Let's Play With Geometry (Purple Comet 2015 High School #5)

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Middle School Math Grades 5-8, Ages 10-13, MATHCOUNTS, AMC 8

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Let's Play With Geometry (Purple Comet 2015 High School #5)

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#1 h Aug 25, 2016, 10:45 AM • 1 Y

Y by Adventure10

The two diagonals of a quadrilateral have lengths $12$ and $9$ , and the two diagonals are perpendicular to each other. Find the area of the quadrilateral.

This post has been edited 2 times. Last edited by djmathman, Jan 17, 2020, 2:55 AM

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#2 h Aug 25, 2016, 1:55 PM • 2 Y

Y by Adventure10, Mango247

The only three quadrilaterals I found that always have perpendicular diagonals are squares, rhombuses and kites. Squares have diagonals of the same length so it could only be a rhombus or a kite. For both the answer would be 12x9/2 which is 54.

This post has been edited 1 time. Last edited by lemon007, Aug 25, 2016, 4:58 PM

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algebra_star1234

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algebra_star1234

#3 h Aug 25, 2016, 2:08 PM • 2 Y

Y by Adventure10, Mango247

@above, you cannot assume that it is a kite or rhombus. You only know that the diagonals are orthodiagonal. We could have a trapezoid that is orthodiagonal or any other quadrilateral.

If they are perpendicular, then we can split it into 4 right triangles which have a total area of $\frac{9 \cdot 12}{2} = 54$

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#4 h Aug 25, 2016, 9:16 PM • 1 Y

Y by Adventure10

algebra_star1234 wrote:

@above, you cannot assume that it is a kite or rhombus. You only know that the diagonals are orthodiagonal. We could have a trapezoid that is orthodiagonal or any other quadrilateral.

If they are perpendicular, then we can split it into 4 right triangles which have a total area of $\frac{9 \cdot 12}{2} = 54$

actually, if this was a competition problem, you CAN assume it.

since there is no specification, we can pick a conveninent shape that fits the restrictions.

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