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If in triangle \(ABC\) angle \(ABC\) is the largest and point \(D\) lies on segment \(AC\), is the area of triangle \(ABD\) larger than that of triangle \(DBC\)?

It becomes easier if you draw a sketch for problems like this one.

I'm attaching a sketch that I draw. So, we have two triangles ABD and CBD. BE is the height of both these smaller triangles as well as of the bigger triangle ABC.

We know that a formula for finding the area of a triangle is \(\frac{1}{2}*base*height\). In our case the height is the same for both triangles (BE on the image below). So we need to know which of the bases is longer in order to tell which triangle has the greater area.

S1 explicitly states that AD<DC. This is sufficient to know that the area of triangle ABD is smaller than that of triangle CBD. Since the height is the same, the longer base gives us the greater area.

S2 is not sufficient because the lengths of AB and BC don't influence the area in this case. Imagine point D sliding to the right closer to the point C - this will increase the area of triangle ABD and decrease the area of triangle CBD. Notice that if we shift point D this way, the lengths of AB and BC don't change. Therefore, we don't really care about these values (AB and BC).

Angle ABC is obtuse in the image I attached above. S2 can't be sufficient, see explanation:

Quote:

S2 is not sufficient because the lengths of AB and BC don't influence the area in this case. Imagine point D sliding to the right closer to the point C - this will increase the area of triangle ABD and decrease the area of triangle CBD. Notice that if we shift point D this way, the lengths of AB and BC don't change. Therefore, we don't really care about these values (AB and BC).

BC can't be the base as it's clearly stated in the question stem that point D lies on AC and we are dealing with triangles ABD and DBC. Can you draw a sketch to demonstrate how BC would be the base under the conditions given in the stem? I hope I'm not missing anything big here .

BarneyStinson wrote:

dzyubam wrote:

It becomes easier if you draw a sketch for problems like this one.

consider, BC is base and angle ABC is obtuse. Because, we can't be sure exactly how the figure looks like, we should take all possibilities in to consideration and I am guessing, both statements can help assert the required condition.\

Stmt1: If the AD = DC then both the triangle are equal and as shown in the picture both share the ame altitude and area is determined by which base is larger among the two. - Sufficient.

Stmt2: if AB < BC : the deciding factor still will be the proximity of D to A. If it is farther away from A, triangle ABD is greater.. Since no information about D is given - not Sufficient

If in triangle \(ABC\) angle \(ABC\) is the largest and point \(D\) lies on segment \(AC\), is the area of triangle \(ABD\) larger than that of triangle \(DBC\)?

(1) \(AD<DC\) (2) \(AB<BC\)

Consider the diagram below:

Attachment:

m18-04.PNG [ 4.77 KiB | Viewed 3755 times ]

Notice that \(BE\) is the height of the triangle \(ABC\). Now, the area of triangle \(ABD\) is \(\frac{1}{2}*height*base=\frac{1}{2}*BE*AD\) and the area of triangle \(DBC\) is \(\frac{1}{2}*height*base=\frac{1}{2}*BE*DC\). So, we can see that the area of triangle \(ABD\) will be greater than the area of triangle \(DBC\) if \(AD\) is greater than \(DC\).

(1) \(AD<DC\). Sufficient. (2) \(AB<BC\). Not sufficient.

If in triangle \(ABC\) angle \(ABC\) is the largest and point \(D\) lies on segment \(AC\), is the area of triangle \(ABD\) larger than that of triangle \(DBC\)?

(1) \(AD<DC\) (2) \(AB<BC\)

Consider the diagram below:

Attachment:

m18-04.PNG

Notice that \(BE\) is the height of the triangle \(ABC\). Now, the area of triangle \(ABD\) is \(\frac{1}{2}*height*base=\frac{1}{2}*BE*AD\) and the area of triangle \(DBC\) is \(\frac{1}{2}*height*base=\frac{1}{2}*BE*DC\). So, we can see that the area of triangle \(ABD\) will be greater than the area of triangle \(DBC\) if \(AD\) is greater than \(DC\).

(1) \(AD<DC\). Sufficient. (2) \(AB<BC\). Not sufficient.

Answer: A.

Thanks Bunuel, everyone, very helpful.

One question, however, can you discuss the logic in creating line BE? Unfortunately, I don't "see the light", not fully-understanding, why we need to add an extra line there? Is it because we can't assume that BD is the height (and stemming from not completely grasping the principles of a triangle)?

It becomes easier if you draw a sketch for problems like this one.

consider, BC is base and angle ABC is obtuse. Because, we can't be sure exactly how the figure looks like, we should take all possibilities in to consideration and I am guessing, both statements can help assert the required condition.\

I sketched the diagram of the triangle and was able to arrive at a solution using the first statement, but the second statement alone is insufficient to arrive at an answer no matter how you draw it. I'm guessing the answer is A.

It becomes easier if you draw a sketch for problems like this one.

consider, BC is base and angle ABC is obtuse. Because, we can't be sure exactly how the figure looks like, we should take all possibilities in to consideration and I am guessing, both statements can help assert the required condition.\

My answer to the question is D.

You have a good point but even with angle ABC obtuse, the area of triangles ABD and BDC cannot be compared if we do not get any info on D's location. S1 clarifies the location and compares the length of the bases of these triangles, so only answer is A!

We need to figure out if Triangle ABD > Triangle DBC??

Based on s1 we know that base of Triangle DBC (i.e. DC) > base of Triangle ABD (i.e. AD). Now Area of triangle DBC is 1/2*BD*DC----(1) and Area of triangle ABD is 1/2*BD*AD-----(2) As in both equations DC and AD are the only variables and based on s1 DC > AD therefore triangle ABD is NOT GREATER than triangle DBC. Hence S1 is SUFFICIENT.

Based on s2 AB < BC. This makes no difference in the formula of the area of both the triangles (i.e. triangle ABD and triangle DBC). Therefore S2 is NOT SUFFICIENT

I solved by median approach- if d was mid point of side ac then the position of D decides whether smaller triangles are equal in area or unequal. There is an official guide 12- DS-109 is based on similar lines.

I had a small question if anyone can answer- regarding the OA here - for any triangle abc, which has d as the point on BC, will the height of all 3 triangles be equal ?? {2 smaller and 1 larger triangle} _________________

Cheers !!

Quant 47-Striving for 50 Verbal 34-Striving for 40

My answer is D Statement 1: DC>AD since area of a triangle is 1/2*base*ht. Since both the traingles have the same height. So the triangle which has the larger base will have the largest area. Statement 2: A continuation of what is stated in the stimulus. Does provide any info for the area of triangle.

If in triangle \(ABC\) angle \(ABC\) is the largest and point \(D\) lies on segment \(AC\), is the area of triangle \(ABD\) larger than that of triangle \(DBC\)?

(1) \(AD<DC\) (2) \(AB<BC\)

Consider the diagram below:

Attachment:

m18-04.PNG

Notice that \(BE\) is the height of the triangle \(ABC\). Now, the area of triangle \(ABD\) is \(\frac{1}{2}*height*base=\frac{1}{2}*BE*AD\) and the area of triangle \(DBC\) is \(\frac{1}{2}*height*base=\frac{1}{2}*BE*DC\). So, we can see that the area of triangle \(ABD\) will be greater than the area of triangle \(DBC\) if \(AD\) is greater than \(DC\).

(1) \(AD<DC\). Sufficient. (2) \(AB<BC\). Not sufficient.

Answer: A.

Thanks Bunuel, everyone, very helpful.

One question, however, can you discuss the logic in creating line BE? Unfortunately, I don't "see the light", not fully-understanding, why we need to add an extra line there? Is it because we can't assume that BD is the height (and stemming from not completely grasping the principles of a triangle)?

Thanks in advance for your insight

Yes, we cannot assume that BD is the height.

We are asked to compare the areas of the two triangles and for that we need the height. _________________