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.
Advertisements
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 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.
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.
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. 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.
In the given figure, the diagonals AC and BD intersect at point O. If OB = OD and AB//DC,
show that:
(i) Area (Δ DOC) = Area (Δ AOB).
(ii) Area (Δ DCB) = Area (Δ ACB).
(iii) ABCD is a parallelogram.
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.
Show that:
The ratio of the areas of two triangles on the same base is equal to the ratio of their heights.
