Question
P is a point in the interior of parallelogram ABCD. If 
, what is the value of 
?
Hint:
Given , ABCD is a parallelogram with area 18 sq.cm
BC// AD the let distance between BC and AD be FG = h (height in cm)
Now find the area of parallelogram = base × height
Taking base a BC find the area and get the length of side BC
Now find the area triangles individually by ½ base × height formula and add them.
The correct answer is: 9 cm2
Ans :- 9 cm2
Explanation :-
Step 1:- Find the area of parallelogram ABCD by taking BC as base and FG as height h .
Given area of parallelogram = 18 sq.cm
Area of parallelogram = length of base BC × height =BC × h = 18 sq.cm
Step 2:-find the areas of APD
In
, we get height drawn to vertex P to base AD is PF
Let PF be a
Then area of = ½ Base × height = ½ AD × a
As we know AD = BC ( opposite sides of parallelogram are equal)
area of APD = ½ × BC × a
Step 3:-find the areas of CPB
In 
, we get height drawn to vertex P to base CB is PG
PG = FG -FP = h-a
Then area of = ½ Base × height = ½ BC × (h-a)
Step 4:- find area of + area of
area of +area of = ½ × BC × a + ½ BC × (h-a) = ½ × BC ×h
We know BC × h = 18 sq.cm
Therefore ,area of APD+area of = ½ × BC × h = ½ × 18 = 9 sq.cm.
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