English

In the Given Below Fig. Abcd, Abfe and Cdef Are Parallelograms. Prove that Ar (δAde) = Ar (δBcf) - Mathematics

Advertisements
Advertisements

Question

In the given below fig. ABCD, ABFE and CDEF are parallelograms. Prove that ar (ΔADE)
= ar (ΔBCF)

Solution

Given that,
ABCDis a parallelogram  ⇒  AD = BC
CDEF is a parallelogram  ⇒  DE  = CF
ABFE is a parallelogram  ⇒   AE = BF
Thus, in Δs ADE and BCF,we have

AD = BC,DE = CF and AE = BF
So, by SSS criterion of congruence, we have

      ΔADE  ≅ ΔABCF
  ∴ ar ( Δ ADE )  = ar ( BCF )

shaalaa.com
  Is there an error in this question or solution?
Chapter 14: Areas of Parallelograms and Triangles - Exercise 14.3 [Page 45]

APPEARS IN

RD Sharma Mathematics [English] Class 9
Chapter 14 Areas of Parallelograms and Triangles
Exercise 14.3 | Q 8 | Page 45

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD. Show that ar (APB) = ar (BQC).


In the given figure, P is a point in the interior of a parallelogram ABCD. Show that

(i) ar (APB) + ar (PCD) = 1/2ar (ABCD)

(ii) ar (APD) + ar (PBC) = ar (APB) + ar (PCD)

[Hint: Through. P, draw a line parallel to AB]


Parallelogram ABCD and rectangle ABEF are on the same base AB and have equal areas. Show that the perimeter of the parallelogram is greater than that of the rectangle.


In the following figure, ABCD is parallelogram and BC is produced to a point Q such that AD = CQ. If AQ intersect DC at P, show that

ar (BPC) = ar (DPQ).

[Hint: Join AC.]


In the below fig. ABCD and AEFD are two parallelograms. Prove that
(1) PE = FQ
(2) ar (Δ APE) : ar (ΔPFA) = ar Δ(QFD) : ar (Δ PFD)
(3) ar (ΔPEA) = ar (ΔQFD)


Two parallelograms are on equal bases and between the same parallels. The ratio of their areas is ______.


ABCD is a parallelogram in which BC is produced to E such that CE = BC (Figure). AE intersects CD at F. If ar (DFB) = 3 cm2, find the area of the parallelogram ABCD.


In trapezium ABCD, AB || DC and L is the mid-point of BC. Through L, a line PQ || AD has been drawn which meets AB in P and DC produced in Q (Figure). Prove that ar (ABCD) = ar (APQD)


The diagonals of a parallelogram ABCD intersect at a point O. Through O, a line is drawn to intersect AD at P and BC at Q. Show that PQ divides the parallelogram into two parts of equal area.


ABCD is a trapezium in which AB || DC, DC = 30 cm and AB = 50 cm. If X and Y are, respectively the mid-points of AD and BC, prove that ar (DCYX) = `7/9` ar (XYBA)


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×