What is the area of the trapezoid shown? (1) angle a = 120 degrees (2) the perimeter of trapezoid abcd = 36?

The length of a rectangle is 4 yd more than twice the width x. The area is 390

masya89 [10]

Step-by-step explanation:

step 1. let l be the length of the rectangle

step 2. l = 2x + 4

step 3. lx = 390 yd^2

step 4. (2x + 4)x = 390

step 5. 2x^2 +4x = 390

step 6. 2x^2 + 4x - 390 = 0

step 7. -4 +- sqr(4^2 -4(2)(-390))/2(2) quadratic formula

step 8. (-4 +- sqr(3136))/4

step 9. (-4 +- 56)/4

step 10. x = 13yd, l = 2x + 4 = 2(13) + 4 = 30yd.

3 0

3 years ago

In the rectangular trapezoid ABCD AC is driven by a CD (see figure).

eimsori [14]

Answer:

solution given:

let's see only in a right-angled triangle Δ ACD.

AC=3 units

AD=5 units

since Δ ACD is a right-angled triangle. It satisfies Pythagoras law

$AD^{2}=AC^{2}+CD^{2}$

25=9+ $CD^2$

$CD^2$ =25-9=16

$CD=\sqrt{16} =4 units$

Now

Area of rectangle Δ ACD= $\frac{1}{2} CD*AC=\frac{1}{2}*4*3=6\: square \:units$

similarly,

$\frac{1}{2} AD*CE=6\: square \:units\\ 5*CE=6*2\\CE=\frac{12}{5}=2.4 units$

since AB=CE=2.4 units

Δ ACE is a right-angled triangle. It satisfies Pythagoras law.

$AC^{2}=AE^{2}+CE^{2}\\3^2=AE^2+2.4^2\\9-5.76=AE^2\\AE and BC=\sqrt{1.24} =1.11 units$

now

area of trapezoid=area ofΔACD+Area of ΔABC

=6+ $\frac{1}{2}AB*BC=6+\frac{1}{2}*2.4*1.11=6+1.33 =7.33\: square \:units$

6 0

2 years ago

Find the value of x³+ y³ + z³ – 3xyz if x² + y² + z² = 83 and x + y + z = 15

raketka [301]

identity used is

x³ + y³+ z³– 3xyz = (x + y + z) (x² + y² + z² – xy – yz – zx).

then use

(x+y+z)² = x²+y²+z²+2(xy+yz+xz)

225= 83 + 2(xy+yz+xz)

xy+yz+xz = (225-83)/2

xy+yz+xz= 142/2

xy+yz+xz= 71

ok

now use identity

x³ + y³+ z³– 3xyz = (x + y + z) (x² + y² + z² – xy – yz – zx).

now

x³ + y³+ z³– 3xyz = 15 (83 – xy – yz – zx).

= 15[83 - (71)]

= 15×12

=180

4 0

3 years ago

Use 3.1 < sqrt 10 <3.2 to find the possible values of each expression. 2+sqrt 10

lilavasa [31]

The inequality which represents possible values of the expression 2+sqrt 10 by virtue of the given inequality; 3.1 < sqrt 10 < 3.2 as in the task content is; 5.1 < 2 + sqrt 10 < 5.2.

<h3>Which inequality correctly expresses the possible values of the expression; 2 + √10 as required in the task content?</h3>

It follows from the task content that the expression given is;

3.1 < sqrt 10 < 3.2

Since the given premises is an inequality, it follows that adding the same number to all parts of the inequality stills holds the inequality true.

Hence by adding 2 to all parts of the inequality, we have;

2 + 3.1 < 2 + sqrt 10 < 2 + 3.2

Therefore, we have;

5.1 < 2 + sqrt 10 < 5.2

Ultimately, 5.1 < 2 + sqrt 10 < 5.2 represent the possible values of the expression 2+sqrt 10 as given by the inequality 3.1 < sqrt 10 < 3.2.