If $\frac{AD}{DC} = \frac{3}{4}$ and $AB = 7.5$,then $BC = \ldots$

The perimeter of rhombus $ABCD$ is $116$. If $AC = 42$, find $BD$.

In $\Delta PQR$,$m \angle Q = 90^{\circ}$ and $\overline{QS}$ is an altitude. If $PS - SR = 10$ and $PQ^2 - QR^2 = 260$,then $PR = \dots$

In $\Delta ABC$,$A-M-B$,$A-N-C$ and $\overline{MN} \parallel \overline{BC}$. If $AM = 3x - 1$,$MB = 2x + 1$,$AN = 3x + 1$ and $NC = 5x + 1$,find the value of $x$.

In $\Delta ABC$ and $\Delta PQR$,$\angle A \cong \angle R$ and $\angle B \cong \angle P$. If $AB = 6$,$BC = 8$ and $PR = 9$,then $PQ = \ldots$

Vedclass Test Series

Exam Paper Generator

Online Exam Module