Question

In a square PQRS of side 5 cm, A, B, C and D are points on sides PQ, QR, RS and SP respectively such as PA = PD = RB = RC = 2 cm. Prove that ABCD is a rectangle. Also, find the area and perimeter of the rectangle.

Answer 1

In ΔAPD, ∠P = 90°
∴ AD² = AP² + PD²
= 2² + 2²
= 4 + 4
= 8
⇒ AD = `2sqrt(2)"cm"`
Similarly, we can prove that in ΔBRC,
BC = `2sqrt(2)"cm"`
∴ AD = BC    ....(i)
In ΔAQB, ∠Q = 90°
∴ AB² = AQ² + BQ²
= 3² + 3²
= 9 + 9
= 18
⇒ AB = `3sqrt(2)"cm"`
Similarly, we can prove that in ΔCSD,
CD = `3sqrt(2)"cm"`
∴  AB = CD     ....(ii)
|Again, in ΔAPD,
AP = PD
⇒ ∠PAD = ∠PDA = 45°
Also, in ΔAQB,
AQ = BQ
⇒ ∠QAB = ∠QBA = 45°
Now, ∠PAD + ∠DAB + ∠QAB = 180°
⇒ 45° + ∠DAB + 45° = 180°
⇒ ∠DAB = 90°
Similarly, we can prove that ∠ABC, ∠BCD and ∠ADC are 90° each.
Thus, ABCD is a rectangle as opposite as opposite sides are equal and each angle is 90°.
Now,
Area of a rectangle ABCD
= AD x AB
= `2sqrt(2) xx 3sqrt(2)`
= 12cm²
Perimeter of a rectangle ABCD
= AB + BC + CD + AD
= `2sqrt(2) + 3sqrt(2) + 2sqrt(2) + 3sqrt(2)`
= `10sqrt(2)"cm"`.