# 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) f

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 ​

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3 weeks 2023-01-15T00:57:50+00:00 2 Answers 0 views 0

i)  (9 + 5√3)  cm²
ii)  (10 + 6√3)  cm
Step-by-step 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.
$$x=\dfrac{-b \pm \sqrt{b^2-4ac} }{2a}\quad\textsf{when }\:ax^2+bx+c=0$$
$$\implies a=1,\quad b=-2, \quad c=-11$$
Therefore:
$$\implies x=\dfrac{-(-2) \pm \sqrt{(-2)^2-4(1)(-11)} }{2(1)}$$
$$\implies x=\dfrac{2 \pm \sqrt{4+44}}{2}$$
$$\implies x=\dfrac{2 \pm \sqrt{48}}{2}$$
$$\implies x=\dfrac{2 \pm \sqrt{16 \cdot 3}}{2}$$
$$\implies x=\dfrac{2 \pm \sqrt{16}\sqrt{3}}{2}$$
$$\implies x=\dfrac{2 \pm 4\sqrt{3}}{2}$$
$$\implies x=1 \pm 2\sqrt{3}$$
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
$$\sf Area = \dfrac{1}{2}bh$$
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 \dfrac{1}{2}bh\\\\\implies \sf Area & = \dfrac{1}{2}(2+2\sqrt{3})(3+2\sqrt{3})\\\\& = \dfrac{1}{2}(6+10\sqrt{3}+12)\\\\ & = \dfrac{1}{2}(18+10\sqrt{3})\\\\ & = 9+5\sqrt{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\sqrt{3})+(3+2\sqrt{3})+(5+2\sqrt{3})\\& = 2+3+5+2\sqrt{3}+2\sqrt{3}+2\sqrt{3}\\& = 10+6\sqrt{3}\:\: \sf cm\end{aligned} 2. Step-by-step 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