square ABCD is a rectangle. If AB^{2} + BC^{2 | vedclass

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

Question

(D) In a rectangle $\square ABCD$,opposite sides are equal,so $AB = CD$ and $BC = DA$.Given the equation: $AB^{2} + BC^{2} + CD^{2} + DA^{2} = 128$.Su

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(D) In a rectangle $\square ABCD$,opposite sides are equal,so $AB = CD$ and $BC = DA$.Given the equation: $AB^{2} + BC^{2} + CD^{2} + DA^{2} = 128$.Substituting the equal sides,we get: $AB^{2} + BC^{2} + AB^{2} + BC^{2} = 128$.This simplifies to: $2(AB^{2} + BC^{2}) = 128$.Dividing by $2$,we get: $AB^{2} + BC^{2} = 64$.In the right-angled triangle $	riangle ABC$,by the Pythagorean theorem: $AC^{2} = AB^{2} + BC^{2}$.Therefore,$AC^{2} = 64$.Taking the square root of both sides,$AC = \sqrt{64} = 8$.

$\square ABCD$ is a rectangle. If $AB^{2} + BC^{2} + CD^{2} + DA^{2} = 128$,find the length of diagonal $\overline{AC}$.

Similar Questions

In $\Delta ABC$,$AB + BC = 23$,$BC + AC = 32$,and $AB + AC = 25$. Show that $\Delta ABC$ is a right-angled triangle.

In $\Delta ABC$,$\overline{AD}$,$\overline{BE}$ and $\overline{CF}$ are medians. Prove that $3(AB^2 + BC^2 + AC^2) = 4(AD^2 + BE^2 + CF^2)$.

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

It is given that $\triangle ABC \sim \triangle DFE$,$\angle A = 30^{\circ}$,$\angle C = 50^{\circ}$,$AB = 5 \, cm$,$AC = 8 \, cm$ and $DF = 7.5 \, cm$. Then,the following is true:

In $\Delta PQR$,$m\angle Q = 90^{\circ}$,$PR = a^{2} + b^{2}$ and $QR = a^{2} - b^{2}$. Then,$PQ = \ldots$ $(a, b \in R^{+}; a > b)$

Vedclass Test Series

Exam Paper Generator

Online Exam Module