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2 years ago

Find the length of ad in the diagram shown

Mathematics

2 answers:

kvv77 [185]2 years ago

4 0

The answer to this is 7.5.

Ede4ka [16]2 years ago

4 0

Let as consider two triangles ABE and ACD

As we can see these two triangles ABE and ACD are similar triangles

Hence the ratio of sides must be equal

$\frac{AB}{AC} = \frac{AE}{AD}$

Now let AE = x

So plugging all the values we get

AB= 5 , AC= 11 , AE =x , AD = x+9

$\frac{5}{11} =\frac{x}{x+9}$

Cross multiplying we get

5(x+9) = 11(x)

Simplifying

5x +45 = 11x

Subtract 5x from both sides

45 = 6x

Divide both sides by 6

x= 45/6 = 7.5

Hence AE = 7.5

But we have to find the length of AD

AD = AE+ED = 7.5 + 9 = 16.5

AD= 16.5

Option D is the answer

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damaskus [11]

<u>ANSWER: </u>

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<u>SOLUTION:</u>

Given, $f(x)=x^{2}+12 x+24$ -- eqn 1

x-intercepts of the function are the points where function touches the x-axis, which means they are zeroes of the function.

Now, let us find the zeroes using quadratic formula for f(x) = 0.

$X=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

Here, for (1) a = 1, b= 12 and c = 24

$X=\frac{-(12) \pm \sqrt{(12)^{2}-4 \times 1 \times 24}}{2 \times 1}$

$\begin{array}{l}{X=\frac{-12 \pm \sqrt{144-96}}{2}} \\\\ {X=\frac{-12 \pm \sqrt{48}}{2}} \\\\ {X=\frac{-12 \pm \sqrt{16 \times 3}}{2}} \\\\ {X=\frac{-12 \pm 4 \sqrt{3}}{2}} \\ {X=\frac{2(-6+2 \sqrt{3})}{2}, \frac{2(-6-2 \sqrt{3})}{2}} \\\\ {X=(-6+2 \sqrt{3}),(-6-2 \sqrt{3})}\end{array}$

Hence the x-intercepts of $\mathrm{x}^{2}+12 \mathrm{x}+24=0 \text { are }(-6+2 \sqrt{3}),(-6-2 \sqrt{3})$

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goldenfox [79]

What do you mean by this The area of the paralleogram is 184 units squaredgive me your full question then i can help you

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Sergio [31]

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