NCERT Exemplar Class 9 Maths Solutions Chapter 9 Areas of Parallelograms and Triangles

NCERT Exemplar Class 9 Maths Solutions Chapter 9-Exercise 9.1

Question:1

The median of a triangle divides it into two
(A) triangles of equal area
(B) congruent triangles
(C) right triangles
(D) isosceles triangles

Question:2

In which of the following figures (Fig. 9.3), you find two polygons on the same base and between the same parallels?

Question:3

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 cm ²
(B) a square of area 25 cm ²
(C) a trapezium of area 24 cm ²
(D) a rhombus of area 24 cm ²

Question:4

In Fig. 9.4, the area of parallelogram ABCD is :


(A) AB × BM
(B) BC × BN
(C) DC × DL
(D) AD × DL

Question:5

In Fig. 9.5, if parallelogram ABCD and rectangle ABEM 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 = (Perimeter of ABEM)

Answer:

Question:6

The mid-point of the sides of a triangle along with any of the vertices as the fourth point make a parallelogram of area equal to
(A) $\frac{1}{2}ar(ABC)$
(B) $\frac{1}{3}ar(ABC)$
(C) $\frac{1}{4}ar(ABC)$
(D) $ar(ABC)$

Question:7

Two parallelograms are on equal bases and between the same parallels. The ratio of their areas is
(A) 1 : 2
(B) 1 : 1
(C) 2 : 1
(D) 3 : 1

Question:8

ABCD is a quadrilateral whose diagonal AC divides it into two parts, equal in area, then ABCD
(A) is a rectangle
(B) is always a rhombus
(C) is a parallelogram
(D) need not be any of (A), (B) or (C)

Question:9

If a triangle and a parallelogram are on the same base and between same parallels, then the ratio of the area of the triangle to the area of parallelogram is
(A) 1 : 3
(B) 1 : 2
(C) 3 : 1
(D) 1 : 4

Question:10

ABCD is a trapezium with parallel sides AB = a cm and DC = b cm (Fig.). E and F are the mid-points of the non-parallel sides. The ratio of ar (ABFE) and ar (EFCD) is


(A) a : b
(B) (3a + b) : (a + 3b)
(C) (a + 3b) : (3a + b)
(D) (2a + b) : (3a + b)

Answer:B

(Mid-segment Theorem on Trapezium: the line segment joining the midpoints of the nonparallel sides of a trapezium is half the sum of the lengths of the parallel sides and is also parallel to them.)
So, $EF=\frac{1}{2}(a+b)$
Now, ABFE and EFCD are also trapeziums.
$ar(ABFE)=\frac{1}{2}(EF+AB)h$
$=\frac{1}{2}[\frac{1}{2}(a+b)+a]\times h$
$=\frac{h}{4}(3a+b)$
$ar(EFCD)=\frac{1}{2}(CD+EF)h$
$=\frac{1}{2}[b+\frac{1}{2}(a+b)]\times h$
$=\frac{h}{4}(a+3b)$
Required ratio
$\therefore \frac{ar(ABEF)}{ar(EFCD)}=\frac{\frac{h}{4}(3a+b)}{\frac{h}{4}(a+3b)}=\frac{(3a+b)}{(a+3b)}$
So, the required ratio is $(3a+b):(a+3b)$
Hence, (B) is the correct answer.

NCERT Exemplar Class 9 Maths Solutions Chapter 9-Exercise 9.2

Question:1

Write true or false and justify your answer.
ABCD is a parallelogram and X is the mid-point of AB. If ar (AXCD) = 24 cm², then ar (ABC) = 24 cm².

= (AB) (GX) …(2)

Area of trapezium AXCD = $\frac{1}{2}$ (Sum of parallel sides) (Distance between them)

= $\frac{1}{2}(DC+AX)(GX)$

Now, $AX=\frac{1}{2}AB$ and $AB=CD$ (Given)

Area of trapezium $AXCD=\frac{1}{2}(AB+\frac{1}{2}AB)(GX)$…(3)

Area of $\mathrm{△}ABC=\frac{1}{2}$ (base) (height)

$=\frac{1}{2}(AB)(GX)$

$=\frac{1}{2}(AB)(GX)$ …(4)

Now,
Area of trapezium $AXCD=\frac{1}{2}(AB+\frac{1}{2}AB)(GX)=24c{m}^2$
(given)

$\frac{3}{4}(AB)(GX)=24c{m}^2$
$\frac{1}{4}(AB)(GX)=8c{m}^2$
$\frac{1}{2}(AB)(GX)=16c{m}^2$

From equation (4),
$\frac{1}{2}(AB)(GX)=$ Area of $\mathrm{△}ABC=16c{m}^2$

Hence, the given statement is false.

Question:2

Write true or false and justify your answer.
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.

PS = 5 cm, PR = 13 cm
In D PSR, using Pythagoras theorem,
$P{S}^2+S{R}^2=P{R}^2$
$5^2+S{R}^2={13}^2$
$S{R}^2=169-25=144$
SR = 12 cm = PQ (opposite sides of a rectangle are equal)
Area of a triangle is given as $\frac{1}{2}\times base\times height$
Now, $ar(\mathrm{△}PQR)=\frac{1}{2}\times PQ\times QR=\frac{1}{2}\times 12\times 5=30c{m}^2$
As, $PA<PQ$

$ar(\mathrm{△}PAS)<ar(\mathrm{△}PQR)$
or $ar(\mathrm{△}PAS)<30c{m}^2$
But it is given that $ar(PAS)=30c{m}^2$

Hence, the given statement is false.

Question:3

Write true or false and justify your answer.
PQRS is a parallelogram whose area is 180 cm² and A is any point on the diagonal QS. The area of DASR = 90 cm².

Answer:

Given, PQRS is a parallelogram whose area is $180c{m}^2$

We know that diagonal of a parallelogram divides it into two triangles of equals area, so
$\therefore ar(\mathrm{△}QRS)=\frac{1}{2}ar(\parallel gmPQRS)$
$=\frac{1}{2}\times 180=90c{m}^2$

Given that A is any point on SQ
So, $ar(\mathrm{△}ASR)<ar(\mathrm{△}QRS)$
Hence,$ar(\mathrm{△}ASR)<90c{m}^2$
But given that area of $\mathrm{△}ASR=90c{m}^2$

Hence, the given statement is false.

Question:4

Write true or false and justify your answer.
ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Then ar (BDE) = $\frac{1}{4}$ ar ( ABC).

Answer:

Given $\mathrm{△}ABC$ and $\mathrm{△}BDE$ are two equilateral triangles.
D is the mid-point of BC, so BD = CD = $\frac{1}{2}$ BC

Hence, $ar(\mathrm{△}BDE)=\frac{1}{4}ar(\mathrm{△}ABC)$

Therefore the given statement is true.

Question:5

Write true or false and justify your answer.
In Fig., ABCD and EFGD are two parallelograms and G is the mid-point of CD. Then ar (DPC) = $\frac{1}{2}$ar ( EFGD).

Answer:[False]

NCERT Exemplar Class 9 Maths Solutions Chapter 9-Exercise 9.3

Question:1

In Fig., PSDA is a parallelogram. Points Q and R are taken on PS such that PQ = QR = RS and $PA\parallel QB\parallel RC$. Prove that ar (PQE) = ar (CFD).

Answer:

Question:2

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)

Answer:

Question:3

The area of the parallelogram ABCD is $90c{m}^2$ (see Fig.9.13). Find

(i) ar (ABEF)
(ii) ar (ABD)
(iii) ar (BEF)

Answer:

Question:4

In $\mathrm{△}ABC$, D is the mid-point of AB and P is any point on BC. If $CQ\parallel PD$ meets AB in Q (Fig.), then prove that $ar(BPQ)=\frac{1}{2}ar(ABC)$.

Answer:

In $\mathrm{△}ABC$, D is the mid-point of AB and P is any point on BC.
$CQ\parallel PD$
Join D and C

Consider $\mathrm{△}ABC$,
Since D is the mid-point of AB
So CD is the median of $\mathrm{△}ABC$
Using mid-point theorem,
$ar(\mathrm{△}BCD)=\frac{1}{2}ar(\mathrm{△}ABC)$…(i)
Now, $CQ\parallel PD$
Since $\mathrm{△}PDQ$ and $\mathrm{△}PDC$ are on the same base PD and between the same parallel lines PD & QC. So we have:
$\therefore ar(\mathrm{△}PDQ)=ar(\mathrm{△}PDC)$ …(ii)

Question:5

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

Answer:

Question:6

O is any point on the diagonal PR of a parallelogram PQRS (Fig.). Prove that ar (PSO) = ar (PQO).

Answer:

Let diagonals PR and QS intersect each other at T.

Question:7

ABCD is a parallelogram in which BC is produced to E such that CE = BC (Fig.). AE intersects CD at F.
If ar (DFB) = $3c{m}^2$, find the area of the parallelogram ABCD.

Answer:

Question:8

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

Answer:

Question:9

If the mid-points of the sides of a quadrilateral are joined in order, prove that the area of the parallelogram so formed will be half of the area of the given quadrilateral (Fig.).

Answer:

NCERT Exemplar Class 9 Maths Solutions Chapter 9-Exercise 9.4

Question:1

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)

Answer:

Hence proved.

Question:2

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.

Answer: