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Difficult geometry problem
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anshumishra
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Should be Bgaruhape wrote:I'm not sure how to solve it. The explanation is very difficult. I hope that someone can give an easy one.
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Check out the solution :-
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Anshu
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Night reader
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hi, let me attach less attractive pic here 
take your ABC square as height*base/2 where height is x (unknown), base is BC --> x*(2+1)/2
3x/2=30 (this one is given as square of ABC), x=20 ok? so we know the height -HUREY
now switch to ADC - we need height here, and it doesn't matter where the height falls on line (i.e. at which point, what matters is the direction of height towards our base in ADC, base here is DC), we have height=20
20*1/2=10
answer B oppps
take your ABC square as height*base/2 where height is x (unknown), base is BC --> x*(2+1)/2
3x/2=30 (this one is given as square of ABC), x=20 ok? so we know the height -HUREY
now switch to ADC - we need height here, and it doesn't matter where the height falls on line (i.e. at which point, what matters is the direction of height towards our base in ADC, base here is DC), we have height=20
20*1/2=10
answer B
oppps
garuhape wrote:I'm not sure how to solve it. The explanation is very difficult. I hope that someone can give an easy one.
Thanks
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GMATGuruNY
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Since the only conditions are that BD=2, that DC=1, and that the area of ABC = 30, we can redraw ABC as a right triangle that satisfies all the conditions given:garuhape wrote:I'm not sure how to solve it. The explanation is very difficult. I hope that someone can give an easy one.
Thanks
Area of ADC = 1/2*b*h = 1/2*1*20 = 10.
The correct answer is B.
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GMATGuruNY
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A helpful take-away:
When a triangle is split into two smaller triangles by drawing a line segment from one vertex to the opposite side:
The ratio of the areas of the two smaller triangles = the ratio of the lengths of the two line segments created in the opposite side.
Here's proof:
The figure above shows that Area of ADB: Area of ADC = BD:DC = x:y.
When a triangle is split into two smaller triangles by drawing a line segment from one vertex to the opposite side:
The ratio of the areas of the two smaller triangles = the ratio of the lengths of the two line segments created in the opposite side.
Here's proof:
The figure above shows that Area of ADB: Area of ADC = BD:DC = x:y.
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Night reader
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my small takeways:
- since the square of triangle is the function of height*base/2, the triangles positioned in the same plan (XY) can be measured by the height property and the baseline angles;
- for the triangles positioned inf different planes (XYZ), the projections of the height of triangles and their baseline angles -90` could be helpful for comparing their squares.
- since the square of triangle is the function of height*base/2, the triangles positioned in the same plan (XY) can be measured by the height property and the baseline angles;
- for the triangles positioned inf different planes (XYZ), the projections of the height of triangles and their baseline angles -90` could be helpful for comparing their squares.
My knowledge frontiers came to evolve the GMATPill's methods - the credited study means to boost the Verbal competence. I really like their videos, especially for RC, CR and SC. You do check their study methods at https://www.gmatpill.com
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Night reader
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i think it's quite possible since the vertices from where the heights fall onto bases indicate that the heights are equal, so we just keep the ratio of bases for deducing the square of triangle ADC
garuhape wrote:Thanks Guys,
I had the following idea: we know that the base is split into two parts (1/3 and 2/3). Can't we therefore just split the are by this ratio?
Cheers
My knowledge frontiers came to evolve the GMATPill's methods - the credited study means to boost the Verbal competence. I really like their videos, especially for RC, CR and SC. You do check their study methods at https://www.gmatpill.com
