Question

Can anyone solve the 19th one?

Answer 1

$If\ DE\ ||\ BC\ then\ \Delta ACB\sim\Delta AED.$

$\trxt{We know}\ AD:DB=4:1\to AD=4x,\ DB=1x\ (x-unit\ of\ length)\\\\\text{therefore}\ AB=4x+1x=5x\to AB:AD=5:4.\\\\\Delta ACB\sim\Delta AED\ in\ scale\ k=\dfrac{5}{4}.\\\\\text{We know}:\\\text{The ratio of the area of similar figures is equal to the square of the scale.}\\\\\dfrac{A_{\Delta ACB}}{A_{\Delta AED}}=\left(\dfrac{5}{4}\right)^2=\dfrac{25}{16}\\\\A_{\Delta ACB}=A_{\Delta AED}+A_{BCED}\\\\A_{BCED}=225\ cm^2\\\\\text{Therefore}$

$\dfrac{A_{\Delta ACB}}{A_{\Delta AED}}=\dfrac{A_{\Delta AED}+225}{A_{\Delta AED}}\\\\\text{The equation}:\\\\\dfrac{A_{\Delta AED}+225}{A_{\Delta AED}}=\dfrac{25}{16}\qquad|\text{cross multiply}\\\\16(A_{\Delta AED}+225)=25A_{\Delta AED}\qquad|\text{use distributive property}\\\\16A_{\Delta AED}+(16)(225)=25A_{\Delta AED}\\\\16A_{\Delta AED}+3600=25A_{\Delta AED}\qquad|-16A_{\Delta AED}\\\\3600=9A_{\Delta AED}\qquad|:8\\\\A_{\Delta AED}=400\ (cm^2)\\\\Answer:\ Area\ of\ \triangle ADE=400\ cm^2$

Answer 2

Answer:

ADE = 400

Step-by-step explanation:

I did this a slightly different way.

Every part of triangle ADE is in a 4 to 5 ratio with every corresponding part of Triangle ABC.  Let us suppose that the underlying length of AD is 4*x and the length of AB = 5x

That means that the height of Triangle ADE is in the ratio of 4x  to 5x for Triangle ABC

The base of ADE is 4y to 5y for Triangle ABC

The area of Triangle ADE = 1/2 4x * 4y

The area of Triangle ABC = 1/2 5x * 5y

The difference between these two triangles is

1/2 * 5x * 5y - 1/2 * 4x * 4y = 225   Multiply through by 2 to get rid of the 1/2s

25xy - 16xy = 450                          Combine the like terms on the left

9xy = 450                                       Divide by 9 on both sides.

9xy/9 = 450/9

xy = 50

The area of ADE = 1/2 * 16 xy = 1/2 * 16 * 50 = 400