हिंदी

In the Given Figure, Ad // Be // Cf. Prove that Area (δAec) = Area (δDbf)

Advertisements
Advertisements

प्रश्न

In the given figure, AD // BE // CF.
Prove that area (ΔAEC) = area (ΔDBF)

योग
Advertisements

उत्तर

We know that the area of triangles on the same base and between the same parallel lines are equal.

Consider ABED quadrilateral; AD || BE.
With the common base, BE and between AD and BE parallel lines, we have
Area of ΔABE = Area of ΔBDE

Similarly, in BEFC quadrilateral, BE || CF
With common base BC and between BE and CF parallel lines, we have
Area of ΔBEC = Area of ΔBEF

Adding both equations, we have
Area of ΔABE + Area of ΔBEC = Area of ΔBEF + Area of ΔBDE
⇒ Area of AEC = Area of DBF

Hence Proved.

shaalaa.com
Figures Between the Same Parallels
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 15: Area Theorems [Proof and Use] - Exercise 16 (A) [पृष्ठ १९७]

APPEARS IN

सेलिना Concise Mathematics [English] Class 9 ICSE
अध्याय 15 Area Theorems [Proof and Use]
Exercise 16 (A) | Q 16 | पृष्ठ १९७

संबंधित प्रश्न

In the following, AC // PS // QR and PQ // DB // SR.

Prove that: Area of quadrilateral PQRS = 2 x Area of the quad. ABCD.


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 the given figure, diagonals PR and QS of the parallelogram PQRS intersect at point O and LM is parallel to PS. Show that:

(i) 2 Area (POS) = Area (// gm PMLS)
(ii) Area (POS) + Area (QOR) = Area (// gm PQRS)
(iii) Area (POS) + Area (QOR) = Area (POQ) + Area (SOR).


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 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.


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.


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.


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.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×