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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.
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Solution
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"`.
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