Question

The diagram shows a rectangle ABCD where AB:AD is 1:2 point E on AC is such that DE is perpendicular to AC. What ia the ratio of the area of triangle DCE to the rectangle ABCD ? whoever does it will get 50 points from me so plz help

Answer 1

Answer: 1 : 10

Step-by-step explanation:

Here, In triangles, CED and CDA,

$\angle CED \cong \angle CDA$ ( Right angles )

And, $\angle ECD\cong \angle ACD$ ( Reflexive angles )

Thus, By AA similarity postulate,

$\triangle CED\cong \triangle CDA$

Thus, By the property of similar triangle,

$\frac{\text{ Area of triangle }CED}{\text{ Area of triangle } CDA } = (\frac{CD}{CA})^2$

= $(\frac{x}{\sqrt{x^2+(2x)^2} })^2$

= $(\frac{1}{\sqrt{1+4} })^2$

= $\frac{1}{5}$

Thus the Area of triangle CED = 1/5 of area of triangle CDA,

But Area of triangle CDA = 1/2 of area of rectangle ABCD ( By the property of rectangle)

⇒ Area of triangle CED = 1/5 of 1/2 of Area of rectangle ABCD

⇒ Area of triangle CED = 1/10 of Area of rectangle ABCD

Answer 2

Answer:

1/10 is answer dear friend