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# Geometry

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In triangle ABC, angle C is a right angle. Point M is the midpoint of AB, point N is the midpoint of AC, and point O is the midpoint of AM. The perimeter of triangle ABC is 112 and ON = 15. What is the area of triangle ABC?

Nov 25, 2021

#1
+22572
+2

Since N is the midpoint of AC and O is the midpoint of BC, NO  =  ½AB.

Since M is the midpoint of AB, MB = NO = 15.

Therefore, AB = 2·MB = 30.

Since the perimeter of the triangle is 112 and AB = 30, AC + BC = 82.

Since AC2 + BC2 = AB2, AC2 + BC2 = 302​ = 900

Since  AC2 + BC2 = 900     --->     AC2 + 2·AC·BC + BC2  =  900 + 2·AC·BC

--->     (AC + BC)2  =   2·AC·BC

Since AC + BC = 52,          --->     (82)2  =  2·AC·BC

6724  =  2·AC·BC

Dividing by 4,                                 1681  =  ½·AC·BC

But, ½·AC·BC is the area of triangle(ABC).

Nov 26, 2021
#2
+1

AB is NOT equal to 30!!!     AB is the hypotenuse of triangle ABC.

btw, point O is the midpoint of AM.

Guest Nov 26, 2021
#4
+115901
+1

Thanks Gino,

You have done a lot of work there but unfortunately, you did label  O incorrectly.

$$15^2=m^2+t^2-2mt\cdot cos\alpha\\ Cos\alpha = \frac{2t}{4m}=\frac{t}{2m}\\ so\\ 225=m^2+t^2-2mt \cdot \frac{t}{2m}\\ 225=m^2\\ m=15$$

This means that  the hypotenuse is  60

and the sum of the other 2 sides is  112-60 =52

As guest has already told us very loudly  ,  this is impossible.

So such a triangle cannot exist.

Melody  Nov 27, 2021
#3
+1

This question is wrong!!!

The sum of legs MUST be greater than the length of the hypotenuse!!!

Nov 27, 2021