In the given figure , ABCD is a parallelogram. DL is perpendicular to AB and AB =AD =13 cm. If the area of parallelogram is $156 c m^{2}$ then find the length of AL. [2 MARKS]

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Solution

Formula: 0.5 Marks
Solution 1 Mark

Area of parallelogram $= B a s e \times c o r r e s p o n d i n g a l t i t u d e$

$156 = 13 \times D L$

$D L = 12 c m$

Apply Pythagoras theorem in Triangle DLA

$A D^{2} = D L^{2} + A L^{2}$

$13^{2} = 12^{2} + A L^{2}$

$A L^{2} = 25$

$A L = 5 c m$

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In the given figure , ABCD is a parallelogram. DL is perpendicular to AB and AB =AD =13 cm. If the area of parallelogram is 156 cm2 then find the length of AL. [2 MARKS]

Formula: 0.5 Marks Solution 1 Mark Area of parallelogram =Base×corresponding altitude 156=13×DL DL=12cm Apply Pythagoras theorem in Triangle DLA AD2=DL2+AL2 132=122+AL2 AL2=25 AL=5cm

In the given figure , ABCD is a parallelogram. DL is perpendicular to AB and AB =AD =13 cm. If the area of parallelogram is $156 c m^{2}$ then find the length of AL. [2 MARKS]

Formula: 0.5 Marks
Solution 1 Mark

Area of parallelogram $= B a s e \times c o r r e s p o n d i n g a l t i t u d e$

$156 = 13 \times D L$

$D L = 12 c m$

Apply Pythagoras theorem in Triangle DLA

$A D^{2} = D L^{2} + A L^{2}$

$13^{2} = 12^{2} + A L^{2}$

$A L^{2} = 25$

$A L = 5 c m$