A triangle has a height that is 4 /3 the length of its base. If the length of its base is 24 cm, what is its area?
2 answers:
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Answer:
- 60°
Step-by-step explanation:
<em>Refer to attached picture</em>
Given rectangle ABCD. Its folded shape is given as well.
It has sides a and b and diagonal d, with the angle BCD marked as α.
We need to find the value of angle AOC or BOD.
This is going to be 2α, which is easy to find out. BOD is cut by the angle bisector, each of the angles is α.
Let the area of the rectangle is A, then the folded shape has area 2/3A.
<u>The area of rectangle ABCD:</u>
- A = ab
<u>From the triangle CBD we get:</u>
- a/d = sin α ⇒ a = d sin α
- b/d = cos α ⇒ b = d cos α
<u>Then</u>
- A = d sin α * d cos α = d²sin α cos α
Now, lets find the area of the folded shape.
It is the area of the rectangle minus the area of the triangle CMB as this part is lost as part of overlap.
<u>Area of the triangle:</u>
- A = 1/2dh
<u>Since h/(d/2) = tan α ⇒ h = 1/2d tan α, the area is:</u>
- A = 1/2d*1/2d tan α = 1/4d² tan α
<u>We have the difference of the areas which is 1/3 of the area of the rectangle:</u>
- 1/3(d²sin α cos α) = 1/4d² tan α
- sin α cos α = 3/4 tan α
- sin α cos α = 3/4 sin α / cos α
- cos² α = 3/4
- cos α = √(3/4)
- cos α = √3/2
- α = arccos (√3/2)
- α = 30°
<u>The acute angle formed by the two diagonals of the rectangle is:</u>
- 2α = 2*30° = 60°
Answer: the system has no solution.
Step-by-step explanation:
\displaystyle\\
\left \{ {{x^2y=16\ \ \ \ \ (1)} \atop {x^2+4y+16=0\ \ \ \ \ (2)}} \right. .\\
Multiply\ both\ sides\ of\ the\ equation\ (2)\ by\ y\ (y\neq 0):\\
x^2y+4y^2+16y=0\\
We\ substitute\ equation\ (1)\ into\ equation\ (2):\\
16+4y^2+16y=0\\
4y^2+16y+16=0\\
4*(y^2+4y+4)=0\\
4*(y^2+2*y*2+2^2)=0\\
4*(y+2)^2=0\\
Divide\ both\ sides\ of\ the \ equation\ by\ 4:\\
(y+2)^2=0\\
(y+2)*(y+2)=0\\
So,\ y+2=0\\
y=-2.\\