In a triangle ABC, a straight line parallel to BC intersects AB and AC at point D and E respectively. If the area of ADE is one-fifth of the area of ABC and BC=10 cm, then DE equals
2√5
2 cm
4 cm
4√5
A
2 cm
B
4 cm
C
2√5
D
4√5
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Solution
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It is given that area of △ADE=15× area of △ABC
Let the height of the △ABC be h and height of the △ADC be x as shown in the diagram. Therefore, we have:
12×x×DE=15×12×h×10⇒12xDE=1010h⇒12xDE=h⇒xh=2DE
Area of trapezium DECB=Area of △ABC−Area of△ADE. Thus,
h−x2(10+DE)=45×12×h×10
⇒(1−xh)(10+DE)=8
⇒(1−2DE)(10+DE)=8
⇒10+DE−20DE−2=8
⇒DE−20DE+8=8
⇒DE−20DE=8−8
⇒DE−20DE=0⇒DE2−20DE=0
⇒DE2−20=0⇒DE2=20⇒DE=√20⇒DE=√2×2×5⇒DE=2√5
Hence, DE=2√5.
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