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The area of a triangle within a rectangle is half the area of the rectangle if they share the same base. If the triangle is required to be isosceles, the point defining the triangle's third vertex must be at the midpoint of the rectangle's other side. If the triangle must also be a right triangle, the third vertex point must be at the corner of the rectangle opposite the shared base.
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Other Link :
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Welcome1
- 2. TRIANGLE
- 3. RECTANGLE
- 4. AREA AREA OF TRIANGLE IN A RECTANGLE WITH SAME BASE
- 5. The area of the rectangle = AB×AD D A C B P Q
- 6. The area of the triangle =½bh =½AB×PQ =½AB×AD =½×area of rectangle
- 7. So where ever be the point P on CD, the area of the triangle APB is half the area of the rectangle ABCD
- 8. Problem To get an isosceles triangle of area half that of the rectangle, where must be the point P ?
- 9. P should be the mid point of CD such that AP=BP D C A B P
- 10. Problem If ABP is to be right , where must be the position of P ? P must be in the position of C D A B C P