Parallelogramatic

Geometry Level 2

In the given figure, ABCD is a parallelogram. If the area of the parallelogram is 156 sq cm, find the length of AL.


The answer is 5.

Arpan Ray by Arpan Ray , 17, India

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1 solution

Arpan Ray
Sep 29, 2016

Area of the parallelogram = 156 sq cm

[Area of a normal parallelogram = b a s e base X h e i g h t height ]

Base = 13 cm

Height (DL) = A r e a B a s e \frac{Area}{Base} = 156 13 \frac{156}{13} = 12 cm

DLA is a right- angle triangle.

Hypotenuse (DA) = 13cm

Altitude (DL) = 12 cm

Base (AL) = ??

{According to Pythagoras Theorem}

= ( A l t i t u d e ) (Altitude) 2 \frac{2}{} + ( B a s e ) (Base) 2 \frac{2}{} = ( H y p o t e n u s e ) (Hypotenuse) 2 \frac{2}{} [Solving it]

= ( 12 ) (12) 2 \frac{2}{} + ( A L ) (AL) 2 \frac{2}{} = ( 13 ) (13) 2 \frac{2}{}

= 144 + 144 + ( A L ) (AL) 2 \frac{2}{} = 169 = 169 [Shifting 144 from L.H.S. to R.H.S.]

= ( A L ) (AL) 2 \frac{2}{} = 169 − 144 = 169 - 144

= ( A L ) (AL) 2 \frac{2}{} = 25 = 25 [Square root on both sides]

= √ √ ( A L ) (AL) 2 \frac{2}{} = √ 25 = √25

= √ √ A L AL X A L AL = √ = √ 5 5 X 5 5

= A L = 5 c m AL = 5 cm = A n s Ans \

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