Geometry Problem. Post your solution in the comments box below.

Level: Mathematics Education, High School, Honors Geometry, College.

This entry contributed by Ajit Athle.

Click the figure below to see the complete problem 876.

## Saturday, May 18, 2013

### Problem 876: Equilateral Triangle, any Point, Perpendicular, Right Triangle Area, Sum of Areas

Labels:
area,
equilateral,
perpendicular,
right triangle,
triangle

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From the point P trace the lines parallel to the three sides of the triangle.

ReplyDeleteThe equilateral triangle is divided into three isosceles triangles and three parallelograms.

Draw AP, BP and CP. So in the equilateral triangle are formed 12 triangles congruent two by two.

S1 and S2 each contain only one of the six different triangles.Then: S1 = S2 = S / 2.

Do ponto P trace segmentos de retas paralelas aos lados passando pelo ponto P, dessa forma ficam determinados três triângulos equiláteros e três paralelogramos.

ReplyDeleteSejam respectivamente: G, H, (lado AB), I, J (lado AC) e K, L (lado BC) os pontos de interseção dos lados com os segmentos de reta.

Assim é fácil ver que a áreas de:

GDP = HDP = A1;

IFP = JFP = A2;

KEP = LEP = A3;

Dos paralelogramos temos:

HAIP => HAP = IAP = A4

JCKP => JCP = KCP = A5

LBHP => LBP = HBP = A6

S1 + S3 + S5 = A1 + A4 + A5 + A3 + A2 + A6 = A1 + A2 + A3 + A4 + A5 + A6 = S/2

Um abraço,

prof. Carlos Loureiro.