Difference between revisions of "1966 AHSME Problems/Problem 14"
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Draw the rectangle <math>ABCD</math> with <math>AB</math> = <math>5</math> and <math>AD</math> = <math>3</math>. We created our diagonal, <math>AC</math> and use the Pythagorean Theorem to find the length of <math>AC</math>, which is <math>\sqrt34</math>. Since <math>EF</math> breaks the diagonal into <math>3</math> equal parts, the lenght of <math>EF</math> is <math>\frac {\sqrt34}{3}</math>. The only other thing we need is the height of <math>BEF</math>. Realize that the height of <math>BEF</math> is also the height of right triangle <math>ABC</math> using <math>AC</math> as the base. The area of <math>ABC</math> is <math>\frac{15}{2}</math> (using the side lengths of the rectangle). The height of <math>ABC</math> to base <math>AC</math> is <math>\frac{15}{2}</math> divided by <math>\sqrt34 \cdot 2</math> (remember, we multiply by <math>2</math> because we are finding the height from the area of a triangle which is <math>\frac{bh}{2}</math>). That simplfies to <math>\frac{15}{\sqrt34}</math> which equal to <math>\frac{15 \cdot \sqrt34}{34}</math>. Now doing all the arithmetic, <math>\frac {\sqrt34}{3} \cdot \frac{15 \cdot \sqrt34}{34}</math> = <math> \frac {5 \cdot 3}{3 \cdot 2}</math> = <math>\frac {5}{2}</math>. | Draw the rectangle <math>ABCD</math> with <math>AB</math> = <math>5</math> and <math>AD</math> = <math>3</math>. We created our diagonal, <math>AC</math> and use the Pythagorean Theorem to find the length of <math>AC</math>, which is <math>\sqrt34</math>. Since <math>EF</math> breaks the diagonal into <math>3</math> equal parts, the lenght of <math>EF</math> is <math>\frac {\sqrt34}{3}</math>. The only other thing we need is the height of <math>BEF</math>. Realize that the height of <math>BEF</math> is also the height of right triangle <math>ABC</math> using <math>AC</math> as the base. The area of <math>ABC</math> is <math>\frac{15}{2}</math> (using the side lengths of the rectangle). The height of <math>ABC</math> to base <math>AC</math> is <math>\frac{15}{2}</math> divided by <math>\sqrt34 \cdot 2</math> (remember, we multiply by <math>2</math> because we are finding the height from the area of a triangle which is <math>\frac{bh}{2}</math>). That simplfies to <math>\frac{15}{\sqrt34}</math> which equal to <math>\frac{15 \cdot \sqrt34}{34}</math>. Now doing all the arithmetic, <math>\frac {\sqrt34}{3} \cdot \frac{15 \cdot \sqrt34}{34}</math> = <math> \frac {5 \cdot 3}{3 \cdot 2}</math> = <math>\frac {5}{2}</math>. | ||
− | <math>\fbox{C}</math> | + | == Solution 2== |
+ | Draw the rectangle <math>ABCD</math> with <math>AB</math> = <math>5</math> and <math>AD</math> = <math>3</math>. Next draw a height from vertex <math>B</math> to diagonal <math>AC</math> intersecting at <math>G</math> and similarly from <math>D</math> to <math>AC</math> intersecting at <math>J</math>. Let's call this height <math>h</math> and we can notice that <math>BG</math> = <math>DJ</math> = <math>h</math>. Since the points <math>E</math> and <math>F</math> are trisecting diagonal <math>AC</math> the bases and heights of triangles <math>BEA</math>, <math>BEF</math>, <math>BEC</math>, <math>DEA</math>, <math>DEF</math>, and <math>DEC</math> have the same areas. Hence the area of one of these triangles such as <math>BEF</math> is <math>\frac{5 \cdot 3}{2}</math> = <math>\frac {5}{2}</math>. So our answer is <math>\fbox{C}</math>. | ||
== See also == | == See also == |
Revision as of 22:58, 12 December 2019
Contents
Problem
The length of rectangle is 5 inches and its width is 3 inches. Diagonal is divided into three equal segments by points and . The area of triangle , expressed in square inches, is:
Solution
Draw the rectangle with = and = . We created our diagonal, and use the Pythagorean Theorem to find the length of , which is . Since breaks the diagonal into equal parts, the lenght of is . The only other thing we need is the height of . Realize that the height of is also the height of right triangle using as the base. The area of is (using the side lengths of the rectangle). The height of to base is divided by (remember, we multiply by because we are finding the height from the area of a triangle which is ). That simplfies to which equal to . Now doing all the arithmetic, = = .
Solution 2
Draw the rectangle with = and = . Next draw a height from vertex to diagonal intersecting at and similarly from to intersecting at . Let's call this height and we can notice that = = . Since the points and are trisecting diagonal the bases and heights of triangles , , , , , and have the same areas. Hence the area of one of these triangles such as is = . So our answer is .
See also
1966 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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All AHSME Problems and Solutions |
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