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

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

i think the answer is b.

step-by-step explanation:

not 100% sure though.