Try to work out why this method works. You will find that it is a geometrical translation of the formula (a + b2)2 − (a − b2)2 = ab.
Step-by-step solution
Idea: The side of the big square AFGH is the average a + b2, and the arc turns it into the hypotenuse of a right triangle whose other side is a − b2.
- AE = AB = b, and F is the midpoint of ED, so AF = AE + AD2 = a + b2. AFGH is a square, so HG = HA = a + b2, and HK = HG (radii of the arc).1 mark
- BH = AH − AB = a + b2 − b = a − b2. BKPH is a rectangle (KP ∥ AH), so KP = BH = a − b2.1 mark
- Triangle HPK is right-angled at P. Baudhāyana–Pythagoras: HP2 = HK2 − KP2 = (a + b2)2 − (a − b2)2.1 mark
- = a2 + 2ab + b24 − a2 − 2ab + b24 = 4ab4 = ab. So the square HPQS (area HP2) equals the rectangle ABCD (area ab).1 mark
Check: With a = 9, b = 4 (as in the picture): AF = 6.5, BH = 2.5, HP = √(6.52 − 2.52) = √36 = 6, and 62 = 36 = 9 × 4 ✓.
Answer to write in the exam
AF = AE + AD2 = a + b2 ⇒ HK = HG = a + b2 (radii)
KP = BH = AH − AB = a + b2 − b = a − b2
In right △HPK: HP2 = HK2 − KP2
= (a + b2)2 − (a − b2)2 = 4ab4 = ab
∴ ar(HPQS) = ar(ABCD)
Common mistakes that cost marks
- Taking AF = a2. F is the midpoint of ED, not of AD.
- Using HP2 = HK2 + KP2. HK is the hypotenuse, so subtract.
- Thinking KP equals AB; it equals BH.
How this can come in the exam
Using (a + b2)2 − (a − b2)2 = ab, a rectangle 32 cm × 8 cm has the same area as a square of side
- 8 cm
- 12 cm
- 20 cm
- 16 cm
Show answer
(D) 16 cm
202 − 122 = 400 − 144 = 256 = 32 × 8, so the side is √256 = 16 cm.
Try one yourself
For a rectangle 25 cm × 9 cm, find AF, BH and the side of the square in Baudhāyana’s construction.
Show answer
AF = 17, BH = 8, side = √(172 − 82) = √225 = 15 cm (152 = 225 = 25 × 9).
More questions like this
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- The sides of a triangular plot are in the ratio 3: 5: 7; its perimeter is 300 m. Find its area.