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zalisa [80]
3 years ago
5

A rectangle and a square have the same area. The width of the rectangle is 2 in less than the side of the square and the length

of the rectangle is 3 in less than twice the side of the square. What are the dimensions of the rectangle?
Mathematics
1 answer:
lina2011 [118]3 years ago
5 0
The area of a rectangle is A=LW, the area of a square is A=S^2.

W=S-2 and L=2S-3

And we are told that the areas of each figure are the same.

S^2=LW, using L and W found above we have:

S^2=(2S-3)(S-2)  perform indicated multiplication on right side

S^2=2S^2-4S-3S+6  combine like terms on right side

S^2=2S^2-7S+6  subtract S^2 from both sides

S^2-7S+6=0  factor:

S^2-S-6S+6=0

S(S-1)-6(S-1)=0

(S-6)(S-1)=0, since W=S-2, and W>0, S>2 so:

S=6 is the only valid value for S.  Now we can find the dimensions of the rectangle...

W=S-2 and L=2S-3  given that S=6 in

W=4 in and L=9 in

So the width of the rectangle is 4 inches and the length of the rectangle is 9 inches.


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In right △ABC, the altitude CH to the hypotenuse AB intersects angle bisector AL in point D. Find the sides of △ABC if AD = 8 cm
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Answer:

AC=8\sqrt{3}\ cm\\ \\AB=16\sqrt{3}\ cm\\ \\BC=24\ cm

Step-by-step explanation:

Consider right triangle ADH ( it is right triangle, because CH is the altitude). In this triangle, the hypotenuse AD = 8 cm and the leg DH = 4 cm. If the leg is half of the hypotenuse, then the opposite to this leg angle is equal to 30°.

By the Pythagorean theorem,

AD^2=AH^2+DH^2\\ \\8^2=AH^2+4^2\\ \\AH^2=64-16=48\\ \\AH=\sqrt{48}=4\sqrt{3}\ cm

AL is angle A bisector, then angle A is 60°. Use the angle's bisector property:

\dfrac{CA}{CD}=\dfrac{AH}{HD}\\ \\\dfrac{CA}{CD}=\dfrac{4\sqrt{3}}{4}=\sqrt{3}\Rightarrow CA=\sqrt{3}CD

Consider right triangle CAH.By the Pythagorean theorem,

CA^2=CH^2+AH^2\\ \\(\sqrt{3}CD)^2=(CD+4)^2+(4\sqrt{3})^2\\ \\3CD^2=CD^2+8CD+16+48\\ \\2CD^2-8CD-64=0\\ \\CD^2-4CD-32=0\\ \\D=(-4)^2-4\cdot 1\cdot (-32)=16+128=144\\ \\CD_{1,2}=\dfrac{-(-4)\pm\sqrt{144}}{2\cdot 1}=\dfrac{4\pm 12}{2}=-4,\ 8

The length cannot be negative, so CD=8 cm and

CA=\sqrt{3}CD=8\sqrt{3}\ cm

In right triangle ABC, angle B = 90° - 60° = 30°, leg AC is opposite to 30°, and the hypotenuse AB is twice the leg AC. Hence,

AB=2CA=16\sqrt{3}\ cm

By the Pythagorean theorem,

BC^2=AB^2-AC^2\\ \\BC^2=(16\sqrt{3})^2-(8\sqrt{3})^2=256\cdot 3-64\cdot 3=576\\ \\BC=24\ cm

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