Question

Two polygons are congruent and the perimeter of the first polygon is 24 cm. If the sides of the second polygon are consecutive integers (x, x+1, x+2, x+3, etc.), what value of x makes the polygons into congruent triangles? Use x as the smallest side.

Answer 1

Since we know that the polygons are triangle we know that a triangle has three sides. The problem states that the polygons are congruent which means that they have the same area and same perimeter. So the second polygon must have a perimeter of 24cm. We know that the lengths of the sides of the second polygon are x,x+1,x+2,x+3... We know that we have a triangle so we know that it has three sides of length x, x+1, and x+2, which have to equal 24 cm. So add up the sides x+(x+1)+(x+2)=24 cm. We must simplify the left hand side of the equation, which simplifies to 3x+3=24. Subtract 3 from both sides. 3x=21. Divide both sides by 3. x=7 cm. So the shortest side x is 7cm. The next side is x+1=7+1=8. And the third side is 9cm. I hope this helps.

Answer 2

thank you soo much that has help