Only one of two squares is known its length

Geometry Level pending

Based from figure above, □ ABCD \square \text{ABCD} and □ BEFG \square \text{BEFG} are two squares with point C \text{C} is lying on side BG ‾ \overline{\text{BG}} , point B \text{B} is lying on side AE ‾ \overline{\text{AE}} , ∣ AB ‾ ∣ = 8 cm \lvert \overline{\text{AB}} \rvert = 8 \text{ cm} , and ∣ AB ‾ ∣ < ∣ BE ‾ ∣ \lvert \overline{\text{AB}} \rvert < \lvert \overline{\text{BE}} \rvert .

Find [ △ ACF ] [\triangle \text{ACF}] in cm 2 \text{cm}^{2} .

Note that

  • ∣ AB ‾ ∣ \lvert \overline{\text{AB}} \rvert denotes the length of side AB ‾ \overline{\text{AB}} ,

  • [ △ ABC ] [\triangle \text{ABC}] denotes the area of triangle △ ABC \triangle \text{ABC} , and

  • the diagram isn't drawn to scale.


The answer is 32.

Skye Rzym by SKYE RZYM , 20, Indonesia

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1 solution

Skye Rzym
Apr 27, 2017

Let ∣ BE ‾ ∣ = x cm \lvert \overline{\text{BE}} \rvert = x \text{ cm}

Then [ △ ACF ] = [ □ ABCD ] + [ □ BEFG ] − ( [ △ ADC ] + [ △ CFG ] + [ △ AEF ] ) [\triangle \text{ACF}] = [\square \text{ABCD}] + [\square \text{BEFG}] - ([\triangle \text{ADC}] + [\triangle \text{CFG}] + [\triangle \text{AEF}]) = [ □ ABCD ] + [ □ BEFG ] − [ △ ADC ] − [ △ CFG ] − [ △ AEF ] = [\square \text{ABCD}] + [\square \text{BEFG}] - [\triangle \text{ADC}] - [\triangle \text{CFG}] - [\triangle \text{AEF}] = ∣ AB ‾ ∣ 2 + ∣ BE ‾ ∣ 2 − ∣ CD ‾ ∣ . ∣ DA ‾ ∣ 2 − ∣ FG ‾ ∣ . ∣ GC ‾ ∣ 2 − ∣ AE ‾ ∣ . ∣ EF ‾ ∣ 2 = \lvert \overline{\text{AB}} \rvert^{2} + \lvert \overline{\text{BE}} \rvert^{2} - \frac{\lvert \overline{\text{CD}} \rvert . \lvert \overline{\text{DA}} \rvert}{2} - \frac{\lvert \overline{\text{FG}} \rvert . \lvert \overline{\text{GC}} \rvert}{2} - \frac{\lvert \overline{\text{AE}} \rvert . \lvert \overline{\text{EF}} \rvert}{2} = ( 8 cm ) 2 + ( x cm ) 2 − 8 cm . 8 cm 2 − x cm . ( x − 8 ) cm 2 − ( x + 8 ) cm . x cm 2 = (8 \text{ cm})^{2} + (x \text{ cm})^{2} - \frac{8 \text{ cm } . 8 \text{ cm}}{2} - \frac{x \text{ cm } . (x-8) \text{ cm}}{2} - \frac{(x+8) \text{ cm } . x \text{ cm}}{2} = 64 cm 2 + x 2 cm 2 − 32 cm 2 − x 2 2 cm 2 + 4 x cm 2 − x 2 2 cm 2 − 4 x cm 2 = 64 \text{ cm}^{2} + x^{2} \text{ cm}^{2} - 32 \text{ cm}^{2} - \frac{x^{2}}{2} \text{ cm}^{2} + 4x \text{ cm}^{2} - \frac{x^{2}}{2} \text{ cm}^{2} - 4x \text{ cm}^{2} = 32 cm 2 + x 2 cm 2 − x 2 cm 2 = 32 \text{ cm}^{2} + x^{2} \text{ cm}^{2} - x^{2} \text{ cm}^{2} = 32 cm 2 = \boxed{32 \text{ cm}^{2}}

0 Helpful 0 Interesting 0 Brilliant 0 Confused

I get that the answer is independent of x x .

Is there a good way of phrasing the problem such that people do not solve it by fixing x = 8 x= 8 or x = 0 x = 0 ?

I like this problem, am thinking of featuring it.

Calvin Lin Staff - 4 years, 1 month ago

Maybe with a new statement : |AB|<|BG|.

SKYE RZYM - 4 years, 1 month ago

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