Mastering Important Concepts in Triangles - Exercise Question #2

Let's look at the detailed solution of the above problem



Steps 1 & 2: Understand Question and Draw Inferences
Given a quadrilateral ABCD with right angles at B and D.
We need to find if AD > BC is true.

Step 3: Analyze Statement 1 independently
Given that \(Angle CAB = Angle BCA\)

Therefore, in \(ΔABC\),
\(Angle CAB = Angle BCA = 45^o\)

Therefore, \(BC = AB\)
This doesn’t give us any relation between AD and BC.

So statement 1 is not sufficient to arrive at a unique answer.

Step 4: Analyze Statement 2 independently
Given that \(CD > AB\).
In the right angled triangle ABC, we have
\(AC^2 = AB^2 + BC^2\)
Similarly, in the right angled triangle ADC, we have
\(AC^2 = AD^2 + CD^2\)
Therefore,
\(AD^2 + CD^2 = AB^2 + BC^2\)……………. (I)

Given that CD > AB
• \(CD^2 > AB^2\)
In other words, a term in the LHS of (I) is greater than another term in the RHS of (I)
So for (I) to be true (the equality to hold true), the remaining terms on either sides of LHS and RHS should compensate for the imbalance created by \(CD^2 and AB^2\).

From (I),
\(CD^2 -AB^2 = BC^2 -AD^2\)

Let’s say \(CD^2 – AB^2 = k\)(a positive number since \(CD^2 > AB^2\))
Then
\(BC^2 – AD^2 = k\) (the same positive number).

Therefore,\(BC^2 > AD^2\)

• \(BC > AD\)

Therefore, statement 2 is sufficient to arrive at a unique answer.

Hence the correct Answer is B.