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Areas

1) The area of the trapezoid ABCD is equal to the sum of the areas of the two triangles adjacent to the parallel sides, a and b, plus twice their product. 2) The area of triangle S in terms of a, b, c can be expressed as (a + b + c) + (a + b + c)2 + 4ac. 3) The area of the quadrilateral AEFD, in terms of a, b, c, is ac(a + 2b + c) divided by b2 - ac.

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Marvin Kalngan
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208 views3 pages

Areas

1) The area of the trapezoid ABCD is equal to the sum of the areas of the two triangles adjacent to the parallel sides, a and b, plus twice their product. 2) The area of triangle S in terms of a, b, c can be expressed as (a + b + c) + (a + b + c)2 + 4ac. 3) The area of the quadrilateral AEFD, in terms of a, b, c, is ac(a + 2b + c) divided by b2 - ac.

Uploaded by

Marvin Kalngan
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
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Area

Question 1

A trapezium ABCD with AB//DC is A B


divided into four triangles by its
diagonals. a

Let the triangles adjacent to the parallel


sides have areas a and b. b

D C
Find the area of the trapezoid in terms of
a and b.

Question 2

A F B
ABCD is a rectangle.
b
The areas of the right angled triangles a
are a, b, c, as in the figure.
S
E
Find the area of the triangle, S,
in terms of a, b, c. c

D C

Question 3

A
ABC is a triangle.
BD and CE cut at F.
If area of ∆BEF = a, area of ∆BFC = b,
area of ∆CFD = c, find the area of the
E D
quadrilateral AEFD.
a c
F
b

C
B
Answers

Question 1 A B

Let the two diagonals AC and BD meets at E. a y


c E d
Let DE = x, BE = y
x b
c : a = x : y, b : d = x : y
D C
∴c:a=b:d
∴ ab = cd (1)
Area of ∆ACD = Area of ∆BCD
∴ c+b=d+b
∴ c=d (2)
(2) ↓ (1), c = d = ab

∴ Area of trapezium ABCD = a + b + c + d = a + b + 2 ab = ( a+ b )


2

Question 2
A F x B
Let
p = DC, q = AD b
a y
x = FB, y = BE
q S
2a 2c E
x = p− , y=q−
q p
c

1 1 2a   2c  D p C
b= xy = p −   q − 
2 2 q  p 
4ac
2b = pq − 2a − 2c −
pq
4ac
pq − 2(a + b + c) − =0
pq
(pq)2 – 2(a + b + c)(pq) – 4ac = 0

Consider only the positive root, we have:


2(a + b + c) + [ 2(a + b + c)]2 − 4(1)( −4ac)
pq = = (a + b + c) + (a + b + c) 2 + 4ac
2(1)

∴ S = pq – (a + b + c) = (a + b + c) + 4ac
2
Question 3

a+x BF b
= = A
y FD c
∴ by = ca + cx (1)

c+y CF b x y
= = E D
x FE a
∴ bx = ac + ay (2) c
a F
Solve (1), (2), we have: b

C
ac(a + b) ac( b + c) B
x= , y=
b 2 − ac b 2 − ac
ac(a + 2b + c)
∴ Area of quad. AEFD = x + y =
b 2 − ac

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