ABCD is a parallelogram in which BC is produced to E such that CE = BC (Fig. 9.17). AE intersects CD at F. If ar (DFB) = 3 cm2, find the area of the parallelogram ABCD.
Solution:
Given, ABCD is a parallelogram
BC is produced to E such that CE = BC
C is the midpoint of BE
AE intersects CD at F
ar(DFB) = 3 cm²
We have to find the area of parallelogram ABCD
We know that the area of triangles on the same base and between the same parallel lines are equal.
Triangles ADF and DFB are on the same base DF and between the same parallel lines CD and AB.
So, ar(ADF) = ar(DFB)
ar(ADF) = 3 cm²
Considering triangle ABE,
Since C is the midpoint of BE
By converse of midpoint theorem,
EF = AF ------------------------ (2)
Considering triangles ADF and ECF,
We know that the vertically opposite angles are equal.
∠AFD = ∠CFE
EF = AF
Since BE || AD and AE is the transversal.
We know that the alternate interior angles are equal.
∠DAF = ∠CEF
By ASA criteria, the triangles ADF and ECF are congruent.
We know that congruent triangles have equal area
So, ar(CFE) = ar(ADF)
ar(CFE) = 3 cm²
Considering triangle BFE,
Since C is the midpoint of BE
CF is the median of triangle BFE.
We know that the median of a triangle divides it into two triangles of equal areas.
ar(CEF) = ar(BFC)
So, ar(BFC) = 3 cm²
From the figure,
ar(BDC) = ar(BDF) + ar(BFC)
ar(BDC) = 3 + 3
ar(BDC) = 6 cm²
We know that the diagonal of a parallelogram divides it into two congruent triangles of equal areas.
Area of parallelogram ABCD = 2(Area of triangle BDC)
= 2(6)
= 12 cm²
Therefore, the area of the parallelogram ABCD is 12 cm².
✦ Try This: D, E and F are the mid-points of the sides BC, CA and AB, respectively of an equilateral ΔABC. Show that ΔDEF is also an equilateral triangle.
☛ Also Check: NCERT Solutions for Class 9 Maths Chapter 9
NCERT Exemplar Class 9 Maths Exercise 9.3 Problem 7
ABCD is a parallelogram in which BC is produced to E such that CE = BC (Fig. 9.17). AE intersects CD at F. If ar (DFB) = 3 cm2, find the area of the parallelogram ABCD
Summary:
ABCD is a parallelogram in which BC is produced to E such that CE = BC (Fig. 9.17). AE intersects CD at F. If ar (DFB) = 3 cm2 , the area of the parallelogram ABCD is 12 cm²
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