Find the area of the square ABCD | A Nice Geometry Problem | Important Geometry and Algebra Skills

[@jimlocke9320]

Drop perpendicular from point M to DC. Call the intersection point N. This line segment passes through the midpoint of DQ, which is also the center of the circle, call that point O. As Math Booster did, let length AM = a, which is also the length of MB, DN and CN. Let length ON = f. Consider right ∆ODN. OD is a radius of the circle so has length 5. Using the Pythagorean theorem, a² + f² = 5². OM is also a radius of the circle, so has length 5. MN = f + ON = f + 5. MN = side length of the square, which also equals length DC = 2a. So, f + 5 = 2a or f = (2a - 5). Some viewers may recognize that if ∆ODN is the special 3-4-5 right triangle. With f = 3, a = 4, the equations f + 5 = 2a and a² + f² = 5² will both balance. However, we can solve it by substituting f = (2a - 5) in a² + f² = 5². a² + (2a - 5)² = 5² a² + 4a² -20a +25 = 25 5a² -20a = 0 a² -4a = 0 a(a - 4) = 0 a = 0 or a = 4 We discard a = 0 as invalid, assuming that the square must not have a 0 length side. If a = 4, f = 3 and we do indeed have a special 3-4-5 right triangle. The square has sides of length 8, so its area is 64, as Math Booster also found.

[@bekaluu1]

Did it same exact way

[@marioalb9726]

Being S the side of square: Calling BQ = "c" Right triangle DQC: (S - c)² = 10²-S² S-c = √(10²-S²) c= S - √(10²-S²) Right triangle DMQ: DM² + MQ² = 10² [ (½S)² + S² ] + [ (½S)²+c² ] = 10² (5/4)S² + [ (½S)²+ (S-√(10²-S²))²) ] =10² Just one variable in this formula, just "S' Clearing S, we obtain: S = 8 cm Area = S² = 64 cm² ( Solved √ )

[@santiagoarosam430]

Si estudiamos el rectángulo CDPQ》La flecha "f" de los lados verticales es la mitad que la de los horizontales》(10-2f)^2 + (10-4f)^2 = 10^2 》 f^2 +6f +5 =0》f=1》CD=8》Área ABCD=8^2=64 Gracias y saludos.

[@harikatragadda]

Chord PM subtends the same angle at D and Q. If R is midpoint of PQ, then ∆DAM is Similar to ∆QRM AM/AD = a/(2a) = MR/QR = MR/a Therefore MR = a/2 and PD = 2a - a/2 = 3a/2 The Right Triangle PDC has sides (3a/2, 2a, 20), giving a = 4 Square Area = (2a)² = 64

[@hongningsuen1348]

The simplest solution, I think: 1. Noted quadrilateral CDPQ is a cyclic quadrilateral. DQ is the diameter of the circle (Thales theorem. Hence radius of circle = DQ/2 = 10/2 = 5. Centre of circle O is at midpoint of DQ. 2. Let side of the square be S. 3. In triangle CDQ, DQ^2 = CD^2 + CQ^2 (Pythagoras theorem) Hence 10^2 = S^2 +CQ^2 and CQ = SQR (100 - S^2) 4. Construct a line from M to centre of circle O and then P on DC. It is parallel to CQ and bisects DC at P. (radius to tangent at 90) 5. OP = 1/2 x CQ = 1/2 SQR (100 - S^2) (Midpoint theorem) 6. MP = S = MO + OP = 5 + 1/2 SQR(100 - S^2) (MO radius of circle) (S - 5)^2 = 1/4 (100 - S^2) 5S^2 = 40S S = 8 7. Area of square = S^2 = 8^2 = 64.

[@EnnioPiovesan]

Observe that the triangles MAD and MBQ are similar because they are both right triangles and because they have equal angles ADM and QMB. The equality of the angles derives from the fact that they are both circumferential angles subtended by equal arcs (ADM subtended by PM and QMB subtended by MQ). Consequently the ratio between MB and BQ is 2 (like that between AD and AM) and therefore that between QC and BC is 3:4. But then also the legs QC and DC of the right triangle DQC are in the ratio 3:4. So the sides of the right triangle DQC form the Pythagorean triple (3,4,5) and therefore their values are 6,8,10. The side of the square DC is therefore 8 and the area 64.

[@hongningsuen1348]

Your method is excellent. The reason for similarity of triangles MAD and MBQ is easier to understand when angle DMQ is noted to be a right angle by Thales theorem.

[@LyudmilYovkov]

I will follow your notations. Let AM = BM = x and QH is perpendicular to AD. Then (from triangle DHQ) we have HQ = 2x and DH = sqrt(100-4x^2) => AH = BQ = 2x - sqrt(100-4x^2). Obviously 0 < x < 5. From the equality BM^2 = BQ * BC we obtain the equation x^2 = [2x - sqrt(100-4x^2)] * 2x whose unique positive root is x = 4 => S = (2x)^2 = 64.

[@BabisKoutroulis]

In the general case, it is enough

[@quigonkenny]

Let AM = BM = x. As AB is a side of the square, AB = BC = CD = DA = 2x. As DQ is the diameter of the circle, r = 5. Let O be the midpoint of DQ, at the center of the circle. Draw radius OM. As AB is tangent to vircle O at M, ∠OMA = ∠BMO = 90°. Draw ON, where N is the midpoint of DC. As DC is parallel to AB, ∠ONC = ∠DNO = 90°. As AD and MN are parallel, AM = DN = x and MN = AD = 2x. As OM = r = 5, ON = 2x-5. Triangle ∆DNO: ON² + DN² = OD² (2x-5)² + x² = 5² 4x² - 20x + 25 + x² = 25 5x² = 20x x = 20x/5x = 4 Square ABCD: A = (2x)² = 4x² = 4(4²) = 64 sq units

[@marioalb9726]

Being S the side of square: Being BQ a fraction 'r' of BC Tangent theorem: (½S)² = S. rS r = ¼ S²/S² ratio r = 1/4 Every time we have this configuration, the ratio between BQ and BC is 1/4, independently of dimensions. Taking the appropriate right triangle: S² + (¾ S)² = 10² 25/16 . S² = 100 S² = 100 . 16/25 S² = 64 cm² Area = 64 cm² (Solved √ )

[@charlesbromberick4247]

Nice problem. The key to so many of these problems is to take a deep breath, hold it 10 seconds and start with, "let x =..."

[@Clock_Tune]

Sir can you please try physics problems as well

[@CanalMiTube]

Nice exercise. (1/2 DC)^2 = (2R-X)X. X = 2R - DC. A system of two, DC and X.

[@abdulhadial-omari7482]

could someone explain this method?

[@RahulKumar-id5cq]

Let angle CDQ=∆ •QB=BC-QC=CD-QC=10cos∆-10sin∆ •angle DQP=∆ Let cente of circle O an mid point of PQ be R. Now MR=MO-OR=5-5sin∆ now equating MR=QB 5-5sin∆=10cos∆-10sin∆ sin∆=3/5 cos∆=4/5 side of square ABCD=10cos∆ =8 Hence,Area=64

[@luigipirandello5919]

Beautiful problem. Thank you, sir.

[@saddatsabanature4307]

❤ Thanks you sir ❤

[@voltalimwabbit2351]

Could ee figure out the radius of the circle? Thank you

[@정명주-r6v]

재미로 풀어봤습니다.

[@JEE2026IIT]

The Q was really good duh!

[@giuseppemalaguti435]

64

[@JEE2026IIT]

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