In the diagram above, triangle ABC has a right angle at B and a perime

Was a bit tricky for me, tried to assume what the triangle could be and was able to find possible answer.

Since it is a right angled triangle assuming 3,4,5 as triplets of the triangle we find that a perimeter of 120 can be obtained by multiplying each side by 10.
So sides of triangle become AB =40; BC=30 (because AB>BC); and AC=50

Triangle ABD is similar to triangle BDC because BD is perpendicular to AC. We get,
AD*CD = BD^2 = 24^2
CD = AC-AD = 50-AD
AD*(50-AD) = 24^2
Let, AD=x
x^2 - 50x + 24^2 = 0
x^2 - 32x - 18x + 576 =0
(x-32)(x-18) = 0
=> AD= 32 or AD=18
Because AB>BC we have AD>CD
=> AD=32 and CD=18

A(△ABD) / A(△BDC) = AD*BD / CD*BD = AD/CD = 32/18 = 16/9

Answer: D

Since, the perimeter of the triangle ABC is 120

AB=x, BC=y, AC=z and AB>BC (x>y)

BD=24

x+y+z=120

Pythagorean Triplet (3,4,5 has a sum of 12) meaning thereby that we can replace x,y,z by a multiple of this triplet.

x=40, y=30 and z=50

Now that Tr (ABD) is similar to Tr (BCD)

Ratio of Area of similar triangles = ratio of square of corresponding sides
Ar (ABD)/Ar (BCD)=AB*AB/BC*BC=40*40/30*30=16/9

Therefore option D is the answer.

Let us see what we are asked.
Area of ABD=\(\frac{1}{2}*AD*BD\)
Area of CBD=\(\frac{1}{2}*CD*BD\)
Thus their ratio = Area of ABD : Area of CBD =\(\frac{1}{2}*AD*BD\): \(\frac{1}{2}*CD*BD\)=AD:CD
So what we have in options is AD/CD

Next, triangles ABD and CBD, also triangle ABC, are similar, and we will get AD*CD=BD^2.
That is so because, triangles ABD and CBD give us \(\frac{AD}{BD}=\frac{BD}{CD}\)

Thus, \(AD*CD=BD^2=24^2\)

Various ways to solve further

1) AD and CD have to be factors of 24^2, so AD/CD too should have only factors of 24^2. ONLY 16/9 possible.

2) The ratio AD/CD should come out as some square/some square. Again 16/9, which is 4^2/3^2 fits in.

3) We can take the help of perimeter given to solve.


D
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in #3 how can we use the perimeter to get the ratio?

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