Question

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. 

Given: PQRS is a rhombus. Diagonals PR and SQ intersect each other at point T. 
To prove: PS\(^2\) + SR\(^2\) + QR\(^2\) + PQ\(^2\) = PR\(^2\) + QS\(^2\) 
Activity: Diagonals of a rhombus bisect each other. 
In \(\triangle\)PQS, PT is the median and in \(\triangle\)QRS, RT is the median. 
\(\therefore\) by Apollonius theorem, 
 

\[\begin{aligned} PQ^2 + PS^2 &= \boxed{\phantom{X}} + 2QT^2 \quad \dots \text{(I)} \\ QR^2 + SR^2 &= \boxed{\phantom{X}} + 2QT^2 \quad \dots \text{(II)} \\ \text{Adding (I) and (II),} \quad PQ^2 + PS^2 + QR^2 + SR^2 &= 2(PT^2 + \boxed{\phantom{X}}) + 4QT^2 \\ &= 2(PT^2 + \boxed{\phantom{X}}) + 4QT^2 \quad (\text{RT = PT}) \\ &= 4PT^2 + 4QT^2 \\ &= (\boxed{\phantom{X}})^2 + (2QT)^2 \\ \therefore \quad PQ^2 + PS^2 + QR^2 + SR^2 &= PR^2 + \boxed{\phantom{X}} \\ \end{aligned}\]

Hint:

An easier way to prove this property is using the Pythagorean theorem. Since the diagonals of a rhombus bisect each other at right angles, consider any one of the four right-angled triangles (e.g., \(\triangle\)PTQ). Here, \(PQ^2 = PT^2 + QT^2\). Since all four sides are equal, the sum of squares of sides is \(4PQ^2\). Substituting gives \(4(PT^2 + QT^2) = 4PT^2 + 4QT^2 = (2PT)^2 + (2QT)^2 = PR^2 + QS^2\).

Answer 1

Step 1: Understanding the Concept: 
This activity uses Apollonius's theorem on triangles formed by the diagonals of a rhombus to prove a key property of the rhombus. Apollonius's theorem relates the length of a median of a triangle to the lengths of its sides. 
 

Step 2: Key Formula or Approach: 
Apollonius's Theorem: In a triangle ABC, if AD is a median to side BC, then \( AB^2 + AC^2 = 2(AD^2 + BD^2) \). 
Properties of a Rhombus: 1. All sides are equal. 2. Diagonals bisect each other at right angles. 
 

Step 3: Detailed Explanation: 
Here is the completed activity with the blanks filled in. 
Given: PQRS is a rhombus. Diagonals PR and SQ intersect each other at point T. 
To prove: PS\(^2\) + SR\(^2\) + QR\(^2\) + PQ\(^2\) = PR\(^2\) + QS\(^2\) 
Activity: Diagonals of a rhombus bisect each other. So T is the midpoint of PR and SQ. 
In \(\triangle\)PQS, PT is the median to side QS. In \(\triangle\)QRS, RT is the median to side QS. (The activity incorrectly states it's median to PR, but from the equations it's clear it's the median to QS). 
\(\therefore\) by Apollonius theorem, 
 

\[\begin{aligned} \\ \text{In } \triangle PQS \text{ with median } PT: \quad PQ^2 + PS^2 &= \boxed{2PT^2} + 2QT^2 \quad \dots \text{(I)} [6pt] \\ \text{In } \triangle QRS \text{ with median } RT: \quad QR^2 + SR^2 &= \boxed{2RT^2} + 2ST^2 [4pt] \\ \text{Since } T \text{ is the midpoint of } SQ, \text{ we have } ST = QT. [4pt] \\ \Rightarrow \quad QR^2 + SR^2 &= 2RT^2 + 2QT^2 \quad \dots \text{(II)} [8pt] \\ \text{Adding (I) and (II):} [4pt] \\ PQ^2 + PS^2 + QR^2 + SR^2 &= 2PT^2 + 2QT^2 + 2RT^2 + 2QT^2 [4pt] \\ &= 2(PT^2 + \boxed{RT^2}) + 4QT^2 [4pt] \\ \text{Since diagonals bisect, } T \text{ is the midpoint of } PR, \text{ so } RT = PT. [4pt] \\ \Rightarrow PQ^2 + PS^2 + QR^2 + SR^2 &= 2(PT^2 + \boxed{PT^2}) + 4QT^2 [4pt] \\ &= 2(2PT^2) + 4QT^2 [4pt] \\ &= 4PT^2 + 4QT^2 [4pt] \\ &= (2PT)^2 + (2QT)^2 [4pt] \\ &= (\boxed{2PT})^2 + (2QT)^2 [6pt] \\ \text{Since } 2PT = PR \text{ and } 2QT = QS, [4pt] \\ \therefore \quad PQ^2 + PS^2 + QR^2 + SR^2 &= PR^2 + \boxed{QS^2}. \\ \end{aligned}\]
 

Step 4: Final Answer: 
The activity is completed, proving that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.