A box is to be constructed with a rectangular base and a height of h cm. The length, l, of the box is 10 cm, and the width, w, i
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<span>The <u>correct answer</u> is:
x=64/15, y=136/15.
Explanation:
We will use the Pythagorean theorem to solve this.
For the smaller triangle, the legs of the triangle are x and 8, and the hypotenuse is 8. In the Pythagorean theorem, this gives us:
x</span>²<span>+8</span>²<span>=y</span>²
<span>
Simplifying, we have:
x</span>²<span>+64=y</span>²<span>.
For the large triangle, the legs of the triangle are y and 17, and the hypotenuse is 15+x. In the Pythagorean theorem, this gives us:
y</span>²<span>+17</span>²<span>=(15+x)</span>²<span>.
Simplifying the left hand side we have:
y</span>²<span>+289 = (15+x)</span>²<span>.
In order to simplify the right hand side, we write (15+x)</span>²<span> as the product of two binomials:
y</span>²<span>+289=(15+x)(15+x).
Multiplying the right hand side, we have:
y</span>²<span>+289=15*15+15*x+x*15+x*x
y</span>²<span>+289=225+15x+15x+x</span>²
<span>y</span>²<span>+289=225+30x+x</span>²<span>.
We can now substitute the value from the smaller triangle in place of y</span>²<span>:
x</span>²<span>+64+289=225+30x+x</span>²<span>.
Combining like terms<span>, we have:
x</span></span>²<span><span>+353=225+30x+x</span></span>²<span><span>.
Subtracting x</span></span>²<span><span> from both sides, we have:
x</span></span>²<span><span>+353-x</span></span>²<span><span>=225+30x+x</span></span>²<span><span>-x</span></span>²
<span><span>353=225+30x.
Subtract 225 from both sides:
353-225=225+30x-225
128=30x.
Divide both sides by 30:
128/30=30x/30
128/30 = x.
Both the numerator and denominator of this fraction are divisible by 2:
(128/2)/(30/2)=64/15=x.
Now we use this to find y; in our Pythagorean theorem equation from earlier, we have:
x</span></span>²<span><span>+64=y</span></span>²<span><span>.
Using our value for x, we have:
(64/15)</span></span>²<span><span>+64=y</span></span>²<span><span>.
Squaring the fraction, we have:
4096/225+64=y</span></span>²<span><span>.
In order to write 64 as a fraction using 225 as the denominator, we multiply:
(64*225)/225=14400/225.
This gives us:
4096/225+14400/225=y</span></span>²
<span><span>
Combining like terms,
18496/225=y</span></span>²<span><span>.
Taking the square root of both sides, we have:
</span></span>
<span><span>
</span></span>
x=64/15, y=136/15.
Explanation:
We will use the Pythagorean theorem to solve this.
For the smaller triangle, the legs of the triangle are x and 8, and the hypotenuse is 8. In the Pythagorean theorem, this gives us:
x</span>²<span>+8</span>²<span>=y</span>²
<span>
Simplifying, we have:
x</span>²<span>+64=y</span>²<span>.
For the large triangle, the legs of the triangle are y and 17, and the hypotenuse is 15+x. In the Pythagorean theorem, this gives us:
y</span>²<span>+17</span>²<span>=(15+x)</span>²<span>.
Simplifying the left hand side we have:
y</span>²<span>+289 = (15+x)</span>²<span>.
In order to simplify the right hand side, we write (15+x)</span>²<span> as the product of two binomials:
y</span>²<span>+289=(15+x)(15+x).
Multiplying the right hand side, we have:
y</span>²<span>+289=15*15+15*x+x*15+x*x
y</span>²<span>+289=225+15x+15x+x</span>²
<span>y</span>²<span>+289=225+30x+x</span>²<span>.
We can now substitute the value from the smaller triangle in place of y</span>²<span>:
x</span>²<span>+64+289=225+30x+x</span>²<span>.
Combining like terms<span>, we have:
x</span></span>²<span><span>+353=225+30x+x</span></span>²<span><span>.
Subtracting x</span></span>²<span><span> from both sides, we have:
x</span></span>²<span><span>+353-x</span></span>²<span><span>=225+30x+x</span></span>²<span><span>-x</span></span>²
<span><span>353=225+30x.
Subtract 225 from both sides:
353-225=225+30x-225
128=30x.
Divide both sides by 30:
128/30=30x/30
128/30 = x.
Both the numerator and denominator of this fraction are divisible by 2:
(128/2)/(30/2)=64/15=x.
Now we use this to find y; in our Pythagorean theorem equation from earlier, we have:
x</span></span>²<span><span>+64=y</span></span>²<span><span>.
Using our value for x, we have:
(64/15)</span></span>²<span><span>+64=y</span></span>²<span><span>.
Squaring the fraction, we have:
4096/225+64=y</span></span>²<span><span>.
In order to write 64 as a fraction using 225 as the denominator, we multiply:
(64*225)/225=14400/225.
This gives us:
4096/225+14400/225=y</span></span>²
<span><span>
Combining like terms,
18496/225=y</span></span>²<span><span>.
Taking the square root of both sides, we have:
</span></span>
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