मराठी

Abcd is a Trapezium with Ab Parallel to Dc. a Line Parallel to Ac Intersects Ab at X and Bc at Y. Prove that Area of ∆Adx = Area of ∆Acy. - Mathematics

Advertisements
Advertisements

प्रश्न

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.

बेरीज
Advertisements

उत्तर

Join CX, DX and AY.
Now, triangles ADX and ACX are on the same base AX and between the parallels AB and DC.
∴ A( ΔADX ) = A( ΔACX )                            ….(i)

Also, triangles ACX and ACY are on the same base AC and between the parallels AC and XY.
∴ A( ΔACX ) = A( ΔACY )                            ….(ii)

From (i) and (ii), we get
A( ΔADX ) = A( ΔACY )



shaalaa.com
Figures Between the Same Parallels
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 16: Area Theorems [Proof and Use] - Exercise 16 (C) [पृष्ठ २०२]

APPEARS IN

सेलिना Concise Mathematics [English] Class 9 ICSE
पाठ 16 Area Theorems [Proof and Use]
Exercise 16 (C) | Q 11 | पृष्ठ २०२

संबंधित प्रश्‍न

The given figure shows the parallelograms ABCD and APQR.
Show that these parallelograms are equal in the area.
[ Join B and R ]


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.


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


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 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:

  1. ΔEAC ≅ ΔBAF
  2. Area of the square ABDE
  3. 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.


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 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 )


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×