हिंदी

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

Advertisements
Advertisements

प्रश्न

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

उत्तर

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
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 14: Areas of Parallelograms and Triangles - Exercise 14.3 [पृष्ठ ४५]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 9
अध्याय 14 Areas of Parallelograms and Triangles
Exercise 14.3 | Q 8 | पृष्ठ ४५

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

If E, F, G and H are respectively the mid-points of the sides of a parallelogram ABCD show that ar (EFGH) = 1/2ar (ABCD)

 


In the given figure, PQRS and ABRS are parallelograms and X is any point on side BR. Show that

(i) ar (PQRS) = ar (ABRS)

(ii) ar (AXS) = 1/2ar (PQRS)


In the following figure, ABCD, DCFE and ABFE are parallelograms. Show that ar (ADE) = ar (BCF).


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)


In which of the following figures, you find two polygons on the same base and between the same parallels?


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)


If the mid-points of the sides of a quadrilateral are joined in order, prove that the area of the parallelogram so formed will be half of the area of the given quadrilateral (Figure).

[Hint: Join BD and draw perpendicular from A on BD.]


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.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×