मराठी

In the Following Figure, Ce is Drawn Parallel to Diagonals Db of the Quadrilateral Abcd Which Meets Ab Produced at Point E. Prove that δAde and Quadrilateral Abcd Are Equal in Area - Mathematics

Advertisements
Advertisements

प्रश्न

In the following figure, CE is drawn parallel to diagonals DB of the quadrilateral ABCD which meets AB produced at point E.
Prove that ΔADE and quadrilateral ABCD are equal in area.

बेरीज

उत्तर

Since ΔDCB and ΔDEB are on the same base DB and between the same parallels i.e. DB // CE, therefore we get

Ar. ( ΔDCB) = Ar. ( ΔDEB )
Ar. ( ΔDCB + ΔADB ) = AR. (ΔDEB + ΔADB )
Ar. ( ABCD ) = Ar. ( ΔADE )
Hence proved.

shaalaa.com
Figures Between the Same Parallels
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 16: Area Theorems [Proof and Use] - Exercise 16 (A) [पृष्ठ १९६]

APPEARS IN

सेलिना Concise Mathematics [English] Class 9 ICSE
पाठ 16 Area Theorems [Proof and Use]
Exercise 16 (A) | Q 6 | पृष्ठ १९६

संबंधित प्रश्‍न

In the given figure, if the area of triangle ADE is 60 cm2, state, given reason, the area of :
(i) Parallelogram ABED;
(ii) Rectangle ABCF;
(iii) Triangle ABE.


The given figure shows a rectangle ABDC and a parallelogram ABEF; drawn on opposite sides of AB.
Prove that: 
(i) Quadrilateral CDEF is a parallelogram;
(ii) Area of the quad. CDEF
= Area of rect. ABDC + Area of // gm. ABEF.


In the given figure, D is mid-point of side AB of ΔABC and BDEC is a parallelogram.

Prove that: Area of ABC = Area of // gm BDEC.


In parallelogram ABCD, P is a point on side AB and Q is a point on side BC.
Prove that:
(i) ΔCPD and ΔAQD are equal in the area.
(ii) Area (ΔAQD) = Area (ΔAPD) + Area (ΔCPB)


In the following figure, DE is parallel to BC.
Show that: 
(i) Area ( ΔADC ) = Area( ΔAEB ).
(ii) Area ( ΔBOD ) = Area( ΔCOE ).


Show that:

A diagonal divides a parallelogram into two triangles of equal area.


ABCD is a parallelogram in which BC is produced to E such that CE = BC and AE intersects CD at F.

If ar.(∆DFB) = 30 cm2; find the area of parallelogram.


In a parallelogram ABCD, point P lies in DC such that DP: PC = 3:2. If the area of ΔDPB = 30 sq. cm.
find the area of the parallelogram ABCD.


ABCD is a trapezium with AB parallel to DC. A line parallel to AC intersects AB at X and BC at Y.
Prove that the area of ∆ADX = area of ∆ACY.


In the following figure, BD is parallel to CA, E is mid-point of CA and BD = `1/2`CA
Prove that: ar. ( ΔABC ) = 2 x ar.( ΔDBC )


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×