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# i need help

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A rectangle contains a strip of width $1,$ as shown below.  Find the area of the strip.

Apr 30, 2024

#1
+9665
+1

Label the diagram as shown. Let $$\angle FCE = \theta$$ and $$EC = x$$.

Let H be a point on AC such that FH is perpendicular to AC. Then $$FH = 8$$.

Note that $$CE = FD = HA = x$$. Then $$CH = 10 - x$$

$$CF = \sqrt{8^2 + (10 - x)^2} = \sqrt{x^2 - 20x + 164}$$ by using Pythagoras theorem in $$\triangle CFH$$.

In $$\triangle ECG$$$$\sin \theta = \dfrac{EG}{EC} = \dfrac1x$$.

In $$\triangle FHC$$$$\sin \theta = \dfrac{FH}{CF} = \dfrac8{\sqrt{x^2 -20x+164}}$$.

So we have $$\dfrac1x = \dfrac8{\sqrt{x^2 - 20x + 164}}$$. Squaring gives $$\dfrac1{x^2} = \dfrac{64}{x^2 - 20x + 164}$$.

Rearranging,

$$64x^2 = x^2 - 20x + 164\\ 63x^2 + 20x - 164 = 0\\ x = \dfrac{-20 \pm \sqrt{20^2 - 4(63)(-164)}}{2(63)}\\ x = \dfrac{8 \sqrt{163} - 10}{63}\text{ (reject negative root)}$$

Hence, the area is $$\dfrac{8x}2 = \dfrac{32 \sqrt{163} - 40}{63}$$.

May 2, 2024

#1
+9665
+1

Label the diagram as shown. Let $$\angle FCE = \theta$$ and $$EC = x$$.

Let H be a point on AC such that FH is perpendicular to AC. Then $$FH = 8$$.

Note that $$CE = FD = HA = x$$. Then $$CH = 10 - x$$

$$CF = \sqrt{8^2 + (10 - x)^2} = \sqrt{x^2 - 20x + 164}$$ by using Pythagoras theorem in $$\triangle CFH$$.

In $$\triangle ECG$$$$\sin \theta = \dfrac{EG}{EC} = \dfrac1x$$.

In $$\triangle FHC$$$$\sin \theta = \dfrac{FH}{CF} = \dfrac8{\sqrt{x^2 -20x+164}}$$.

So we have $$\dfrac1x = \dfrac8{\sqrt{x^2 - 20x + 164}}$$. Squaring gives $$\dfrac1{x^2} = \dfrac{64}{x^2 - 20x + 164}$$.

Rearranging,

$$64x^2 = x^2 - 20x + 164\\ 63x^2 + 20x - 164 = 0\\ x = \dfrac{-20 \pm \sqrt{20^2 - 4(63)(-164)}}{2(63)}\\ x = \dfrac{8 \sqrt{163} - 10}{63}\text{ (reject negative root)}$$

Hence, the area is $$\dfrac{8x}2 = \dfrac{32 \sqrt{163} - 40}{63}$$.

MaxWong May 2, 2024
#2
+129803
0

Very nice, Max  ....this one is  somewhat difficult  !!!

Could you explain how the area is  8x / 2  ???......my small mind doesn't understand (LOL!!!)

CPhill  May 2, 2024