Q14 of 109 Page 330

In the given figure ABCD, DCFE and ABFE are parallelograms. Show that ar(ΔADE) = ar(ΔBCF).

Now, here in ∆ADE and ∆BCF,

AD = BC (Opposite sides of parallelogram ABCD


DE = CF (Opposite sides of parallelogram DCEF)


AE = BF ​(Opposite sides of parallelogram ABFE)​


​​∴ By SSS axiom,


∆ADE ≅​ ∆BCF


And,


ar(∆ADE) = ar(∆BCF) (By cpct)


More from this chapter

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12

Look at the statements given below:

I. If AD, BE and CF be the altitudes of a ΔABC such that AD = BE = CF, then ΔABC is an equilateral triangle.


II. If D is the mid-point of hypotenuse AC of a right ΔABC, then BD = AC.


III. In an isosceles ΔABC in which AB = AC, the altitude AD bisects BC.


Which is true?


A. I only B. II only


C. I and III D. II and III

 13

In the given figure, D and E are two points on side BC of ΔABC such that BD = DE = EC.

Prove that


ar (ΔABD) = ar (ΔADE) = ar (ΔAEC).


 15

In the given figure, ABCD is a trapezium in which AB || DC and diagonals AC and BD intersect at O. Prove that ar(ΔAOD) = ar(ΔBOC).

 16

Show that a diagonal divides a parallelogram into two triangles of equal area.