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

Answers ( 2 )

    0
    2023-01-15T00:59:35+00:00
    Answer:
    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
    [tex]a^2+b^2=c^2[/tex]
    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:
    [tex]\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}[/tex]
    To find the value of x use the quadratic formula.
    Quadratic Formula
    [tex]x=\dfrac{-b \pm \sqrt{b^2-4ac} }{2a}\quad\textsf{when }\:ax^2+bx+c=0[/tex]
    [tex]\implies a=1,\quad b=-2, \quad c=-11[/tex]
    Therefore:
    [tex]\implies x=\dfrac{-(-2) \pm \sqrt{(-2)^2-4(1)(-11)} }{2(1)}[/tex]
    [tex]\implies x=\dfrac{2 \pm \sqrt{4+44}}{2}[/tex]
    [tex]\implies x=\dfrac{2 \pm \sqrt{48}}{2}[/tex]
    [tex]\implies x=\dfrac{2 \pm \sqrt{16 \cdot 3}}{2}[/tex]
    [tex]\implies x=\dfrac{2 \pm \sqrt{16}\sqrt{3}}{2}[/tex]
    [tex]\implies x=\dfrac{2 \pm 4\sqrt{3}}{2}[/tex]
    [tex]\implies x=1 \pm 2\sqrt{3}[/tex]
    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
    [tex]\sf Area = \dfrac{1}{2}bh[/tex]
    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:
    [tex]\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}[/tex]

    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.
    [tex]\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}[/tex]

    the-sides-of-a-right-angled-triangle-has-sides-1-cm-2-cm-and-4-cm-i-find-the-area-of-the-triangl-28277792-57

    0
    2023-01-15T00:59:35+00:00
    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

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