Pythagoras Theorem and it's important questions

Chapter 6 Class 10 Triangles
Serial order wise

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### Transcript

Question7 Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals. Given:- Rhombus ABCD with diagonals AC & BD intersecting at O To prove: Sum of square of all sides = Sum of the squares of it s diagonals AB2 + BC2 + CD2 + AD2 = AC2 + BD2 Proof: Since sides of a rhombus are equal AB = BC = CD = AD We know that, diagonals of a rhombus bisect each other a right angles . Therefore, = = = =90 Also AO = CO = 1/2 AC & BO = DO = 1/2 BD Now , AOB is a right angle triangle Using Pythagoras theorem (Hypotenuse)2 = (Height)2 + (Base)2 (AB)2 = (OA)2 + (OB)2 (AB)2 = (1/2 )2+(1/2 )2 (AB)2 = 2/4+ 2/4 (AB)2 = ( 2 + 2)/4 4AB2 = AC2 + BD2 AB2 + AB2 + AB2 + AB2 = AC2 + BD2 AB2 + BC2 + CD2 + AD2 = AC2 + BD2 Hence proved

#### Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.