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08 Apr 2020, 10:57
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E
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Difficulty:
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80% (01:27) correct
20%
(01:33)
wrong
based on 116
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If ABCD is a rectangle, then what is the area of triangle ALC?
A. 2.5
B. 3
C. 3.5
D. 4
E. 4.5
Attachment:
GRE Math Q1.png [ 1.77 KiB | Viewed 7962 times ]
Archit3110
08 Apr 2020, 11:33
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Bunuel
∆ ADL = 3:4:5 ; AREA = 1/2*3*4 = 6
AND ∆ ABC = 1/2 * 7*3= 11.5
rectangle area= 7*3 =21
∆ALC = 21-(6+11.5) ; 4.5
OPTION E
11 Apr 2020, 06:13
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ABCD is a rectangle, hence angle ADL is a right angle and triangle ADL is a right angled triangle.
In triangle ADL, hypotenuse = AL = 5 and leg = AD = 3. Therefore, DL = base = 4.
ABCD is a rectangle, therefore AB = DC = 7
DC = DL + LC, so 7 = 4 + LC which gives us LC = 3.
Area of triangle ALC = ½ X LC X AD {since AD is the height of the triangle ALC} = ½ X 3 X 3 = 4.5 sq.units.
The correct answer option is E.
Remember – if the base of a triangle lies on one of the lengths of a rectangle and the third vertex lies on the other length, the width of the rectangle represents the height of the triangle.
Hope that helps!
In triangle ADL, hypotenuse = AL = 5 and leg = AD = 3. Therefore, DL = base = 4.
ABCD is a rectangle, therefore AB = DC = 7
DC = DL + LC, so 7 = 4 + LC which gives us LC = 3.
Area of triangle ALC = ½ X LC X AD {since AD is the height of the triangle ALC} = ½ X 3 X 3 = 4.5 sq.units.
The correct answer option is E.
Remember – if the base of a triangle lies on one of the lengths of a rectangle and the third vertex lies on the other length, the width of the rectangle represents the height of the triangle.
Hope that helps!
