I'm working on problem number 63, which involves finding the area of a specific triangle. My strategy was to derive equations for each side of the triangle, labeled as 2b and b. I then expressed these equations in terms of 'y'. For the rectangle's area calculation, I set 'length' equal to the expression for the 2b side and 'width' equal to the b side. After multiplying these expressions and integrating the result over the bounds [0,h], I substituted 'h' for 'y' and simplified, arriving at an answer of 4/3(b^2h). However, the textbook's solution is 2/3(b^2h). I'm trying to understand if I made an arithmetic error or took a wrong approach. The Bartleby walkthrough uses similar triangles for this problem. I'm curious if my approach could also work but maybe with a correction in my calculations.
I have attached pictures of my work and problem in the textbook.
![(3) A PIRAMID w height, in + RECTANGULAR BASE W/ Dinsusions b. +26) k
wint
Azka
(h)
kont
C
26=2
m= C-hch
y = h=y
25-€ 25
h24
y=0= = (x-26) 2y ==-2/5 + 12/1²
640
محطان
M-0-6--h
✔
&?
425 = 26-x
*- y²b-2b
y-u = -22 (x_ b) => y = = = 2 ² + 1 2
y =
y= b[b-?) __ yb= hlb-x)
b
di
y = n(26_x) —> 9²3 = h (23-x) p/x= 25-425-/ V- $ 26² [36²
26
#
3 (1+ 4
W = X = 6+ =
نتشار
h(26-2)
26
T
25
23/1-4
A-AN
Aly1=(26-426 / 6+46)
Algl-23² +2139-426² - y²26²
14
ht
14²_6-x_ Aly) = 26² = y +26² = 26 (14)
x=4b-b
2
A
(26,0)
A
5
b
0
hb
S
V. $ 25² (1-22 dy
h
K
401 40
15151
> V = 26²³ [h=37²²0]
h
V = 26² [h
C
*+24² [y=+=+=
0
CY
B
Du
dom
B](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8b313955-89cd-4b2a-a000-54fbe962adf3%2F884913f4-34e3-49b6-8f83-be37317a6965%2F6lb7gdc_processed.jpeg&w=3840&q=75)
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