Question

prove that the sum of the square of the sides of a rhombus is equal to the sum of the square of each diagonals​

Answer 1

Answer Expert Verified

Given :- A rhombus ABCD whose diagonals AC and BD intersect at O. To Prove :- ( AB² + BC² + CD² + DA² ) = ( AC² + BD² ). ➡ We know that the diagonals of a rhombus bisect each other at right angles. ... [ In a rhombus , all sides are equal ] .

Answer 2

Answer:

rhombus ABCD with diagonal AC and BD intersecting at O

sum of squares of all sides is equal to sum of squares of its diagonals

= AB² + BC² + CD² + AD² = AC² + BD²

since sides of rhombus are equal,

AB=BC=CD=AD

Step-by-step explanation:hope this help you please mark me brainliest :)