Prove that the sum of the squares of the diagonals of parallelogram is equal to the sum of the squares of its sides
Solution:
Let's draw a parallelogram ABCD as shown below,
AB = CD, AD = BC [Since opposite sides of a parallelogram are equal]
AC and BD are the diagonals of the parallelogram ABCD
Draw AE ⊥ CD, DF ⊥ AB
We have to prove that, AC² + BD² = AB² + BC² + CD² + AD²
EA = DF (Perpendiculars drawn between same parallel lines)
According to Pythagoras theorem,
In ΔAEC
AC² = AE² + EC²
AC² = AE² + (ED + DC)² [Since EC = ED + DC]
AC² = AE² + DE² + DC² + 2DE.DC [Since (a + b)² = a² + b² + 2ab]
AC² = AD² + DC² + 2DE.DC .....(i) [Since, AD² = AE² + DE² in ΔAED]
In ΔDFB
BD² = DF² + BF²
BD² = DF² + (AB - AF)² [Since BF = AB = AF]
BD² = DF² + AB² + AF² - 2AB.AF [Since (a - b)² = a² + b² - 2ab]
BD² = AD² + AB² - 2AB.AF .....(ii) [Since, AD² = DF² + AF² in ΔAFD]
Adding (i) and (ii)
AC² + BD² = AD² + DC² + 2DE.DC + AD² + AB² - 2AB.AF
AC² + BD² = BC² + DC² + 2AB.AF + AD² + AB² - 2AB.AF [Since AD = BC and DE = AF, CD = AB]
Thus, AC² + BD² = AB² + BC² + CD² + AD²
☛ Check: NCERT Solutions for Class 10 Maths Chapter 6
Video Solution:
Prove that the sum of the squares of the diagonals of parallelogram is equal to the sum of the squares of its sides
NCERT Class 10 Maths Solutions Chapter 6 Exercise 6.6 Question 6
Summary:
Hence proved that the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides.
☛ Related Questions:
- In Fig. 6.61, two chords AB and CD intersect each other at the point P. Prove that: (i) ΔAPC ~ ΔDPB (ii) AP.PB = CP.DP.
- In Fig. 6.62, two chords AB and CD of a circle intersect each other at the point P (when produced) outside the circle. Prove that (i) ΔPAC ~ ΔPDB (ii) PA.PB = PC.PD.
- In Fig. 6.63, D is a point on side BC of ∆ABC such that BD/CD = BA/CA. Prove that: AD is the bisector of ∠BAC.
- Nazima is fly fishing in a stream. The tip of her fishing rod is 1.8 m above the surface of the water and the fly at the end of the string rests on the water 3.6 m away and 2.4 m from a point directly under the tip of the rod. Assuming that her string (from the tip of her rod to the fly) is taut, how much string does she have out (see Fig. 6.64)? If she pulls in the string at the rate of 5 cm per second, what will be the horizontal distance of the fly from her after 12 seconds?