Difference between revisions of "2001 AMC 10 Problems/Problem 24"
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==Solution 2== | ==Solution 2== | ||
Simpler is just drawing the trapezoid and then using what is given to solve. | Simpler is just drawing the trapezoid and then using what is given to solve. | ||
− | Draw a line | + | Draw a line parallel to <math>\overline{AD}</math> that connects the longer side to the corner of the shorter side. Name the bottom part <math>x</math> and top part <math>a</math>. |
− | By the Pythagorean theorem, it is obvious that <math>a^{2} + 49 = (2x+a)^{2}</math> (the RHS is the fact the two sides added together equals that). Then, we get <math>a^2 + 49 = 4x^2 + 4ax + a^2</math>, cancel out and factor and we get <math>49 = 4x(x+a)</math>. Notice that <math>x(x+a)</math> is what the question asks, so the answer is <math>boxed{\textbf{(B)}\ 12.25} </math>. | + | By the Pythagorean theorem, it is obvious that <math>a^{2} + 49 = (2x+a)^{2}</math> (the RHS is the fact the two sides added together equals that). Then, we get <math>a^2 + 49 = 4x^2 + 4ax + a^2</math>, cancel out and factor and we get <math>49 = 4x(x+a)</math>. Notice that <math>x(x+a)</math> is what the question asks, so the answer is <math>\boxed{\textbf{(B)}\ 12.25} </math>. |
Solution by IronicNinja | Solution by IronicNinja | ||
+ | |||
+ | ==Solution 3== | ||
+ | We know it is a trapezoid and that <math>\overline{AB}</math> and <math>\overline{CD}</math> are perpendicular to <math>\overline{AD}</math>. If they are perpendicular to <math>\overline{AD}</math> that means this is a right-angle trapezoid (search it up if you don't know what it looks like or you can look at the trapezoid in the first solution). We know <math>\overline{AD}</math> is <math>7</math>. We can then set the length of <math>\overline{AB}</math> to be <math>x</math> and the length of <math>\overline{DC}</math> to be <math>y</math>. <math>\overline{BC}</math> would then be <math>x+y</math>. Let's draw a straight line down from point <math>B</math> which is perpendicular to <math>\overline{DC}</math> and parallel to <math>\overline{AD}</math>. Let's name this line <math>M</math>. Then let's name the point at which line <math>M</math> intersects <math>\overline{DC}</math> point <math>E</math>. Line <math>M</math> partitions the trapezoid into ▭ <math>ADEB</math> and <math>\triangle</math> <math>BEC</math>. We will use the triangle to solve for <math>xy</math> using the Pythagorean theorem. The line segment <math>\overline{EC}</math> would be <math>y-x</math> because <math>\overline{DC}</math> is <math>y</math> and <math>\overline{DE}</math> is <math>x</math>. <math>\overline{DE}</math> is <math>x</math> because it is parallel to <math>\overline{AB}</math> and both are of equal length. Because of the Pythagorean theorem, we know that <math>(EC)^2+(BE)^2=(BC)^2</math>. Substituting the values we have we get <math>(y-x)^2+(7)^2=(x+y)^2</math>. Simplifying this we get <math>(y^2-2xy+x^2)+(49)=(x^2+2xy+y^2)</math>. Now we get rid of the <math>x^2</math> and <math>y^2</math> terms from both sides to get <math>(-2xy)+(49)=(2xy)</math>. Combining like terms we get <math>(49)=(4xy)</math>. Then we divide by <math>4</math> to get <math>(12.25)=(xy)</math>. Now we know that <math>xy\ =\ 12.25</math> which is answer choice <math>\boxed{\textbf{(B)}\ 12.25} </math>. | ||
+ | |||
+ | Solution By: MATHCOUNTSCMS25 | ||
+ | |||
+ | Fixed <math>\text{\LaTeX}</math> - Mliu630XYZ, palaashgang, anshulb, JoyfulSapling | ||
+ | |||
+ | ==Solution 4 (EZ Cheez) == | ||
+ | |||
+ | Choose any value for <math>BC</math>, and then use Pythagorean theorem to get <math>CD - AB</math>, and <math>AB = (BC - (CD - AB))/2</math>). Then multiply <math>AB \cdot CD</math>. | ||
+ | |||
+ | For example: | ||
+ | |||
+ | <math>BC=25</math>. <math>CD - AB = \sqrt{25^2 - 7^2} = 24</math>. <math>AB = (25 - 24)/2=0.5</math>. <math>CD = 0.5 + 24 = 24.5</math>. | ||
+ | <math>AB \cdot CD = (0.5)(24.5)= \boxed{\textbf{(B)}\ 12.25}</math>. | ||
==See Also== | ==See Also== |
Latest revision as of 17:58, 3 March 2024
Problem
In trapezoid ,
and
are perpendicular to
, with
,
, and
. What is
?
Solution
If and
, then
. By the Pythagorean theorem, we have
Solving the equation, we get
.
Solution 2
Simpler is just drawing the trapezoid and then using what is given to solve.
Draw a line parallel to that connects the longer side to the corner of the shorter side. Name the bottom part
and top part
.
By the Pythagorean theorem, it is obvious that
(the RHS is the fact the two sides added together equals that). Then, we get
, cancel out and factor and we get
. Notice that
is what the question asks, so the answer is
.
Solution by IronicNinja
Solution 3
We know it is a trapezoid and that and
are perpendicular to
. If they are perpendicular to
that means this is a right-angle trapezoid (search it up if you don't know what it looks like or you can look at the trapezoid in the first solution). We know
is
. We can then set the length of
to be
and the length of
to be
.
would then be
. Let's draw a straight line down from point
which is perpendicular to
and parallel to
. Let's name this line
. Then let's name the point at which line
intersects
point
. Line
partitions the trapezoid into ▭
and
. We will use the triangle to solve for
using the Pythagorean theorem. The line segment
would be
because
is
and
is
.
is
because it is parallel to
and both are of equal length. Because of the Pythagorean theorem, we know that
. Substituting the values we have we get
. Simplifying this we get
. Now we get rid of the
and
terms from both sides to get
. Combining like terms we get
. Then we divide by
to get
. Now we know that
which is answer choice
.
Solution By: MATHCOUNTSCMS25
Fixed - Mliu630XYZ, palaashgang, anshulb, JoyfulSapling
Solution 4 (EZ Cheez)
Choose any value for , and then use Pythagorean theorem to get
, and
). Then multiply
.
For example:
.
.
.
.
.
See Also
2001 AMC 10 (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
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