NCERT Exemplar Solutions for Class 9 Maths Chapter 9 - Areas Of Parallelograms And Triangles

NCERT Exemplar Solutions Class 9 Maths Chapter 9 – Free PDF Download

NCERT Exemplar Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles are provided here for students who are appearing for their final exams. To encourage easy learning and help them understand the concepts of areas of parallelograms and triangles based on different scenarios, free NCERT exemplars are provided here, which can be downloaded in the form of a PDF. Students can use these solved exemplar questions as a reference tool while practising the NCERT textbook exercise questions, as well. These solutions are prepared by the subject experts at BYJU’S, and the topics covered are with respect to the latest CBSE Syllabus (2023-2024). The main topics covered in Chapter 9, Areas of Parallelograms and Triangles, are as follows:

• Areas of parallelograms on the same base and between the same parallels
• Areas of triangles on the same base and between the same parallels
• Median of a triangle dividing into two triangles

Learn to find the area of parallelogram and triangle by referring to NCERT Exemplar for Class 9 Maths Chapter 9, Areas of Parallelograms and Triangles. Solving the exemplar questions will help students to learn important formulas and theorems of parallelograms and triangles covered in the chapter in an easy and better way.

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Exercise 9.1 Page No: 85

Write the correct answer in each of the following:

1. The median of a triangle divides it into two

(A) triangles of equal area

(B) congruent triangles

(C) right triangles

(D) isosceles triangles

Solution:

(A) triangles of equal area

Explanation:

The median of a triangle divides it into triangles of equal area.

Hence, option (A) is the correct answer.

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

Solution:

(D)

Explanation:

In figure (D), the parallelograms, PQRA and BQRS, are on the same base QR and between the same parallels QR and PS.

Hence, option (D) is the correct answer.

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 cm²

(B) a square of area 25 cm²

(C) a trapezium of area 24 cm²

(D) a rhombus of area 24 cm²

Solution:

(D) a rhombus of area 24 cm²

Explanation:

According to the question,

Let ABCD be the rectangle.

E is the mid-point of the side AB

F is the mid-point of the side BC

G is the mid-point of the side CD

H is the mid-point of the side DA

The figure obtained by joining the midpoints E, F, G and H is a rhombus.

The sides of the rectangle are given,

We know that,

Side of the rectangle AB = diagonal of the rhombus FH = 8cm

Similarly,

Side of the rectangle AD = diagonal of the rhombus EG = 6cm

Then,

Area of the rhombus = ½ × EG × FH

= ½ × 6 × 8

= 24 cm²

Hence, option (D) is the correct answer.

4. In Fig. 9.4, the area of parallelogram

ABCD is:

(A) AB × BM

(B) BC × BN

(C) DC × DL

(D) AD × DL

Solution:

(C) DC × DL

Explanation:

Area of parallelogram = Base × Corresponding altitude

= AB × DL … (eq 1)

Since opposite sides of a parallelogram are equal,

We get,

AB = DC

Substituting this in eq(1), we get,

Area of parallelogram = AB × DL

= DC × DL

Hence, option (C) is the correct answer.

5. In Fig. 9.5, 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)

Solution:

(C) Perimeter of ABCD > Perimeter of ABEM

Explanation:

In rectangle ABEM,

AB = EM …(eq.1) [sides of rectangle]

In parallelogram ABCD,

CD = AB …(eq.2)

Adding equations (1) and (2),

We get

AB + CD = EM + AB …(i)

We know that,

The perpendicular distance between two parallel sides of a parallelogram is always less than the length of the other parallel sides.

BE < BC and AM < AD

On adding both above inequalities, we get

SE + AM <BC + AD or BC + AD> BE + AM

On adding AB + CD on both sides, we get

AB + CD + BC + AD> AB + CD + BE + AM

⇒ AB+BC + CD + AD> AB+BE + EM+ AM    [∴ CD = AB = EM]

Hence,

We get,

The perimeter of parallelogram ABCD > perimeter of rectangle ABEM

Hence, option (C) is the correct answer.

Exercise 9.2 Page No: 88

Write True or False and justify your answer:

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

Solution:

False

Explanation:

In ΔABC, X is the midpoint of AB

Considering the given question,

Hence ar(ΔAXC) = ar(ΔBXC) = ½ ar(ΔABC)

Therefore,

ar(ΔAXC) = ar(ΔBXC) = 12 cm²

Area of ||gm ABCD = 2 x ar(ΔABC)

= 2 x 24 = 48 cm²

But, according to the question,

area of ||gm ABCD = ar(AXCD) + ar(ΔBXC)

= 24 + 12 = 36 cm²

Hence, we find a contradiction here.

So, If ar (AXCD) = 24 cm², then ar (ABC) ≠ 24 cm²

2. 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.

Solution:

True. Since A is any point on PQ, then ar (PAS) ≠ 30 cm²

But, the statement can be true if PA is equal to PS.

Justification:

According to the question,

PQRS is a rectangle inscribed in a quadrant of a circle of radius 13 cm.

Given,

A is any point on PQ

PA<PQ

ar (△PQR) = ½ ×PQ×QR

= ½ ×12×5

= 30cm²

PS=5 cm

Suppose PA<PQ,

ar(△PAS) < ar(△PQR)

ar(△PAS) < 30 cm²

Suppose PA=PQ,

ar (△PAS) = ½ ×PQ×PS

= ½ ×12×5

= 30cm²

Exercise 9.3 Page No: 89

1. In Fig.9.11, 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).

Solution:

According to the question,

PA║QB║RC║SD & PQ=QR=RS

According to the Equal intercept theorem,

We know that if three or more parallel lines make equal intercepts on traversal, then they make equal intercepts on any other form of traversal.

Hence, we get,

PE=EF=FD& AB=BC=CD

From ΔPQE & ΔDCF,

We get,

∠PEQ=∠DFC

PE=DF

∠QPE=∠CDF

So,

ΔPQE ≅ ΔDCF

Since Congruent figures have equal areas,

We get,

ar ΔPQE= ar ΔDCF

Hence proved.

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. 9.12). Prove that

ar (LZY) = ar (MZYX)

Solution:

According to the question,

LX=XY=YN

XZ II LM

We have,

ar(LZX)+(XZY)=ar(LZY)               — (1)

ar(MXZ)+ar(XZY)=ar(MZYX)        — (2)

Both triangles LZX and MXZ are on the same base XZ and between the same parallels LM and XZ

ar(LZX)=ar(MXZ)

Adding equations (1) and (2),

We get,

ar(LZY)=ar(MZYX)

Hence proved

3. The area of the parallelogram ABCD is 90 cm² (see Fig.9.13). Find

(i) ar (ABEF)

(ii) ar (ABD)

(iii) ar (BEF)

Solution:

According to the question,

Area of parallelogram, ABCD = 90 cm²

(i)

We know that,

Parallelograms on the same base and between the same parallel are equal in area.

Here,

The parallelograms ABCD and ABEF are on the same base AB and between the same parallels AB and CF.

Therefore, ar (ABEF) = ar (ABCD) = 90 cm²

(ii)

We know that,

If a triangle and a parallelogram are on the same base and between the same parallels, then the area of the triangle is equal to half of the area of the parallelogram.

Here,

ΔABD and parallelogram ABCD are on the same base AB and between the same parallels AB and CD.

Therefore, ar (ΔABD) = ½ ar (ABCD)

= ½ x 90 = 45 cm²

(iii)

We know that,

If a triangle and a parallelogram are on the same base and between the same parallels, then the area of a triangle is equal to half of the area of the parallelogram.

Here,

ΔBEF and parallelogram ABEF are on the same base EF and between the same parallels AB and EF.

Therefore, ar (ΔBEF) = ½ ar (ABEF)

= ½ x 90 = 45 cm²

4. In △ ABC, D is the mid-point of AB, and P is any point on BC. If CQ || PD meets AB in Q (Fig. 9.14), then prove that ar (BPQ) = ½ ar (ABC).

Solution:

According to the question,

We have the figure,

In △ ABC,

D is the mid-point of AB

P is any point on BC.

If CQ || PD meets AB at Q

To prove: ar (BPQ) = ½ ar (ABC)

Construction: Join DC

Proof:- Since D is the mid-point of AB. so, in △ ABC, CD is the median.

ar(△ BCD) = ½ ar (△ ABC) ….. (1)

We know that,

△ PDQ and △ PDC are on the same base PD and between the same parallel lines PD QC.

Therefore, ar(△ PDQ) = ar(△ PDC) ……………. (2)

From (1) and (2)

ar(△ BCD) = ½ ar(△ ABC)

ar(△ BCD = ar(△ BPD) + ar(△ PDC)

= ½ ar(△ ABC)

Area of △ PDC = PDQ

So,

ar(△ BPQ) = ar(△ BPD) + ar(△ PDQ)

= ½ ar(△ ABC)

Therefore, ar(△BPQ) = ½ ar(△ ABC)

Hence Proved

Exercise 9.4 Page No: 94

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)

Solution:

According to the question,

ABCD is a parallelogram.

E is a point on BC.

AE and DC are produced to meet at F.

To prove: area (ΔADF) = area (ABFC).

Proof:

We know that,

Triangles on the same base and between the same parallels are equal in area.

Here,

We have,

ΔABC and ΔABF are on the same base AB and between the same parallels, AB || CF.

Area (ΔABC) = area (ΔABF) … (1)

We also know that,

The diagonal of a Parallelogram divides it into two triangles of equal area

So,

Area (ΔABC) = area (ΔACD) … (2)

Now,

Area (ΔADF) = area (ΔACD) + area (ΔACF)

∴ Area (ΔADF) = area (ΔABC) + area (ΔACF) … (From equation (2))

⇒ Area (ΔADF) = area (ΔABF) + area (ΔACF) … (From equation (1))

⇒ Area (ΔADF) = area (ABFC)

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.

Solution:

According to the question,

The diagonals of a parallelogram ABCD intersect at point O.

Through O, a line is drawn to intersect AD at P and BC at Q.

To Prove:

Ar (parallelogram PDCQ) = ar (parallelogram PQBA).

Proof:

AC is a diagonal of || gm ABCD

∴ ar(ΔABC) = ar(ΔACD)

= ½ ar (||gm ABCD) …(1)

In ΔAOP and ΔCOQ,

AO = CO

Since, diagonals of a parallelogram bisect each other,

We get,

∠AOP = ∠COQ

∠OAP = ∠OCQ (Vertically opposite angles)

∴ ΔAOP = ΔCOQ (Alternate interior angles)

∴ ar(ΔAOP) = ar(ΔCOQ) (By ASA Congruence Rule)

We know that,

Congruent figures have equal areas

So,

ar(ΔAOP) + ar(parallelogram OPDC) = ar(ΔCOQ) + ar(parallelogram OPDC)

⇒ ar(ΔACD) = ar(parallelogram PDCQ)

⇒ ½ ar(|| gm ABCD) = ar(parallelogram PDCQ)

From equation (1),

We get,

ar(parallelogram PQBA) = ar(parallelogram PDCQ)

⇒ ar(parallelogram PDCQ) = ar(parallelogram PQBA).

Hence Proved.

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

Solution:

According to the question,

We have,

BE &  CF are medians

E is the midpoint of AC

F is the midpoint of AB

∴ ΔBCE = ΔBEA … ( i )

ΔBCF = ΔCAF

Construct:

Join EF,

By the midpoint theorem,

We get FE || BC

We know that,

Δ on the same base and between the same parallels are equal in area

∴ ΔFBC = ΔBCE

ΔFBC – ΔGBC = ΔBCE – ΔGBC

⇒ ΔFBG = ΔCGE (ΔGBC is common)

⇒ ΔCGE = ΔFBG …( ii )

Subtracting equation (ii) from (i)

We get,

ΔBCE – ΔCGE = ΔBEA – ΔFBG

∴ ΔBGC =quadrilateral AFGE.

4. In Fig. 9.24, CD || AE and CY || BA. Prove that ar (CBX) = ar (AXY).

Solution:

According to the question,

From figure,

We get,

CD||AE

and CY || BA

To prove:

ar (ΔCBX) = ar (ΔAXY) .

Proof:

We know that,

Triangles on the same base and between the same parallels are equal in area.

Here,

ΔABY and ΔABC both lie on the same base AB and between the same parallels CY and BA.

ar (ΔABY) = ar (ΔABC)

⇒ ar (ABX) + ar (AXY) = ar (ABX) + ar (CBX)

Solving and cancelling ar (ABX),

We get,

⇒ ar (AXY) = ar (CBX)

5. 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)

Solution:

According to the question,

We have,

ABCD is a trapezium with AB || DC

Construction:  Join DY and produce it to meet AB produced at P.

In ∆BYP and ∆CYD

∠BYP = ∠CYD    (vertically opposite angles)

Since alternate opposite angles of DC || AP and BC are the transversal

∠DCY = ∠PBY

Since Y is the midpoint of BC,

BY = CY

Thus ∆BYP ≅ ∆CYD (by ASA cogence criterion)

So, DY = YP and DC = BP

⇒ Y is the midpoint of AD

∴ XY || AP and XY = ½ AP (by midpoint theorem)

⇒ XY = ½ AP

= ½ (AB + BP)

= ½ (AB + DC)

= ½ (50 + 30)

= ½ × 80 cm

= 40 cm

Since X is the midpoint of AD

And Y is the midpoint of BC

Hence, trapezium DCYX and ABYX are of the same height, h, cm

Now

⇒ 9 ar(DCXY) = 7 ar(XYBA)

⇒ ar(DCXY) = 7/9 ar(XYBA)

Hence Proved.

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