Advertisements
Advertisements
प्रश्न
In parallelogram ABCD, E is a point in AB and DE meets diagonal AC at point F. If DF: FE = 5:3 and area of ΔADF is 60 cm2; find
(i) area of ΔADE.
(ii) if AE: EB = 4:5, find the area of ΔADB.
(iii) also, find the area of parallelogram ABCD.
उत्तर
ΔADF and ΔAFE have the same vertex A and their bases are on the same straight line DE.
∴ `"A(ΔADF)"/"A(ΔAFE)" = "DF"/"FE"`
⇒ `60/"A(ΔAFE)" = 5/3 `
⇒ A(ΔAFE) = `( 60 xx 3 )/5` = 36cm2.
Now, A(ΔADE) = A(ΔADF) + A(ΔAFE) = 60 + 36 = 96 cm2.
ΔADE and ΔEDB have the same vertex D and their bases are on the same straight line AB.\
∴ `"A( ΔADE )"/"A( ΔEDB )" = "AE"/"EB"`
⇒ `96/"A( ΔEDB )" = 4/5`
⇒ A( ΔEDB ) = `( 96 xx 5 )/4` = 120 cm2 .
Now, A( ΔADB ) and ||m ABCD are on the same base AB and between the same parallels AB and DC.
∴ A( ΔADB ) = `1/2` A( ||m ABCD )
⇒ 216 = `1/2` A( ||m ABCD )
⇒ A( ||m ABCD ) = 2 x 216 = 432 cm2 .
APPEARS IN
संबंधित प्रश्न
In the given figure, if the area of triangle ADE is 60 cm2, state, given reason, the area of :
(i) Parallelogram ABED;
(ii) Rectangle ABCF;
(iii) Triangle ABE.
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)
In the given figure, M and N are the mid-points of the sides DC and AB respectively of the parallelogram ABCD.
If the area of parallelogram ABCD is 48 cm2;
(i) State the area of the triangle BEC.
(ii) Name the parallelogram which is equal in area to the triangle BEC.
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 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.
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.
E, F, G, and H are the midpoints of the sides of a parallelogram ABCD.
Show that the area of quadrilateral EFGH is half of the area of parallelogram ABCD.
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 of the same height is equal to the ratio of their bases.
