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

$A$ triangle whose sides measure ............ is a right-angled triangle.

The areas of two similar triangles are $9$ and $16$. Then,the ratio of their corresponding sides is...............

The areas of two similar triangles are $25$ and $16$. Then,the ratio of their perimeters is $\ldots \ldots \ldots \ldots$

.......... are not the measures of sides of a right-angled triangle.

$\Delta ABC \sim \Delta PQR$ for the correspondence $ABC \leftrightarrow PQR$. If $AB : PQ = 3 : 4$ and the perimeter of $\Delta PQR$ is $24$,find the perimeter of $\Delta ABC$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module