In rectangle ABCD,AB^{2} + BC^{2} + CD^{2} + | vedclass

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

Question

(A) In rectangle $ABCD$,we know that opposite sides are equal,so $CD = AB$ and $DA = BC$.Also,$\angle B = 90^{\circ}$.Given the equation: $AB^{2} + BC

Vedclass · Accepted Answer

$13$

Vedclass · Answer

(A) In rectangle $ABCD$,we know that opposite sides are equal,so $CD = AB$ and $DA = BC$.Also,$\angle B = 90^{\circ}$.Given the equation: $AB^{2} + BC^{2} + CD^{2} + DA^{2} = 338$.Substituting the equal sides: $AB^{2} + BC^{2} + AB^{2} + BC^{2} = 338$.$2(AB^{2} + BC^{2}) = 338$.$AB^{2} + BC^{2} = 169$.In $	riangle ABC$,by Pythagoras theorem: $AC^{2} = AB^{2} + BC^{2}$.Therefore,$AC^{2} = 169$.$AC = \sqrt{169} = 13$.

In rectangle $ABCD$,$AB^{2} + BC^{2} + CD^{2} + DA^{2} = 338$,then $AC = \ldots$

Similar Questions

In the figure,$PQR$ is a right triangle right-angled at $Q$ and $QS \perp PR$. If $PQ = 6 \, cm$ and $PS = 4 \, cm$,find $QS$,$RS$,and $QR$.

$\Delta ABC \sim \Delta PQR$ for the correspondence $ABC \leftrightarrow PQR$. If $AB = 3, BC = 4, AC = 5$ and $QR = 6$,find $PQ$ and $PR$.

The length of a diagonal of a square is $6$. Then, the length of its sides is $\ldots \ldots \ldots$ (in $\sqrt{2}$)

The perimeter of square $ABCD$ is $48$. Find the length of its diagonal $\overline{AC}$. (in $\sqrt{2}$)

In $\Delta ABC$,$m\angle B = 90^{\circ}$. $D$ is the midpoint of $\overline{BC}$ and $F$ is the midpoint of $\overline{AB}$. Prove that $AD^{2} + CF^{2} = \frac{5}{4} AC^{2}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module