Question

2. ABCD is a square and P, Q are the midpoints of BC,CD respectively. If AP = a

and AQ = b. find in terms of a and b the directed line segments (i) AB (ii) AD
(iii) BD and (iv) AC.​

Answer 1

The answer is below

Explanation:

A square is a quadrilateral (four sides and angles). All the sides and angles of a square are equal. Each angle of a square is 90° and the diagonals are equal.

In square ABCD:

AB = BC = CD = AC (all the sides of a square are equal)

Given that P is the midpoint of BC and Q is the midpoint of CD. AP = a and AQ = b.

a) Triangle APB is a right angled triangle:

BC = AB (all sides of a triangle are equal)

PB = 1/2 (BC) = 1/2(AB) = 0.5AB

Using Pythagoras theorem:

AB² + PB² = AP²

substituting:

AB² + (0.5AB)² = a²

AB² + 0.25AB² = a²

1.25AB² = a²

AB² = 0.8a²

AB = a√0.8

b) Triangle AQD is a right angled triangle:

AD = DC (all sides of a triangle are equal)

DQ = 1/2 (DC) = 1/2(AD) = 0.5AD

Using Pythagoras theorem:

AD² + DQ² = AQ²

substituting:

AD² + (0.5AD)² = b²

AD² + 0.25AD² = b²

1.25AD² = b²

AD² = 0.8b²

AD = b√0.8

c) In right triangle ADB, Using Pythagoras theorem:

AB² + AD² = BD²

But AB = AD

Substituting:

0.8a² + 0.8b² = BD²

BD = √(0.8a² + 0.8b²)

But AD = AB, hence 0.8a² = 0.8b²

BD = √(1.6a²)

BD = a√1.6

D) AC = AD (diagonals of a square are equal)

AC = √(0.8a² + 0.8b²)

AC = √(1.6b²) = b√1.6