English

In the Given Figure, Abcd is a Parallelogram; Bc is Produced to Point X. Prove That: Area (δ Abx) = Area (Quad. Acxd) - Mathematics

Advertisements
Advertisements

Question

In the given figure, ABCD is a parallelogram; BC is produced to point X.
Prove that: area ( Δ ABX ) = area (`square`ACXD )

Sum

Solution

Given: ABCD is a parallelogram.
We know that
Area of ΔABC = Area of ΔACD
Consider ΔABX,
Area of ΔABX = Area of ΔABC + Area of ΔACX
We also know that the area of triangles on the same base and between the same parallel lines are equal.
Area of ΔACX = Area of ΔCXD
From the above equations, we can conclude that
Area of ΔABX = Area of ΔABC + Area of ΔACX
= Area of ΔACD+ Area of ΔCXD
= Area of ACXD Quadrilateral

Hence Proved.

shaalaa.com
Figures Between the Same Parallels
  Is there an error in this question or solution?
Chapter 16: Area Theorems [Proof and Use] - Exercise 16 (A) [Page 198]

APPEARS IN

Selina Concise Mathematics [English] Class 9 ICSE
Chapter 16 Area Theorems [Proof and Use]
Exercise 16 (A) | Q 17 | Page 198

RELATED QUESTIONS

ABCD is a trapezium with AB // DC. A line parallel to AC intersects AB at point M and BC at point N.
Prove that: area of Δ ADM = area of Δ ACN.


In the following, AC // PS // QR and PQ // DB // SR.

Prove that: Area of quadrilateral PQRS = 2 x Area of the quad. ABCD.


In the given figure, diagonals PR and QS of the parallelogram PQRS intersect at point O and LM is parallel to PS. Show that:

(i) 2 Area (POS) = Area (// gm PMLS)
(ii) Area (POS) + Area (QOR) = Area (// gm PQRS)
(iii) Area (POS) + Area (QOR) = Area (POQ) + Area (SOR).


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 ).


The given figure shows a pentagon ABCDE. EG drawn parallel to DA meets BA produced at G and CF draw parallel to DB meets AB produced at F.

Prove that the area of pentagon ABCDE is equal to the area of triangle GDF.


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.


In ΔABC, E and F are mid-points of sides AB and AC respectively. If BF and CE intersect each other at point O,
prove that the ΔOBC and quadrilateral AEOF are equal in area.


In parallelogram ABCD, P is the mid-point of AB. CP and BD intersect each other at point O. If the area of ΔPOB = 40 cm2, and OP: OC = 1:2, find:
(i) Areas of ΔBOC and ΔPBC
(ii) Areas of ΔABC and parallelogram ABCD.


Show that:
The ratio of the areas of two triangles on the same base is equal to the ratio of their heights.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×