Advertisements
Advertisements
Question
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.
Advertisements
Solution
The ratio of the area of triangles with the same vertex and bases along the same line is equal to the ratio of their respective bases. So, we have
`"Area of DPB"/"Area of PCB" = "DP"/"PC" = 3/2`
Given: Area of ΔDPB = 30 sq. cm
Let 'x' be the area of the triangle PCB
Therefore, We have,
⇒ `30/x = 3/2`
⇒ x = `30/3 xx 2` = 20 sq.cm.
So area of ΔPCB = 20 sq. cm
Consider the following figure.
From the diagram, it is clear that,
Area( ΔCDB ) = Area( ΔDPB ) + Area( ΔCDB )
= 30 + 20 = 50 sq.cm.
The diagonal of the parallelogram divides it into two triangles ΔADB and ΔCDB of equal area.
Therefore,
Area( parallelogram ABCD ) = 2 x ΔCDB = 2 x 50 = 100 sq.cm.
APPEARS IN
RELATED QUESTIONS
In the given figure, AD // BE // CF.
Prove that area (ΔAEC) = area (ΔDBF)
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, D is mid-point of side AB of ΔABC and BDEC is a parallelogram.
Prove that: Area of ABC = Area of // gm BDEC.
In the following figure, DE is parallel to BC.
Show that:
(i) Area ( ΔADC ) = Area( ΔAEB ).
(ii) Area ( ΔBOD ) = Area( ΔCOE ).
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.
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.
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.
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 of the same height is equal to the ratio of their bases.
