English

Two parallelograms are on equal bases and between the same parallels. The ratio of their areas is ______.

Advertisements
Advertisements

Question

Two parallelograms are on equal bases and between the same parallels. The ratio of their areas is ______.

Options

  • 1 : 2

  • 1 : 1

  • 2 : 1

  • 3 : 1

MCQ
Fill in the Blanks
Advertisements

Solution

Two parallelograms are on equal bases and between the same parallels. The ratio of their areas is 1 : 1.

Explanation:

We know that, parallelogram on the equal bases and between the same parallels are equal in area.

So, ratio of their areas is 1 : 1.

shaalaa.com
  Is there an error in this question or solution?
Chapter 9: Areas of Parallelograms & Triangles - Exercise 9.1 [Page 87]

APPEARS IN

NCERT Exemplar Mathematics Exemplar [English] Class 9
Chapter 9 Areas of Parallelograms & Triangles
Exercise 9.1 | Q 7. | Page 87

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

In the given figure, ABCD is parallelogram, AE ⊥ DC and CF ⊥ AD. If AB = 16 cm, AE = 8 cm and CF = 10 cm, find AD.


If E, F, G and H are respectively the mid-points of the sides of a parallelogram ABCD show that ar (EFGH) = 1/2ar (ABCD)

 


P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD. Show that ar (APB) = ar (BQC).


In the given figure, PQRS and ABRS are parallelograms and X is any point on side BR. Show that

(i) ar (PQRS) = ar (ABRS)

(ii) ar (AXS) = 1/2ar (PQRS)


In the following figure, ABCD, DCFE and ABFE are parallelograms. Show that ar (ADE) = ar (BCF).


ABCD is a parallelogram, G is the point on AB such that AG = 2 GB, E is a point of DC
such that CE = 2DE and F is the point of BC such that BF = 2FC. Prove that:

(1)  ar ( ADEG) = ar (GBCD)

 (2)  ar (ΔEGB) = `1/6` ar (ABCD)

 (3)  ar (ΔEFC) = `1/2` ar (ΔEBF)

 (4)  ar (ΔEBG)  = ar (ΔEFC)

 (5)ΔFind what portion of the area of parallelogram is the area of EFG.


In the following figure, PSDA is a parallelogram. Points Q and R are taken on PS such that PQ = QR = RS and PA || QB || RC. Prove that ar (PQE) = ar (CFD).


ABCD is a square. E and F are respectively the mid-points of BC and CD. If R is the mid-point of EF (Figure), prove that ar (AER) = ar (AFR)


The diagonals of a parallelogram ABCD intersect at a point O. Through O, a line is drawn to intersect AD at P and BC at Q. Show that PQ divides the parallelogram into two parts of equal area.


In the following figure, ABCD and AEFD are two parallelograms. Prove that ar (PEA) = ar (QFD). [Hint: Join PD].


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×