Question

14. Sum of the areas of 2 squares is 544 mº if the difference of their perimeter is 32 m find the sides of
two square
15. State and prove The Pythagoras theorem.​

Answer 1

Let the sides of first and second square be X and Y .

Area of first square = (X)²

And,

Area of second square = (Y)²

According to question,

(X)² + (Y)² = 544 m² ------------(1).

Perimeter of first square = 4 × X

and,

Perimeter of second square = 4 × Y

According to question,

4X - 4Y = 32 -----------(2)

From equation (2) we get,

4X - 4Y = 32

4(X-Y) = 32

X - Y = 32/4

X - Y = 8

X = 8+Y ---------(3)

Putting the value of X in equation (1)

(X)² + (Y)² = 544

(8+Y)² + (Y)² = 544

(8)² + (Y)² + 2 × 8 × Y + (Y)² = 544

64 + Y² + 16Y + Y² = 544

2Y² + 16Y - 544 +64 = 0

2Y² + 16Y -480 = 0

2( Y² + 8Y - 240) = 0

Y² + 8Y - 240 = 0

Y² + 20Y - 12Y -240 = 0

Y(Y+20) - 12(Y+20) = 0

(Y+20) (Y-12) = 0

(Y+20) = 0 Or (Y-12) = 0

Y = -20 OR Y = 12

Putting Y = 12 in EQUATION (3)

X = 8+Y = 8+12 = 20

Side of first square = X = 20 m

and,

Side of second square = Y = 12 m.

Answer 2

Question:

The sum of areas of two squares is 544 m². If the difference of their perimeter is 32 m, find the sides of the two squares

Answer:

$\bigstar{\bold{Side\:of\:the\:first\:square=20\:m}}$

$\bigstar{\bold{Side\:of\:second\:square=12\:m}}$

Step-by-step explanation:

$\Large{\underline{\underline{\bf{Given:}}}}$

• Sum of areas of two squares = 544 m²
• Difference of their perimeters = 32 m

$\Large{\underline{\underline{\bf{To\:Find:}}}}$

• The sides of the two squares

$\Large{\underline{\underline{\bf{Solution:}}}}$

→ Let the side of the first square be a m

→ Let the side of the second square be b m

→ The area of a square is given by,

  Area of a square = (side)²

→ Hence,

  Area of first square = a²

  Area of second square = b²

→ By given,

  a² + b² = 544 ---(1)

→ Perimeter of a square is given by

  Perimeter of a square = 4 × side

→ Hence,

  Perimeter of first square = 4a

  Perimeter of second square = 4b

→ By given,

  4a - 4b = 32

→ Dividing the equation by 4,

  a - b = 8

  a = 8 + b ----(2)

→ Substitute equation 2 in 1

  ( 8 + b)² + b² = 544

   64 + 16b + b² + b² = 544

   64 + 16b + 2b² = 544

   2b² + 16b = 480

→ Dividing the equation by 2

  b² + 8b = 240

→ Solving by splitting the middle term,

   b² + 20b - 12b - 240 = 0

   b (b + 20) - 12 (b + 20) = 0

   (b + 20) (b - 12) = 0

   b = 12 or b = -20

→ Here b can't be negative, therefore b = 12

→ Therefore the side of the second square = 12 m

$\boxed{\bold{Side\:of\:second\:square=12\:m}}$

→ Substitute the value of b in equation 2

   a = 8 + b

   a = 8 + 12

   a = 20 m

→ Hence the side of the first square is 20 m

$\boxed{\bold{Side\:of\:the\:first\:square=20\:m}}$

Question:

State and prove the pythagoras theorem

Answer:

The pythagoras theorem states that in a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

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$\Large{\underline{\underline{\bf{Given:}}}}$

• Δ ABC with ∠C = 90°

 

$\Large{\underline{\underline{\bf{To\:Prove:}}}}$

AB² = AC² + BC²

$\Large{\underline{\underline{\bf{Construction:}}}}$

Draw CD ⊥ AB

$\Large{\underline{\underline{\bf{Proof:}}}}$

→ By theorem we know that if a perpendicular is drawn from the vertex of a right angle of a right angled triangle to the hypotenuse, tthe triangles on both sides of the perpendicular are similar to the whole triangle and each other.

→ Consider ΔBDC and ΔBCA

→ Since they are similar,

  BC/AB = BD/BC

→ Cross multiplying,

   BC² = AB × BD-----(1)

→ Consider ΔCDA and Δ BCA

→ Since they are also similar,

   AC/AB = AD/AC

→ Cross multiplying,

  AC² = AB × AD ----(2)

→ Add equation 1 and 2

  BC² + AC² = AB × BD + AB × AD

  BC² + AC² = AB (BD + AD)

  BC² + AC² = AB × AB ( Since BD + AD = AB)

  AB² = BC² + AC²

→ Hence proved.