Question

The sides of a right angled triangle has sides (x+1)cm, (x+2)cm and (x+4) cm.

i) Find the area of the triangle
ii) find the perimeter of the triangle
pls someone answer this. question ASAP. TNX ​

Answer 1

Explanation:

right-angled triangle.

that means Pythagoras applies :

c² = a² + b²

c being the Hypotenuse (the side opposite of the 90° angle and the longest of the 3 sides).

(x+4)² = (x+1)² + (x+2)²

x² + 8x + 16 = x² + 2x + 1 + x² + 4x + 4

8x + 16 = x² + 6x + 5

x² - 2x = 11

remember,

(x-a)² = x² -2ax + a²

compare to what we have

-2a = -2

a = 1

so, the full square on the left side is

(x-1)² = x² - 2x + 1

to complete the square in our equation we need to add 25 on both sides

x² - 2x + 1 = 11 + 1 = 12

(x - 1)² = 12

x - 1 = sqrt(12)

x = sqrt(12) + 1

and so

i)

when we have all 3 sides a, b, c, the area of the triangle is according to Heron's formula

S = (a + b + c)/2

Area = sqrt(S(S-a)(S-b)(S-c))

in our case that is

S = ((x+1) + (x+2) + (x+4))/2 =

= ((sqrt(12)+1+1) + (sqrt(12)+1+2) + (sqrt(12)+1+4))

/2 = (3×sqrt(12) + 10)/2 =

= 10.19615242...

Area = sqrt(S(S- sqrt(12)-2)(S-sqrt(12)-3)(S-sqrt(12)-5)) =

= sqrt(10.19615242... ×

4.732050808... ×

3.732050808... ×

1.732050808...) =

= sqrt(311.8845727...) =

= 17.66025404... cm²

ii)

the perimeter is

(sqrt(12)+1+1) + (sqrt(12)+1+2) + (sqrt(12)+1+4) =

3sqrt(12) + 10 = 20.39230485... cm

Answer 2

i) (9 + 5√3) cm²

ii) (10 + 6√3) cm

Explanation:

Given sides of a right triangle:

• (x + 1) cm
• (x + 2) cm
• (x + 4) cm

The longest side of a right triangle is the hypotenuse.

The two shortest sides of a right triangle are the legs.

To find the value of x use Pythagoras Theorem.

Pythagoras Theorem

`a^2+b^2=c^2`

where:

• a and b are the legs of the right triangle.
• c is the hypotenuse (longest side) of the right triangle.

Substitute the given expressions for the legs and hypotenuse into the formula:

`\begin{aligned}a^2+b^2 & =c^2\\\implies (x+1)^2+(x+2)^2 & =(x+4)^2\\(x+1)(x+1)+(x+2)(x+2) & =(x+4)(x+4)\\x^2+2x+1+x^2+4x+4 & =x^2+8x+16\\2x^2+6x+5 & =x^2+8x+16\\2x^2-x^2+6x-8x+5-16 & =0\\x^2-2x-11 & =0\end{aligned}`

To find the value of x use the quadratic formula.

Quadratic Formula

`x=(-b \pm √(b^2-4ac) )/(2a)\quad\textsf{when }\:ax^2+bx+c=0`

`\implies a=1,\quad b=-2, \quad c=-11`

Therefore:

`\implies x=(-(-2) \pm √((-2)^2-4(1)(-11)) )/(2(1))`

`\implies x=(2 \pm √(4+44))/(2)`

`\implies x=(2 \pm √(48))/(2)`

`\implies x=(2 \pm √(16 \cdot 3))/(2)`

`\implies x=(2 \pm √(16)√(3))/(2)`

`\implies x=(2 \pm 4√(3))/(2)`

`\implies x=1 \pm 2√(3)`

As 1 - 2√3 ≈ -2.46 we can discount this value of x as it would make two of the sides of the triangle negative values. Since length cannot be negative, the only valid value of x is 1 + 2√3.

Therefore the lengths of the sides of the right triangle are:

• (1 + 2√3 + 1) = (2 + 2√3) cm
• (1 + 2√3 + 2) = (3 + 2√3) cm
• (1 + 2√3 + 4) = (5 + 2√3) cm

Part (i)

Area of a triangle

`\sf Area = (1)/(2)bh`

where:

• b = base
• h = height

The base and the height of a right triangle are its legs.

Substitute the found length of the legs into the formula and solve:

`\begin{aligned}\sf Area & = \sf (1)/(2)bh\\\\\implies \sf Area & = (1)/(2)(2+2√(3))(3+2√(3))\\\\& = (1)/(2)(6+10√(3)+12)\\\\ & = (1)/(2)(18+10√(3))\\\\ & = 9+5√(3) \:\:\sf cm^2\end{aligned}`

Part (ii)

The perimeter of a two-dimensional shape is the distance all the way around the outside. Therefore, the perimeter of a triangle is the sum of the lengths of its sides.

`\begin{aligned}\sf Perimeter & = (2+2√(3))+(3+2√(3))+(5+2√(3))\\& = 2+3+5+2√(3)+2√(3)+2√(3)\\& = 10+6√(3)\:\: \sf cm\end{aligned}`