In △ABC, <B=90, AB=12cm, and AC=15cm. D and E are points on AB and AC respectively such that <AED=90 and DE=3cm. Find:
i) Area of △ADE
ii) Area of quadrilateral BCED : Area of △ABC
In △ABC, <B=90, AB=12cm, and AC=15cm. D and E are points on AB and AC respectively such that <AED=90 and DE=3cm. Find:
i) Area of △ADE
ii) Area of quadrilateral BCED : Area of △ABC
$\displaystyle AB=3AD$
$\displaystyle BC=3DE$
Therefore the area of triangle ABC is 3(3)=9 times the area of triangle ADE.
Therefore the quadrilateral DBCE, being triangle ABC with triangle ADE cut off has area 9(6)-6=8(6)