Given: KLMN is a trapezoid, KF =10 MF ║ LK AKLMF = AFMN Find: KN

Mathematics

1 answer:

Nimfa-mama [501]3 years ago

3 0

You can see the trapezoid in the attached picture.

The data given in the problem are:
KF = 10
LK // MF
A(KLMF) = A(FMN)

We know that KLMF is a parallelogram because:
LM // KF (bases of the trapezoid)
LK // MF (hypothesis).
The area of a parallelogram is given by the formula:
A(KLMF) = b × h
 = KF × h
 = 10 × h

The area of a triangle is given by the formula:
A(FMN) = (b × h) / 2
 = (FN × h) / 2

The problem states that the two areas are congruent, therefore:
A(KLMF) = A(FMN)
10 × h = FN × h / 2
10 = FN / 2
FN = 20

Therefore we can calculate:
KN = KF + FN = 10 + 20 = 30

Hence, KN is 30 units long.

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Area of a triangle with points at (-9,5), (6,10), and (2,-10)

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First we are going to draw the triangle using the given coordinates.
Next, we are going to use the distance formula to find the sides of our triangle.
Distance formula: $d= \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

Distance from point A to point B:
$d_{AB}= \sqrt{[6-(-9)]^2+(10-5)^2}$
$d_{AB}= \sqrt{(6+9)^2+(10-5)^2}$
$d_{AB}= \sqrt{(15)^2+(5)^2}$
$d_{AB}= \sqrt{225+25}$
$d_{AB}= \sqrt{250}$
$d_{AB}=15.81$

Distance from point A to point C:
$d_{AC}= \sqrt{[2-(-9)]^2+(-10-5)^2}$
$d_{AC}= \sqrt{(2+9)^2+(-10-5)^2}$
$d_{AC}= \sqrt{11^2+(-15)^2}$
$d_{AC}= \sqrt{121+225}$
$d_{AC}= \sqrt{346}$
$d_{AC}= 18.60$

Distance from point B from point C
$d_{BC}= \sqrt{(2-6)^2+(-10-10)^2}$
$d_{BC}= \sqrt{(-4)^2+(-20)^2}$
$d_{BC}= \sqrt{16+400}$
$d_{BC}= \sqrt{416}$
$d_{BC}=20.40$

Now, we are going to find the semi-perimeter of our triangle using the semi-perimeter formula:
$s= \frac{AB+AC+BC}{2}$
$s= \frac{15.81+18.60+20.40}{2}$
$s= \frac{54.81}{2}$
$s=27.41$

Finally, to find the area of our triangle, we are going to use Heron's formula:
$A= \sqrt{s(s-AB)(s-AC)(s-BC)}$
$A=\sqrt{27.41(27.41-15.81)(27.41-18.60)(27.41-20.40)}$
$A= \sqrt{27.41(11.6)(8.81)(7.01)}$
$A=140.13$

We can conclude that the perimeter of our triangle is 140.13 square units.