The largested square is cut out from a right angled triangular region with length of 3 cm, 4 cm and 5 cm respectively in such a way that the one vertex of square lies on hypotenuse of triangle. Let us write by calculating the length of side of square
Answers
Draw a Right angled triangle ABC with measurements 3 cm, 4 cm, 5 cm inscribing a square such that it touches hypotenuse.
Refer to attachment
Method 1 : Solving using Areas
Let the side of the inscribed square be x cm
Area of a square = Side² sq.units
Area of the inscribed square DEBF = x² cm²
Consider ∆AED
∠AED = 180° - ∠DEB = 180° - 90° = 90°
Area of a triangle = Base × Height / 2 sq.units
- Base = ED = x cm
- Height = AE = AB - EB = ( 3 - x ) cm
Area of the ∆AED = x( 3 - x ) / 2 = ( 3x - x² ) / 2 cm²
Consider ∆DFC
∠DFC = 180° - ∠DFB = 180° - 90° = 90°
Area of a triangle = Base × Height / 2 sq.units
- Base = FC = BC - BF = ( 4 - x ) cm
- Height = DF = x cm
Area of the ∆AED = x( 4 - x ) / 2 = ( 4x - x² ) / 2 cm²
We know that
Sum of areas of 2 triangles and square = Area of ∆ABC
⇒ ar( ∆AED ) + ar( ∆DFC ) + ar( Square DFBE ) = AB × BC / 2
⇒ ( 3x - x² ) / 2 + ( 4x - x² ) / 2 + x² = 3 × 4 / 2
Multiplying every term by 2
⇒ 3x - x² + 4x - x² + 2x² = 12
⇒ 7x - 2x² + 2x² = 12
⇒ 7x = 12
⇒ x = 12 / 7
Method 2 : Solving by Similarity
Consider ∆DFC and ∆AED
∠DFC = ∠AED = 90°
∠FDC = ∠EAD [ AE ∥ EF , AC is transversal , corresponding angles are equal ]
Third angles are also equal by Angle sum property
By AAA similarity ∆AED ~ ∆DFC
Corresponding sides will be in proportion.
⇒ AE / ED = DF / FC
- AE = ( 3 - x ) cm
- ED = x cm
- DF = x cm
- FC = ( 4 - x ) cm
⇒ ( 3 - x ) / x = x / ( 4 - x )
⇒ ( 3 - x )( 4 - x ) = x²
⇒ 3( 4 - x ) - x( 4 - x ) = x²
⇒ 12 - 3x - 4x + x² = x²
⇒ 12 - 7x = 0
⇒ 7x = 12
⇒ x = 12 / 7 = 1.71
Therefore, the length of side of largest square that can be cut out from the right triangle is 12 / 7 cm or 1.71 cm
The length of side of a square is 12/7 cm. Explanation in image
