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In the preceding figure, ABCD ≈ HIJKL and the ratio of corresponding s
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10 Nov 2021, 07:09
10 Nov 2021, 07:09
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In the preceding figure, ABCD ≈ 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 corresponding s
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11 Nov 2021, 06:33
11 Nov 2021, 06:33
Can't post a figure here so I will leave the idea on how to approach it.
Both figures are pentagons, regular or not is not mentioned.
We can give centers to each figures, let us say O.
Then we can split each figure into 5 triangles
ΔAOB, ΔAOE, ΔEOD, and so on. Similar for other figure
Area of a triangle is 0.5 * base * height
Its given that the bases are in ratio of 4/3, same will be for the heights of each triangles
So the whole area will be in ratio of \(\frac{4}{3}*\frac{4}{3} = \frac{16}{9}\)
Both figures are pentagons, regular or not is not mentioned.
We can give centers to each figures, let us say O.
Then we can split each figure into 5 triangles
ΔAOB, ΔAOE, ΔEOD, and so on. Similar for other figure
Area of a triangle is 0.5 * base * height
Its given that the bases are in ratio of 4/3, same will be for the heights of each triangles
So the whole area will be in ratio of \(\frac{4}{3}*\frac{4}{3} = \frac{16}{9}\)
Re: In the preceding figure, ABCD HIJKL and the ratio of corresponding s
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16 Jul 2024, 22:31
16 Jul 2024, 22:31
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