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# help

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In trapezoid ABCD segments AB and CD are parallel. Point P is the intersection of diagonals AC and BD. The area of ▲PAB is 16 square units, and the area of ▲PCD is 25 square units. What is the area of trapezoid ABCD?

Jan 13, 2024
edited by Hamburger  Jan 13, 2024

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A                 B

16

P

25

D                                 C

The area =  81

To see why

Triangles PAB and PDC  are similar  because angle PDC = angle PBA   and angle PCD =angle PAB

Since the area of APB  =16  and  the area of  DPC = 25....the  scale factor is sqrt [ 16/25] =4/5 = 4 : 5

So  AB = (4/5)CD      and   the height of  APB = (4)/(4 + 5)*h =  (4/9) h   (where h is the height of  the trapezoid)

Let the base  of  DPC = b  = DC.....so the base of APB = (4/5)b = AB

Using triangle PAB

Area = (1/2) (AB) ( 4/9) h

16  = (1/2)(4/5)b * (4/9) h

32  = [ 16  / 45 ] bh

32 ( 45  /16)  = bh

90 = bh

b =  90/h = DC

Then the base  of PAB  = (4/5) (DC)    =   (4/5) (90/h)  =  72/h

So  the area  of the trapezoid is

(1/2) (h)  (AB + DC)   =

(1/2) (h) ( 72/h + 90/h)  =

(1/2) ( h) (72 + 90) / h  =

(1/2) (162)   =

81

Jan 13, 2024