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The figure below is made up of Rectangle ABCD and Triangle AED. Shaded part x is 12 square centimetres smaller than shaded part Y. What is the length of AB?

The figure below is made up of Rectangle ABCD and Triangle AED Shaded part x is 12 square centimetres smaller than shaded part Y What is the length of AB class=

Respuesta :

1. Check the picture attached.

2. Triangles OAB and OEC are similar as they have congruent angles, as shown in the figure.

3. Similarity means that there is a scale factor of similarity, say k.

4. 
[tex] \frac{12-t}{t}=k [/tex] (corresponding sides, each in front of angles alpha)

also 

[tex] \frac{9-a}{a}=k[/tex] (corresponding sides in front of the vertec angles AOB and EOC)

so [tex]\frac{12-t}{t}=\frac{9-a}{a}[/tex]

    [tex]\frac{12}{t}-1= \frac{9}{a}-1 [/tex]

    [tex]\frac{12}{t}= \frac{9}{a}[/tex]

    [tex]\frac{4}{t}= \frac{3}{a}[/tex]
     
    [tex]t= \frac{4a}{3} [/tex]

5. 
    x is y-12, so let's find the areas x and y, in terms of a, so that we form an equation and solve for a.

x =  [tex] \frac{1}{2}at= \frac{1}{2}a* \frac{4a}{3}= \frac{ 2a^{2} }{3} [/tex]

To find the area y we need to write OC in terms of a:

OC= [tex]t*k=\frac{4a}{3}* \frac{9-a}{a}= \frac{36-4a}{3} [/tex]

so y=[tex] \frac{1}{2}*(9-a)*\frac{4(9-a)}{3}= \frac{ 2(9-a)^{2} }{3} [/tex]

6. 

x=y-12

[tex]\frac{ 2a^{2} }{3}=\frac{ 2(9-a)^{2} }{3} - \frac{36}{3} [/tex]

[tex]2a^{2}=2(81-18a+ a^{2} )-36[/tex]

[tex]2a^{2}=162-36a+ 2a^{2} -36[/tex]

[tex]36a=126[/tex]

a=126/36=3.5

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