area of a triangle inside a rectangle

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High School Math Grades 9-12, Ages 13-18

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#1 h May 11, 2019, 2:06 AM • 2 Y

Y by Adventure10, Mango247

In the rectangle $ABCD, M, N, P$ and $Q$ are the midpoints of the sides. If the area of the shaded triangle is $1$ , calculate the area of the rectangle $ABCD$ .

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This post has been edited 3 times. Last edited by parmenides51, Jan 21, 2021, 9:21 PM

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#2 h May 11, 2019, 3:56 AM • 2 Y

Y by Adventure10, Mango247

Let $O$ be center of $(ABCD)$ ; $S$ be midpoint of $BP$
Denote $S_P$ is area of polygon $P$
We have: $S_{BOS} = \dfrac{OS . BN}{2} = \dfrac{NQ . BC}{16} = \dfrac{AB . BC}{16} = 1$
Then: $S_{ABCD} = 16$

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#3 h Jan 19, 2021, 2:24 PM

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$[ABCD]=4[MBNO]=8[BSN]=16[BSO]=16$ , where $O$ is the center of the rectangle and $S$ is the midpoint of $\overline{BP}$ .

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#4 h Jan 19, 2021, 3:57 PM

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Let $O$ be the center of the rectangle. It is well-known that $BP\cap MC\cap ON=X$ where $x$ is their midpoint. Then by area ratios, the shaded area is $\frac14 \cdot [MONB]$ and the rest is easy. Ofc, you could WLOG $ABCD$ is a square of sides $4$ and easily coord bash.

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#5 h Jan 21, 2021, 12:27 AM

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The shaded area is $\frac{1}{4}\cdot\frac{1}{2}\cdot\frac{1}{2}$ the area of the rectangle, so the area of the rectangle is 16*1=16

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