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In the preceding figure, ABCD HIJKL and the ratio of correspondin
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26 Jan 2022, 07:26
26 Jan 2022, 07:26
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Question Stats:
66% (01:33) correct
33% (01:44) wrong based on 6 sessions
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GRe In the preceding figure, ABCD HIJKL and the ratio of correspondin.jpg [ 20.61 KiB | Viewed 1341 times ]
In the preceding figure, \(ABCD \approx HIJKL\) and the ratio of corresponding sides is \(\frac{4}{3}\). What is the ratio of the area of ABCDE to the area of HIJKL?
Express your answer as a fraction.
Show: ::
16/9
Re: In the preceding figure, ABCD HIJKL and the ratio of correspondin
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27 Jan 2022, 03:09
27 Jan 2022, 03:09
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When corresponding sides are in the proportion of \(x : y\) , the areas are in the proportion of \(x^2 : y^2\)
The sides ratio = \(4 : 3\)
The ratio of the area of ABCDE to the area of HIJKL \(= 16 : 9\)
Answer 16:9
The sides ratio = \(4 : 3\)
The ratio of the area of ABCDE to the area of HIJKL \(= 16 : 9\)
Answer 16:9
Re: In the preceding figure, ABCD HIJKL and the ratio of correspondin
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07 Oct 2024, 18:44
07 Oct 2024, 18:44
1
Okay here's a more mathematical way to solve this if anyone's wondering. Here we are only given one side for each pentagon it's implied that they're regular even if they're not let's assume they are.
now let's take one corresponding side from each pentagon AE/HL = 4/3
Now in a regular pentagon the area can be divided into 5 equilateral triangles.
let's take ae = 4x and hl =3x
now areas will be 5*equilateral triangle area : 5*root(3)/4 * a^2
ratios = 5*(root(3)/4)*a1^2 : 5*(root(3)/4)*a2^2
5*(root(3)/4)*4x^2 : 5*(root(3)/4)*3x^2
16:9 (because everything else cancels out)
now let's take one corresponding side from each pentagon AE/HL = 4/3
Now in a regular pentagon the area can be divided into 5 equilateral triangles.
let's take ae = 4x and hl =3x
now areas will be 5*equilateral triangle area : 5*root(3)/4 * a^2
ratios = 5*(root(3)/4)*a1^2 : 5*(root(3)/4)*a2^2
5*(root(3)/4)*4x^2 : 5*(root(3)/4)*3x^2
16:9 (because everything else cancels out)
