Advertisements
Advertisements
Question
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.
Solution
Here AD = DB and EC = DB, therefore EC = AD
Again,
∠EFC = ∠AFD .....( Opposite angles )
Since ED and CB are parallel lines and AC cut this line, therefore
∠ECF = ∠FAD
From the above conditions, we have
ΔEFC = ΔAFD
Adding quadrilateral CBDF in both sides, we have
Area of // gm BDEC = Area of ΔABC.
APPEARS IN
RELATED QUESTIONS
The given figure shows the parallelograms ABCD and APQR.
Show that these parallelograms are equal in the area.
[ Join B and R ]
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, AD // BE // CF.
Prove that area (ΔAEC) = area (ΔDBF)
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 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 figure given alongside, squares ABDE and AFGC are drawn on the side AB and the hypotenuse AC of the right triangle ABC.
If BH is perpendicular to FG
prove that:
- ΔEAC ≅ ΔBAF
- Area of the square ABDE
- Area of the rectangle ARHF.
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.
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 )
The given figure shows a parallelogram ABCD with area 324 sq. cm. P is a point in AB such that AP: PB = 1:2
Find The area of Δ APD.
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.
