English

In the Following Figure, Bd is Parallel to Ca, E is Mid-point of Ca and Bd = 1/2ca Prove That: Ar. ( δAbc ) = 2 X Ar.( δDbc )

Advertisements
Advertisements

Question

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 )

Sum
Advertisements

Solution

Here BCED is a parallelogram, Since BD = CE and BD || CE.
ar. ( ΔDBC ) = ar. ( ΔEBC )      ......( Since they have the same base and are between the same parallels )

In ΔABC,
BE is the median,
So, ar. ( ΔEBC ) = `1/2` ar. ( ΔABC )
Now, ar. ( ΔABC ) = ar. ( ΔEBC ) + ar. ( ΔABE)
Also, ar. ( ΔABC ) = 2ar. ( ΔEBC )
⇒ ar. ( ΔABC ) = 2ar. ( ΔDBC )

shaalaa.com
Figures Between the Same Parallels
  Is there an error in this question or solution?
Chapter 15: Area Theorems [Proof and Use] - Exercise 16 (C) [Page 202]

APPEARS IN

Selina Concise Mathematics [English] Class 9 ICSE
Chapter 15 Area Theorems [Proof and Use]
Exercise 16 (C) | Q 8 | Page 202

RELATED QUESTIONS

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


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.


ABCD and BCFE are parallelograms. If area of triangle EBC = 480 cm2; AB = 30 cm and BC = 40 cm.

Calculate : 
(i) Area of parallelogram ABCD;
(ii) Area of the parallelogram BCFE;
(iii) Length of altitude from A on CD;
(iv) Area of triangle ECF.


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


In the given figure, AP is parallel to BC, BP is parallel to CQ.
Prove that the area of triangles ABC and BQP are equal.


ABCD is a parallelogram a line through A cuts DC at point P and BC produced at Q. Prove that triangle BCP is equal in area to triangle DPQ.


ABCD is a parallelogram. P and Q are the mid-points of sides AB and AD respectively.
Prove that area of triangle APQ = `1/8` of the area of 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.


Show that:

The ratio of the areas of two triangles of the same height is equal to the ratio of their bases.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×