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geomatry problem

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geomatry problem

by sana.noor » Tue Jun 04, 2013 7:15 am
Perimeter of triangle ABC = 15. what is the area of triangle?

1) BC = 7
2) Angle ABC = 70 degree.

OA C
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Source: — Data Sufficiency |
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by vivekchandrams » Tue Jun 04, 2013 9:26 am
Can you please post the explanation.

Thanks
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by faraz_jeddah » Wed Jun 05, 2013 1:10 am
A formula I keep in handy in problems like these

Area = Sq.rt {p(p-a)(p-b)(p-c)}

where
P = 1/2 Perimiter
a,b,c are the sides.

Ok so (1) tells us that one of the sides = 7

There can be 3 possible sets of sides according to Triangle sides property

7,7,1
7,6,2
7,5,3

All the above triangles will have varying areas.
Hence (1) is Not Sufficient.
Key Take away a side cannot be greater than 7

(2) tells us that Angle B = 70 ; Not sufficient

Taking both (1) and (2) together

BC = 7
Angle B = 70

You realize that AC should be equal to BC. It cannot be greater or lesser.
Why?
To maintain 2 rules - The sum of all angles should be 180 & the 3 side property.

So BC=AC=7 with AB = 1

Now
You know all 3 sides. Plug in to the formula.

The Answer is C

Not the most convenient way of solving but it works. :lol:
Last edited by faraz_jeddah on Wed Jun 05, 2013 1:30 am, edited 1 time in total.
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by vivekchandrams » Wed Jun 05, 2013 1:28 am
I have a few doubts in what you said

1. Can't the sides be 7,4,4? You didn't mention it from those derived from statement 1

2. Explain why the triangle has to be isosceles. And what's that rule you were talking about.

Pls elaborate
faraz_jeddah wrote:A formula I keep in handy in problems like these

Area = Sq.rt {p(p-a)(p-b)(p-c)}

P = 1/2 Perimiter
a,b,c are the sides.

Ok

(1) tells us that one of the sides = 7

There can be 3 possible sets of sides according to Triangle sides property

7,7,1
7,6,2
7,5,3

All the above triangles will have varying areas.
Key Take away a side cannot be greater than 7

Hence (1) is Not Sufficient.

(2) tells us that Angle B = 70 ; Not sufficient


Taking both (1) and (2) together

BC = 7
Angle B = 70

You realize that AC should be equal to BC
Why?

To maintain 2 rules - The sum of all angles should be 180 & the 3 side property.

You know all 3 sides. Plug in to the formula.

Not the most convenient way of solving but it works.
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by faraz_jeddah » Wed Jun 05, 2013 1:45 am
The sides can be 7,4,4 but not when Angle B =70
This would mean Angle B = Angle C = 70 with Angle A = 180-140=40. Which cannot be possible because it is opposite to the greatest side. Hence A should be greatest angle.
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by vivekchandrams » Wed Jun 05, 2013 2:14 am
How about this combination:
AB = 1.4, BC = 7, CA = 6.6

This results in these angle (approx)

ABC = 70, BAC = 98, ACB = 11.

I feel this also satisfies the given 2 conditions. Please explain
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by Atekihcan » Wed Jun 05, 2013 3:40 am
A very good data sufficiency problem which tests the basic idea behind data sufficiency which is identifying whether we can solve the problem NOT actually solving it.

I think it is obvious that none of the statements are individually sufficient, so I'm directly going to deal both statements together.

Both statements together:
We can determine the area of a triangle uniquely if we can somehow fix the positions of all the three vertices of the triangle. Once all three vertices are fixed, only one such triangle is possible and it will have an unique area.

Now, taking both statements together, we know (AB + BC + AC) = 15, BC = 7, and AB is making an angle of 70° with BC.

So, length of BC is fixed, i.e. positions of B and C are fixed.
We need to determine whether, position of A is fixed or not.
Image
A must be lying on the ray extending from BC which makes an angle of 70° with BC.
Now, consider three example positions of A : A', A'', and A'''
Clearly, the more away A is from B, the more is the value of (AB + AC)
Now, (AB + AC) must be equal to 15 - 7 = 8
So, A must be having an unique position.

So, the triangle will have an unique area.
So, we can determine the area.

So, both statements together is sufficient.

Answer : C

PS : Actual calculation for complete solution of this problem needs an extensive use of properties of triangle which is way beyond the scope of GMAT.
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