1 thought on “Draw triangle of sides 4,5,7 and cm and draw 3 right triangles of the same area”
Answer:
Step-by-step explanation:
Draw a line, let us say, BC of the given length. (Let us say 3 cm)
With B as center, draw an arc with radius 4 (or 6) and with C as center draw another arc with the remaining dimension, that is 6 (or 4). The point where the arcs will intersect would be point A. ABC would be the given triangle.
Through A, draw a line parallel to BC. Draw a line from B perpendicular to BC meeting the parallel line drawn through A at D. If the perpendicular is drawn through C it may meet the parallel line at E. DBC and EBC are the two right angles with the same area as ABC. Each of the right angled triangle will be mirror image of the other.
Repeat the process with BC equal to 4 and the arc lengths of 3 and 6. The same triangle ABC will still be obtained, but the two new right angles will be different since base has changed from 3 to 4 consequently changing the height too.
Repeat the process once more with BC as 6 and arc lengths of 3 and 4. Again there would be the same triangle ABC, but two more right angled triangles.
Thus there will be 3 right angled triangles with the same area and 3 more mirror images of the same.
Answer:
Step-by-step explanation:
Draw a line, let us say, BC of the given length. (Let us say 3 cm)
With B as center, draw an arc with radius 4 (or 6) and with C as center draw another arc with the remaining dimension, that is 6 (or 4). The point where the arcs will intersect would be point A. ABC would be the given triangle.
Through A, draw a line parallel to BC. Draw a line from B perpendicular to BC meeting the parallel line drawn through A at D. If the perpendicular is drawn through C it may meet the parallel line at E. DBC and EBC are the two right angles with the same area as ABC. Each of the right angled triangle will be mirror image of the other.
Repeat the process with BC equal to 4 and the arc lengths of 3 and 6. The same triangle ABC will still be obtained, but the two new right angles will be different since base has changed from 3 to 4 consequently changing the height too.
Repeat the process once more with BC as 6 and arc lengths of 3 and 4. Again there would be the same triangle ABC, but two more right angled triangles.
Thus there will be 3 right angled triangles with the same area and 3 more mirror images of the same.