[ASK] Find the ratio of the area of triangle BCH and triangle EHD

In summary, a parallelogram ABCD with angle A = angle C = 45° is intercepted by circle K with center C through points B and D. The extended line AD also intercepts the circle at point E and the line BE intercepts CD at point H. The ratio of the area of triangle BCH to the area of triangle EHD is 1:2, with the bases of the triangles being equal to the side length of the rhombus and the perpendicular height of the parallelogram and the corresponding bases of the triangles being r√2 and h√2.
  • #1
Monoxdifly
MHB
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A parallelogram ABCD has angle A = angle C = 45°. Circle K with the center C intercept the parallelogram through B and D. AD is extended so that it intercepts the circle at E and BE intercepts CD at H. The ratio of the area of triangle BCH and triangle EHD is ...

Here I got that triangle BCH and triangle EHD is similar with angle BCH = angle HDE = 45°, angle CHB = angle DHE = 112.5°, and angle CBH = angle DEH = 22.5°. The area of triangle BCH is ½ × CH × h, where h is the parallelogram's height. The area of triangle EHD is ½ × (r - CH) × r. I stuck at the ratio is (CH × h) : ((r - CH) × r). How do I simplify that? Problem is, I don't know CD's length, so I can't approximate the ratio between CH and DH.
 
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  • #2
A parallelogram ABCD has angle A = angle C = 45°. Circle K with the center C intercept the parallelogram through B and D. AD is extended so that it intercepts the circle at E and BE intercepts CD at H. The ratio of the area of triangle BCH and triangle EHD is ...

Note parallelogram ABCD is a rhombus with side length $r$ ...

(area of BCH)/(area of EHD) = $\dfrac{r^2}{(r\sqrt{2})^2} = \dfrac{1}{2}$
 
  • #3
How did you get area of BCH and area of EHD?
 
  • #4
Monoxdifly said:
How did you get area of BCH and area of EHD?

I didn't get the areas ... I determined the ratio of their respective areas.

From the diagram, let triangle BCH have base r with a perpendicular height h. Since triangle EHD is similar, its corresponding base is r√2 with corresponding height h√2.

$\dfrac{\text{area of BCH}}{\text{area of EHD}} = \dfrac{\frac{1}{2}r \cdot h}{\frac{1}{2} r\sqrt{2} \cdot h\sqrt{2}}$

now ... simplify the right side of the above equation.
 
  • #5
1 : 2
 

FAQ: [ASK] Find the ratio of the area of triangle BCH and triangle EHD

1. What is the formula for finding the area of a triangle?

The formula for finding the area of a triangle is A = (1/2)bh, where A represents the area, b represents the base, and h represents the height.

2. How do I find the base and height of a triangle?

The base and height of a triangle can be found by measuring the length of the two sides that meet at a right angle. The side opposite the right angle is the height, and the other side is the base.

3. Can the ratio of the area of two triangles be greater than 1?

Yes, the ratio of the area of two triangles can be greater than 1. This would occur when the two triangles have different base and height measurements.

4. How do I find the ratio of the area of two triangles?

To find the ratio of the area of two triangles, divide the area of one triangle by the area of the other triangle. In this case, the ratio of the area of triangle BCH and triangle EHD would be (area of triangle BCH) / (area of triangle EHD).

5. What does the ratio of the area of two triangles represent?

The ratio of the area of two triangles represents the relationship between their sizes. It can also be interpreted as the number of times one triangle can fit into the other. For example, a ratio of 2 would mean that the second triangle is twice as large as the first triangle.

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