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SCORPION-xisa [38]

3 years ago

10

Geometry help please........

1 answer:

svp [43]3 years ago

4 0

Let m = 10 and n = 2

We know:

y^2 = (m+n).m

y^2 = (10+2).10

y^2 = 120

y = √(120)

Then,

m^2 + x^2 = y^2

10^2 + x^2 = ( √( 120) )^2

100 + x^2 = 120

x^2 = 120 - 100

x^2 = 20

x = √(20)

x = √(4×5)

x = √(4) × √(5)

x = 2√(5)

I hope this has helped!

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If P is the orthocenter of △ABC, AB = 13, BF = 9,and FC = 5.6, find each measure & the perimeter of triangle ABC.

harina [27]

Answer:

$Perimeter = 38.6$

Step-by-step explanation:

Given

$A\ B = 13$

$B\ F = 9$

$F\ C = 5.6$

Required

Calculate the perimeter of A B C

To calculate the perimeter of A B C, we sum up the sides of the triangle

i.e.

$Perimeter = A B + B\ C + A\ C$

AB is given in the question already.

So, we need to calculate B F and F C

To solve for BF, we consider triangle B F C

Using Pythagoras Theorem, we have:

$B\ C^2 = B\ F^2 + F\ C^2$

Substitute values for BF and FC;

$B\ C^2 = 9^2 + 5.6^2$

$B\ C^2 = 81 + 31.36$

$B\ C^2 = 112.36$

$B\ C^2= \sqrt{112.36$

$B\ C = 10.6$

Next, we calculate A C.

To calculate A C, we need to calculate A F, considering triangle B F A

Using Pythagoras Theorem, we have:

$A\ B^2 = B\ F^2 + A\ F^2$

Substitute values for B F and A B;

$13^2 = 9^2 + A\ F^2$

$169= 81 + A\ F^2$

$A\ F^2 = 169 - 81$

$A\ F^2 = 88$

$A\ F = \sqrt{88$

$A\ F = 9.4$

Since F lies on line A C:

$A\ C = A\ F + F\ C$

$A\ C = 9.4 + 5.6$

$A\ C = 15.0$

Recall that:

$Perimeter = A\ B + B\ C + A\ C$

$Perimeter = 13 + 10.6 + 15.0$

$Perimeter = 38.6$

<em>Hence, the perimeter is 38.6</em>

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3 years ago