English

In which of the following figures, you find two polygons on the same base and between the same parallels?

Advertisements
Advertisements

Question

In which of the following figures, you find two polygons on the same base and between the same parallels?

Options

MCQ
Advertisements

Solution


Explanation:

In figures (a), (b) and (c) there are two polygons on the same base but they are not between the same parallels.

In figure (d), there are two polygons (PQRA and BQRS) on the same base and between the same parallels.

shaalaa.com
  Is there an error in this question or solution?
Chapter 9: Areas of Parallelograms & Triangles - Exercise 9.1 [Page 85]

APPEARS IN

NCERT Exemplar Mathematics Exemplar [English] Class 9
Chapter 9 Areas of Parallelograms & Triangles
Exercise 9.1 | Q 2. | Page 85

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

A farmer was having a field in the form of a parallelogram PQRS. She took any point A on RS and joined it to points P and Q. In how many parts the field is divided? What are the shapes of these parts? The farmer wants to sow wheat and pulses in equal portions of the field separately. How should she do it?


In the given figure, P is a point in the interior of a parallelogram ABCD. Show that

(i) ar (APB) + ar (PCD) = 1/2ar (ABCD)

(ii) ar (APD) + ar (PBC) = ar (APB) + ar (PCD)

[Hint: Through. P, draw a line parallel to AB]


In the following figure, ABCD is parallelogram and BC is produced to a point Q such that AD = CQ. If AQ intersect DC at P, show that

ar (BPC) = ar (DPQ).

[Hint: Join AC.]


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.


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


PQRS is a rectangle inscribed in a quadrant of a circle of radius 13 cm. A is any point on PQ. If PS = 5 cm, then ar (PAS) = 30 cm2.


ABCD is a parallelogram in which BC is produced to E such that CE = BC (Figure). AE intersects CD at F. If ar (DFB) = 3 cm2, find the area of the parallelogram ABCD.


In trapezium ABCD, AB || DC and L is the mid-point of BC. Through L, a line PQ || AD has been drawn which meets AB in P and DC produced in Q (Figure). Prove that ar (ABCD) = ar (APQD)


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×