What is the area of quadrilateral ABCD shown?

Area of quadrilateral ABCD = Area of Triangle ADC + Area of Triangle ABC

We need to find the hypotenuse of the triangle ADC.

\(AC^2\) = \(16^2\) + \(12^2\)
= 256 + 144
= 400
AC = 20

Area of triangle ADC = \(\frac{1}{2} \)* 16 * 12
Area of triangle ADC = 96

Area of triangle ABC. Here, we have to find the base length, BC.
We already know AC = 20

\(52^2 \)= \(20^2\) + \(BC^2\)
\(BC^2\) = \(52^2\) - \(20^2\)
BC = 48

Area of triangle ABC = \(\frac{1}{2} \)* 48 * 20
Area of triangle ABC = 480

Therefore, total area = 480 + 96
total area = 576

Answer D
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Both \(△ ADC\) and \(△ ACB\) are right angled triangle...

\(△ ADC\) is a expanded form of \(3 - 4 - 5\) triangle , So, \(AC = 20\)

So, \(BC = \sqrt{52^2 - 20^2}=48\)

Thus, Area of the figure is \(\frac{1}{2}*12*16 + \frac{1}{2}*20*48 = 96 + 480 = 576\), Answer must be (D)
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The small triangle, with legs of length 12 and 16, is just four times as big as a 3-4-5 triangle. So its hypotenuse is 20. The triangle at the top is four times as big as a 5-12-13 triangle, so the missing side (BC) is 48. Now we can just add the areas of the two triangles, or you can just use units digits: the triangle with sides 20 and 48 will have an area ending in zero, and the triangle at the bottom has an area of (12)(16)/2 = 6*16, which ends in 6, so the answer will end in 6, and must be D.

AC = (12^2 + 16^2)^1/2 = 20

BC = (52^2 - 20^2)^1/2 = 48

area of ABCD = 16*12/2 + 20*48/2 = 96 + 480 = 576

Answer D
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Is there a quick way to calculate this equation?

this is one way ( not sure if its the quickest )

\(\sqrt{52^2 - 20^2}\)
=> \(\sqrt{(52+20)(52-20)}\)
=> \(\sqrt{72*32}\)

taking the nearest perfect squares, \(\sqrt{72}=8\) and \(\sqrt{32}=6\)

so approximately, \(\sqrt{72*32}= 8*6 = 48\)

Traingle ADC is a right angled triangled which is right angled at B
The sides of the triangle forms pythogorean triplets.

AD = 12, DC = 16, therefore AC = 20

Traingle ABC is a right angled triangled which is right angled at C
The sides of the triangle forms pythogorean triplets.

AC = 20, AB = 52, therefore BC = 48

Area ABCD = Area ABC + Area ADC
= (1/2)*20*48 + (1/2)*12*16
= 480 + 96
= 576

GMAT is all about making use of opportunities that are thrown at us in term of choices, wordings, number properties etc.

Here, 12-16 as two legs of right angled triangle should tell you that you are looking at 3-4-5 triangle, so hypotenuse = 20.
Now, the area of 12-16-20 triangle is 12*16*1/2=96.

The choices have all integer values, so surely all sides of bigger triangle are integers, say a, 20, 52. The area = 20*a*1/2 =10a, where a is an integer.
The units digit of 10a will be 0, so units digit of 96+10a will be 6.
Only D fits in.

Note :- We knew the method but looking at 20 being one of the side, we avoided calculations and got to our answer.
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To visualize smaller triangle as 3-4-5 is easy but it's little difficult for the bigger one.

I just took 4 out of all the sides and then left with 3-4-5 <-- calculated for smaller triangle, so area - 1/2*3*4 = 6

Bigger triangle = 5-__-13 = 5-12- 13 and area = 1/2 5*12 = 30

Total area = (30+6) *16 = 576 (since we have taken 4 out of all sides we now need to bring it back)

With right triangles and numbers this big, the triangles are usually going to be multiples of 3-4-5 or 5-12-13 triangles, so that's the first thing I'd look for. This triangle is just four times bigger than the 5-12-13 triangle, so the missing side is 48. If you did arrive at that square root though, and wanted to compute its value, you can use the difference of squares:

\(\\
\sqrt{52^2 - 20^2} = \sqrt{(52+20)(52-20)} = \sqrt{(72)(32)} = \sqrt{(2)(36)(32)} = \sqrt{(64)(36)} = (8)(6) = 48\\
\)

With answer choices this close together, I wouldn't recommend estimating the roots (which was done in a post above, and which would be fine if the answers were more spread out).

Solution:

We see that the area of quadrilateral ABCD is the sum of the areas of right triangles ADC and ABC. Triangle ADC is a 3-4-5 right triangle with AC = 5(4) = 20 (notice that 12 = 3(4) and 16 = 4(4)). Since AC = 30, triangle ABC is a 5-12-13 right triangle with BC = 12(4) = 48 (notice that 20 = 5(4) and 52 = 13(4)). Therefore, the area of triangle ADC is ½ x 12 x 16 = 96, and the area of triangle ABC is ½ x 20 x 48 = 480. So, the area of quadrilateral ABCD is 96 + 480 = 576.

Answer: D
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