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PeepalTree
What is the area of the quadrilateral ABCD if BD=8, AB=6, and ED=5?

a. 10
b. 12
c. 14
d. 16
e. 18

Source:800 score


I am sure PeepalTree, you mean ABCE and not ABCD..
Now let's solve step wise..

Step I....
Take ∆ABD, a right angled triangle with sides in ratio 3:4:5 so AD = 2*5=10 or
\(AD^2=AB^2+BD^2=6^2+8^2=10^2......AD=10\)
Step II....
Take ∆s ABD and DEC..
AB and CE are perpendicular on BD..
So ABD and DEC are similar ..
Now ED:AD =5:10=1:2
So areas will be in ratio 1^2:2^2=1:4
Step III....
Now area of ∆ABD is AB*BD*1/2=8*6/2=24
Area of ∆DEC is 24*1/4=6.. rule of ratios of similar triangles
Area of quadrilateral / trapezium ABCE is area of ∆ABD-area of ∆DCE=24-6=18

E
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Hi Chetan,
From Triangle ABD, AE = 5.
And since opp. angles in the quadrilateral are 90 degree each, I took AB=FC = 6 and E as the mid pt. (based on the figure). Is this incorrect approach ?

Also I figured Triangle ABD and DEC are similar,

CE/AB = DE/AD

Therefore CE =3 and it satisfies the mid pt. assumption.

Area of Quadrilateral ABCF = 6x4 =24 sq. units. and area of Triangle AEF = 6 sq. units.

Therefore, area of quad. ABCE = 24 -6 = 18 (E)
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src86
Hi Chetan,
From Triangle ABD, AE = 5.
And since opp. angles in the quadrilateral are 90 degree each, I took AB=FC = 6 and E as the mid pt. (based on the figure). Is this incorrect approach ?

Area of Quadrilateral ABCF = 6x4 =24 sq. units. and area of Triangle AEF = 6 sq. units.

Therefore, area of quad. ABCE = 24 -6 = 18 (E)

No, you cannot take E as the midpoint unless it is clearly specified or you prove and then work ahead.
But a big NO to taking these estimates from the figures
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PeepalTree
What is the area of the quadrilateral ABCD if BD=8, AB=6, and ED=5?

a. 10
b. 12
c. 14
d. 16
e. 18

Source:800 score


I am sure PeepalTree, you mean ABCE and not ABCD..
Now let's solve step wise..

Step I....
Take ∆ABD, a right angled triangle with sides in ratio 3:4:5 so AD = 2*5=10 or
\(AD^2=AB^2+BD^2=6^2+8^2=10^2......AD=10\)
Step II....
Take ∆s ABD and DEC..
AB and CE are perpendicular on BD..
So ABD and DEC are similar ..
Now ED:AD =5:10=1:2
So areas will be in ratio 1^2:2^2=1:4
Step III....
Now area of ∆ABD is AB*BD*1/2=8*6/2=24
Area of ∆DEC is 24*1/4=6.. rule of ratios of similar triangles
Area of quadrilateral / trapezium ABCE is area of ∆ABD-area of ∆DCE=24-6=18

E

thank you chetan2u sorry for the typo
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PeepalTree


What is the area of the quadrilateral ABCE if BD = 8, AB = 6, and ED = 5? (Note: Figure not drawn to scale.)

A. 10
B. 12
C. 14
D. 16
E. 18


Source:800 score

Attachment:
img1.jpg


Area of rectangle = width * length = 4*6=24

Area of Right tringle = 3*4*1/2 = 6

Area of quadrilateral = 24-6 = 18
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LordStark


What is the area of the quadrilateral ABCE if BD = 8, AB = 6, and ED = 5? (Note: Figure not drawn to scale.)

A. 10
B. 12
C. 14
D. 16
E. 18


Source:800 score

Attachment:
img1.jpg


Since triangles ABD & CDE are similar

and Since AD = 10 & ED = 5
EC = 3 and BC = 4

Area of quadrilateral ABCE = 1/2 (EC + AB)*BC = 1/2 (3 + 6) * 4 = 18

IMO E
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