Abcd is a square. parallel lines m, n, and p pass through vertices a, b, and c, respectively. the distance between m, and n is 12 units, and the distance between n and p is 17 units. find the number of square units in the area of square abcd.

Area of the square = 433 unit²

Step-by-step explanation:

From the figure attached, m, n, p are three parallel lines.

Distance between m and n, AE = 12 units

Distance between n and p, FC = 17 units

Now in the ΔAED and ΔCFD,

Let m∠ADE = x°

∠AED = 90° [Given]

Then m∠DAE = (90 -x)°[Since ∠EAD + ∠ADE + ∠AED = 180°]

Since ∠ADC = 90° [angle of a square]

∠ADC = ∠ADE + ∠EDC

90° = x° + ∠EDC

∠EDC = (90 - x)°

and ∠FCD = x° [∠FCD = 180° - (90 - x)°]

Therefore, ∠ADE ≅ ∠FCD = x°

∠EAD ≅ ∠FDC = (90 - x)°

And side AD ≅ DC [ Sides of a square]

Therefore, ΔAED and ΔCFD are congruent.

Therefore, Measure of side ED = side FC = 17 units

Now we apply Pythagoras theorem in ΔAED to calculate the measure of side AD.

AD² = 12² + 17²

      = 144 + 289

      = 433

AD = √433 = 20.80 units

Now area of the square = (Side)² = (20.80)²

                                        = 433 unit²

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