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ciaotaylor
19.05.2020 •
Mathematics
In un trapezio ABCD, la base maggiore AB misura 6a, la base minore CD misura 4a e l'area misura 15a^2. Detto P il punto d'intersezione delle diagonali, determina l'area di PDC
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Ответ:
PCD Δ area = 21 a²
Step-by-step explanation:
You should think in the trapezio form.
In the first step, we need to calculate the height, from the area formula:
(B + b) . h = area
(6a + 4a) . h = 15 a²
h = 15a² / 10 a = 1.5 a
So, 1.5 a + (an x value), will be the height of the big triangle (PBA).
The area for the PBA Δ will be: (CD side . h) /2 + 15 a² (trapezio area)
and PBA Δ area is also (AB side . (1.5+h)) / 2
PCD Δ area = PBA Δ area - trapezio area
(CD side . h)/2 = (AB side . (1.5+h)) / 2 - 15 a²
There you have a unique unknown value, h. Let's solve.
(4a . h)/2 = (6a . (1.5a + h)/2 - 15a²
2a . h = (9a² + 6ah) /2 - 15a²
2a .h = 4.5a² + 3a .h - 15a²
15a² - 4.5a² = 3a .h - 2a . h
10.5 a² = a .h
10.5a = h
PCD Δ area = (4a . 10.5a) / 2 = 21 a²
Ответ:
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Step-by-step explanation:
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