Advertisements
Advertisements
प्रश्न
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.
Advertisements
उत्तर
We have to join PD and BD.
BD is the diagonal of the parallelogram ABCD. Therefore it divides the parallelogram into two equal parts.
∴ Area( ΔABD )= Area ( ΔDBC )
=`1/2` Area ( parallelogram ABCD) ...(i)
DP is the median of ΔABD. Therefore it will divide ΔABD into two triangles of equal areas.
∴ Area( ΔAPD )= Area ( ΔDPB )
= `1/2` Area ( ΔABD )
= `1/2 xx 1/2` Area (parallelogram ABCD) ...[from equation (i)]
= `1/4` Area (parallelogram ABCD) ...(ii)
In ΔAPD, Q is the mid-point of AD. Therefore PQ is the median.
∴ Area(ΔAPQ)= Area (ΔDPQ)
= `1/2` Area (ΔAPD)
= `1/2 xx 1/4` Area (parallelogram ABCD)...[from equation (ii)]
Area (ΔAPQ)= `1/8` Area (parallelogram ABCD),
hence proved
APPEARS IN
संबंधित प्रश्न
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.
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 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 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.
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 in which BC is produced to E such that CE = BC and AE intersects CD at F.
If ar.(∆DFB) = 30 cm2; find the area of parallelogram.
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.
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.
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.
