The median CD to the hypotenuse AB cuts a right triangle ABC into two

Question

The median CD to the hypotenuse AB cuts a right triangle ABC into two smaller triangles with the perimeters of 36 and 50 cm, respectively. What is the value of AC + BC ?

A. 24
B. 26
C. 32
D. 34
E. 36

$!vakumar.m · Accepted Answer

The median CD to the hypotenuse AB cuts a right triangle ABC into two smaller triangles (isosceles) with the perimeters of 36 and 50 cm, respectively.
We need to find value of AC + BC =?
Property of median of right angle triangle, \(CD = AD = BD = \frac{AB}{2} i.e. AB=2*CD \)
\(AC+CD+AD = 36\)
\(AC + AB = 36\).....(1)
\(BC+CD+BD = 50\)
\(BC + AB = 50\) ........(2)
Subtract equation 1 from 2 we get \(BC - AC = 14.\).....(3)

Also from Pythagoras theorem, \(AC^2 + BC^2 = AB^2\)
subtract \(2*AC*BC\) from both sides and re write the equation as
\(AB^2 - 2* AC*BC = (AC - BC)^2 \)
substitute the value of AC and BC from equation 1 and 2 and value AC-BC from equation 3 in above equation
\(AB^2 - 2 (36-AB)* (50-AB) = (-14)^2\)
solving the equation we get
\(AB^2 - 172*AB + 3796 = 0\)
=> \((AB-26)*(AB-146) = 0\)
AB = 26 or 146 (Very high value so can be ignored)
Now substitute the value of AB in equation 1 and 2 to get value of AC = 10 and BC = 24
Therefore, \(AC + BC = 34\)
Answer: D

PS: At 0030 hrs with a drained mind I could workout this solution which is pretty lengthy. Will request the experts to post a short and crisp solution

ABHISEKGMAT123 · Answer

I have a little doubt in the question itself. Median from the point on the hypotenuse bisects the hypotenuse. It also happens to be the circumcentre of the triangle meaning thereby the triangles ACD and BCD should have equal areas. This fact contradicts the data given in the question. Kindly explain.

$!vakumar.m · Answer

the question is flawed
because - Each median divides the triangle into two smaller triangles which have the same area.
here the area given are different

Bunuel · Answer

You are both right:

" AB cuts a right triangle ABC into two smaller triangles with the    areas    of 36 and 50 cm " should have been " AB cuts a right triangle ABC into two smaller triangles with the    perimeters    of 36 and 50 cm ".

Fixed the typo. Thank you! +1.

Bunuel · Answer

This one is still unsolved. Any body want to try ?

Bunuel · Answer

Great job! +1.

I will post OE in couple of days.

ABHISEKGMAT123 · Answer

cd=bd=ad=Circumcentre
This gives AC-BC=14
Check with options for AC+BC The correct option will be a Pythagorean triplet.
So, the answer is D
Let R be the circumradius
BC+2R=36------A
AC+2R=50------B

AC+BC=34 gives the three side as 24 10 and 26
Alternatively,
BC=36-2R
AC=50-2R
Apply pythagoras theorem now
(36-2R)^2+(50-2R)^2=(2R)^2
Putting R=13 and therefore the hypotenuse as 26 gives us a triangle 24 10 26
Therefore AC+BC=24+10=24

radhika30 · Answer

Why will areas be same. Areas will be proportional right. Median bisects the opposite side and due to similarity between two triangles the areas will be proportional

Posted from my mobile device

av1901 · Answer

A different approach

CD is the median to hypotenuse AB

So, we know that   The midpoint of a hypotenuse is equidistant from each vertex of the right triangle
=>  AD = BD = CD = x

Two triangles with perimeters 36 and 50

1)  AC + CD + AD = 36 => AC + 2x = 36 
2)  BC + CD + BD = 50 => BC + 2x = 50

Adding 1) and 2) we get

AC + BC + 4x = 86 
 AC + BC = 86 - 4x

Now, if we examine the answer choices, we can eliminate based on the options not obeying the Pythagorean formula:

A. 24: x = 15.5, AC = 5, BC = 19, AB = 31 (Not obeying Pythagorean Theorem)

B. 26: x = 15, AC = 6, BC = 20, AB = 30 (Not obeying Pythagorean Theorem)

C. 32: x = 13.5, AC = 9, BC = 23, AB = 27 (Not obeying Pythagorean Theorem)

D. 34: x = 13, AC = 10, BC = 24, AB = 26 ( AB^2 = AC^2 + BC^2 )

E. 36: x = 12.5, AC = 11, BC = 25, AB = 25 (Not obeying Pythagorean Theorem)

Answer - D

$!vakumar.m · Answer

Hi  radhika30

For any triangle, the median will always divide the triangle into two triangles with equal areas.
Think this way, the area of a triangle is   \frac{1}{2}(base* height) . 
If you consider the two equal sides of triangles as base, the height is common to both the triangles, so their areas are also same.
Hope this clears your doubt.

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The median CD to the hypotenuse AB cuts a right triangle ABC into two

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