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 cm^{2}

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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