Skip to content
doubtly

Q.3 (A)(2) Complete the following activity to prove that the sum of squares of diagonals of a rhombus is equal to the sum of the squares of the sides.[Figure: rhombus PQRS with diagonals PR and QS intersecting at T.]Given : PQRS is a rhombus. Diagona…

Q.3 (A)(2) Complete the following activity to prove that the sum of squares of diagonals of a rhombus is equal to the sum of the squares of the sides.

[Figure: rhombus PQRS with diagonals PR and QS intersecting at T.]

Given : PQRS is a rhombus. Diagonals PR and SQ intersect each other at point T.
To prove : PS2 + SR2 + QR2 + PQ2 = PR2 + QS2

Activity :

Diagonals of a rhombus bisect each other.
In Δ PQS, PT is the median and in Δ QRS, RT is the median.
∴ by Apollonius theorem,
PQ2 + PS2 = ____ + 2QT2 ................. (I)
QR2 + SR2 = ____ + 2QT2 ................. (II)
adding (I) and (II),
PQ2 + PS2 + QR2 + SR2 = 2 (PT2 + ____) + 4QT2
= 2 (PT2 + ____) + 4QT2 .............. (RT = PT)
= 4PT2 + 4QT2
= (____)2 + (2QT)2
∴ PQ2 + PS2 + QR2 + SR2 = PR2 + ____ .

1 answer

Best answer

Answer: the blanks are 2PT², 2RT², RT², PT², 2PT and QS².

Diagonals of a rhombus bisect each other, so T is the midpoint of PR and of QS (PT = RT, QT = ST).

Apollonius theorem in Δ PQS (median PT): PQ² + PS² = 2PT² + 2QT² ... (I)

In Δ QRS (median RT): QR² + SR² = 2RT² + 2QT² ... (II)

Adding: PQ² + PS² + QR² + SR² = 2(PT² + RT²) + 4QT² = 2(PT² + PT²) + 4QT² = 4PT² + 4QT² = (2PT)² + (2QT)².

Since 2PT = PR and 2QT = QS, PQ² + PS² + QR² + SR² = PR² + QS². Hence proved.

Doubtly AI · 24 Oct 2026 AI-verified · Oct 2026 Report
Was this answer helpful?

Your answer

Log in or create a free account to answer this question.