Question

Find the area of quadrilateral PQRS where ∠Q = ∠S = 90° and PQ = 7 cm, PR = 25 cm and RS = 20 cm.

Answer 1

Given:

• Quadrilateral PQRS with ∠Q = ∠S = 90°
• PQ = 7 cm
• PR = 25 cm
• RS = 20 cm

Stepwise calculation:

1. Draw height QT perpendicular to PR in triangle PQR and height SW perpendicular to PR in triangle RSP.

2. Calculate height QT for ΔPQR:

Area of ΔPQR = `1/2` × PR × height QT

Let height QT = h₁.

3. Calculate height SW for ΔRSP:

Area of ΔRSP = `1/2` × PR × height SW 

Let height SW = h₂.

4. Use Pythagoras Theorem in triangle PQT where ∠Q = 90°:

PQ = 7 cm

PR = 25 cm 

`PT = sqrt(PR^2 - PQ^2)` 

= `sqrt(25^2 - 7^2)` 

= `sqrt(625 - 49)`

= `sqrt(576)`

= 24 cm

So height QT = PT = 24 cm as QT is perpendicular from Q to PR.

5. Use Pythagoras Theorem in triangle RSW where ∠S = 90°:

RS = 20 cm

PR = 25 cm

`RW = sqrt(PR^2 - RS^2)` 

= `sqrt(25^2 - 20^2)` 

= `sqrt(625 - 400)`

= `sqrt(225)`

= 15 cm

So height SW = RW = 15 cm.

6. Area of quadrilateral PQRS

= Area of triangle PQR + Area of triangle RSP 

= `1/2` × PR × height QT + `1/2` × PR × height SW

= `1/2 xx 25 xx 24 + 1/2 xx 25 xx 15` 

= (12.5 × 24) + (12.5 × 15) 

= 300 + 187.5 

= 487.5 cm²

But this is incorrect because both heights QT and SW are along PR; this means QT and SW cannot be summed this way.

The correct approach is to consider the quadrilateral as two triangles sharing diagonal PR:

`"Area of"  ΔPQR = 1/2 xx PQ xx QR`   ...(Since ∠Q = 90°)

`"Area of"  ΔRSP = 1/2 xx RS xx SP`   ...(Since ∠S = 90°)

However, QR and SP are not given.

But using Pythagoras theorem in triangles:

Triangle PQR with right angled at Q:

PQ = 7 cm

PR = 25 cm

So, `QR = sqrt(PR^2 - PQ^2)`

= `sqrt(625 - 49)`

= 24 cm

Triangle RSP with right angled at S:

RS = 20 cm

PR = 25 cm

So, `SP = sqrt(PR^2 - RS^2)`

= `sqrt(625 - 400)`

= 15 cm

Now, 

`"Area of"  ΔPQR = 1/2 xx 7 xx 24 = 84  cm^2`

`"Area of " ΔRSP = 1/2 xx 20 xx 15 = 150  cm^2`

Hence, area of PQRS = 84 + 150 = 234 cm².