There is a triangle with a perimeter of 63 cm, one side of which is 21 cm. Also, one of the medians is perpendicular to one of the angle bisectors. Then what you've got to do is find the side lengths of the triangle

21cm; 28cm; 14cm

Step-by-step explanation:

There is no info in the problem/s  text which one of the triangle's  side is 21 cm. That is why we have to try all possible variants.

Let the triangle is ABC . Let the AK is the angle A bisector and BM is median.

Let O is AK and BM cross point.

Have a look to triangle ABM.  AO is the bisector and AOB=AOM=90 degrees (means that AO is as bisector as altitude)

=> triangle ABM is isosceles => AB=AM  (1)

1. Let AC=21   So AM=21/2=10.5 cm

So AB=10.5 cm as well.  So BC= P-AB-AC=63-21-10.5=31.5 cm

Such triangle doesn' t exist ( is impossible), because the triangle's inequality doesn't fulfill.  AB+AC>BC ( We have 21+10.5=31.5 => AB+AC=BC)

2. Let AB=21 So AM=21 and AC=42 .So  BC= P-AB-AC=63-21-42=0 cm- such triangle doesn't exist.

3. Finally let BC=21 cm. So AB+AC= 63-21=42 cm

We know (1) that AB=AM so AC=2*AB.  So AB+AC=AB+2*AB=3*AB

=>3*AB=42=> AB=14 cm => AC=2*14=28 cm.

Let check if this triangle exists ( if the triangle's inequality fulfills).

BC+AB>AC    21+14>28 - correct=> the triangle with the sides' length 21cm,14 cm, 28cm exists.

This variant is the only possible solution of the given problem.

his first public project he is asked to create a small scale drawing. the garden is a right triangle with base 25m and height 30m. are you trying to solve for the last side/hypotenuse? the irrational form of the answer would be 5\sqrt{61}

i don't see any

step-by-step explanation: