Question

I’m so confused and neeeed help

Answer 1

Answer:

√40 + 6 + 4√2 + √28

Step-by-step explanation:

See the diagram attached.

Δ BCD is a right triangle and BC ² = DC² + DB² {Applying Pythagoras Theorem}

⇒ y² = 2² + 6² = 40

⇒ y = 2√10 units,

Now, Δ ADC is a right triangle and AC² = AD² + DC² {Applying Pythagoras Theorem}

⇒ (4√2)² = x² + 2²  

⇒ x² = 32 - 4 = 28

⇒ x = 2√7.

Therefore, the perimeter of the triangle Δ ABC will be  

= x + 6 + y + 4√2

= 2√7 + 6 + 2√10 + 4√2

= √40 + 6 + 4√2 + √28 (Answer)

Answer 2

Answer:

Perimeter = √40 + 6 + 4√2 + √28

Step-by-step explanation:

As we know that the perimeter of a triangle is the sum of the length of sides.

Perimeter =  4√2 + x + 6 + y

Now for x and y-

I. By Pythagorean theorem in left side right angle triangle-

( 4√2)² = 2² + x²

32 - 4 = x²

x² = 28

x = √28

II. By Pythagorean theorem in Right angle triangle-

y² = 2² + 6²

y² = 4 + 36

y² = 40

y = √40

Hence Perimeter =  4√2 + x + 6 + y

P = 4√2 + √28 + 6 + √40

∴ P = √40 + 6 + 4√2 + √28