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Verdich [7]
3 years ago
14

A right triangle has a perimeter of 24 cm and a hypotenuse of 10 cm. Find the length of the other two sides.

Mathematics
2 answers:
jekas [21]3 years ago
7 0

Answer:

One of the sides is 6 cm and the other is 8 cm

Step-by-step explanation:

Let's call the unknown sides a and b. From the perimeter information (24 cm) we have:

a + b + hypotenuse = 24

a + b + 10 = 24

a + b = 14

b = 14 - a

So now we can right the Pythagorean theorem as follows:

a^2 + b^2 = hypotenuse^2\\a^2 + (14-a)^2=10^2\\a^2+ 14^2-28\,a+a^2=100\\2\,a^2-28\,a +96=0\\2\,(a^2-14\,a+48)=0\\2\,(a^2-6\,a-8\,a+48)=0\\2(a\,(a-6)-8\,(a-6))=0\\2\,(a-6)\,(a-8)=0

and from this expression in factor form to be zero a must be 6 or a must be 8.

Therefore the solutions are a = 6 (and therefore b = 14 - 6 = 8)

or a = 8 (and therefore b = 14 - 8 = 6)

Norma-Jean [14]3 years ago
7 0

Perimeter of triangle = Sum of all sides.

One side of triangle = 10 cm

Measure of other two sides will be → 24-10

Other two sides sum will be = 14 cm

Let's assume one side as x and other as 14-x

Now hypotenuse is of 10 cm

According to Pythagorean theoram

Hypotenuse² = Perpendicular ² + base²

→ 10² = (x)² + (14-x)²

→ 100 = x² + (14-x)²

(a-b)² = a²+b² -2ab

→ 100 = x² + 196 + x² -28 x

→ 100 = 2x² + 196 - 28x

Taking 2 commen both sides

→ 50 = x² -14x + 98

→ 0 = x² -14x +98-50

→ x² -14 x +48 = 0

Factorising :-

→ x² - 6x - 8x + 48 = 0

→ x( x -6) -8(x -6) = 0

→ (x-6)(x-8) = 0

→ x = 6 or x = 8

So x is 6 or 8

Other two sides of triangle are 6 cm and 8 cm

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Given the functionf ( x ) = x^2 + 7 x + 10/ x^2 + 9 x + 20
vladimir1956 [14]

<em>x = -4 is a vertical asymptote for the function.</em>

<h2>Explanation:</h2>

The graph of y=f(x) is a vertical has an asymptote at x=a if at least one of the following statements is true:

1) \ \underset{x\rightarrow a^{-}}{lim}f(x)=\infty\\ \\ 2) \ \underset{x\rightarrow a^{-}}{lim}f(x)=-\infty \\ \\ 3) \ \underset{x\rightarrow a^{+}}{lim}f(x)=\infty \\ \\ 4) \ \underset{x\rightarrow a^{+}}{lim}f(x)=\infty

The function is:

f(x)=\frac{x^2+7x+10}{x^2+9x+20}

First of all, let't factor out:

f(x)=\frac{x^2+5x+2x+10}{x^2+5x+4x+20} \\ \\ f(x)=\frac{x(x+5)+2(x+5)}{x(x+5)+4(x+5)} \\ \\ f(x)=\frac{(x+5)(x+2)}{(x+5)(x+4)} \\ \\ f(x)=\frac{(x+2)}{(x+4)}, \ x\neq  5

From here:

\bullet \ When \ x \ approaches \ -4 \ on \ the \ right: \\ \\ \underset{x\rightarrow -4^{+}}{lim}\frac{(x+2)}{(x+4)}=? \\ \\ \underset{x\rightarrow -4^{+}}{lim}\frac{(-4^{+}+2)}{(-4^{+}+4)} \\ \\ \\ The \ numerator \ is \ negative \ and \ the \ denominator \\ is \ a \ small \ positive \ number. \ So: \\ \\ \underset{x\rightarrow -4^{+}}{lim}\frac{(x+2)}{(x+4)}=-\infty

\bullet \ When \ x \ approaches \ -4 \ on \ the \ left: \\ \\ \underset{x\rightarrow -4^{-}}{lim}\frac{(x+2)}{(x+4)}=? \\ \\ \underset{x\rightarrow -4^{-}}{lim}\frac{(-4^{-}+2)}{(-4^{-}+4)} \\ \\ \\ The \ numerator \ is \ a \ negative \ and \ the \ denominator \\ is \ a \ small \ negative \ number \ too. \ So: \\ \\ \underset{x\rightarrow -4^{-}}{lim}\frac{(x+2)}{(x+4)}=+\infty

Accordingly:

x=-4 \ is \ a \ vertical \ asymptote \ for \\ \\ f(x)=\frac{x^2+5x+2x+10}{x^2+5x+4x+20}

<h2>Learn more:</h2>

Vertical and horizontal asymptotes: brainly.com/question/10254973

#LearnWithBrainly

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