Ncert exemplar solutions mathematics class 9 Solutions for Chapter 9 areas of parallelograms and triangles

The median of a triangle divides it into two

(A) triangles of equal area

(B) congruent triangles

(C) right triangles

(D) isosceles triangles

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

The figure obtained by joining the mid-points of the adjacent sides of a rectangle of sides 8 cm and 6 cm is :

(A) a rectangle of area 24 cm²

(B) a square of area 25 cm²

(C) a trapezium of area 24 cm²

(D) a rhombus of area 24 cm²

In Fig. , the area of parallelogram

ABCD is:

(A) AB × BM

(B) BC × BN

(C) DC × DL

(D) AD × DL

In Fig. , if parallelogram ABCD and rectangle ABEF are of equal area, then :

(A) Perimeter of ABCD = Perimeter of ABEM

(B) Perimeter of ABCD < Perimeter of ABEM

(C) Perimeter of ABCD > Perimeter of ABEM

(D) Perimeter of ABCD = ½ (Perimeter of ABEM)

ABCD is a parallelogram and X is the mid-point of AB. If ar (AXCD) = 24 cm² then ar (ABC) = 24 cm² .

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.

In Fig., 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).

X and Y are points on the side LN of the triangle LMN such that LX = XY = YN. Through X, a line is drawn parallel to LM to meet MN at Z (See Fig. ). Prove that ar (LZY) = ar (MZYX)

The area of the parallelogram ABCD is 90 cm2 (see Fig. ). Find

(i) ar (ABEF)

(ii) ar (ABD)

(iii) ar (BEF)

In △ ABC, D is the mid-point of AB and P is any point on BC. If CQ || PD meets AB in Q (Fig.), then prove that ar (BPQ) = ½ ar (ABC).

A point E is taken on the side BC of a parallelogram ABCD. AE and DC are produced to meet at F. Prove that ar (ADF) = ar (ABFC)

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.

The medians BE and CF of a triangle ABC intersect at G. Prove that the area of △ GBC = area of the quadrilateral AFGE.

In Fig. , CD || AE and CY || BA. Prove that ar (CBX) = ar (AXY).

ABCD is a trapezium in which AB || DC, DC = 30 cm and AB = 50 cm. If X and Y are, respectively the mid-points of AD and BC, prove that ar (DCYX) = 7/9 ar (XYBA)