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Chapters
1: Rational and Irrational Numbers
UNIT-II: COMMERCIAL MATHEMATICS
2: Compound Interest
UNIT-III: ALGEBRA
3: Expansions
4: Factorisation
5: Simultaneous Linear Equations
6: Indices
7: Logarithms
UNIT-IV: GEOMETRY
8: Triangles
9: Inequalities
10: Mid-point Theorem
11: Pythagoras Theorem
12: Rectilinear Figures (Theorems on Parallelograms and Construction of Polygons)
▶ 13: Theorems on Area
14: Circles (Chord and Arc Properties)
UNIT-V: STATISTICS
15: Statistics
16: Graphical Representation of Statistical Data
UNIT-VI: MENSURATION
17: Mensuration
18: Surface Area and Volume of Solids
UNIT-VII: TRIGONOMETRY
19: Trigonometry
20: Simple 2-D Problems in Right Triangle
UNIT-VIII: COORDINATE GEOMETRY
21: Coordinate Geometry
![B Nirmala Shastry solutions for मॅथेमॅटिक्स [इंग्रजी] इयत्ता ९ आयसीएसई chapter 13 - Theorems on Area B Nirmala Shastry solutions for मॅथेमॅटिक्स [इंग्रजी] इयत्ता ९ आयसीएसई chapter 13 - Theorems on Area - Shaalaa.com](/images/mathematics-english-class-9-icse_6:a927b361d63845f4b2afea4ec6bbe35a.jpg)
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Solutions for Chapter 13: Theorems on Area
Below listed, you can find solutions for Chapter 13 of CISCE B Nirmala Shastry for मॅथेमॅटिक्स [इंग्रजी] इयत्ता ९ आयसीएसई.
B Nirmala Shastry solutions for मॅथेमॅटिक्स [इंग्रजी] इयत्ता ९ आयसीएसई 13 Theorems on Area EXERCISE 13 [Pages 161 - 163]
Find the area of parallelogram ABCD, if AD = 15 cm and P is a point on BC such that AP = 9 cm. Also find the area of parallelogram ABQR, where Q and R are points on CD produced.

M is a point on side PQ of rectangle PQRS. SR is produced to a point N and MSRT is a parallelogram. PQ = 8.5 cm, PS = 6 cm. Find the area of
- parallelogram MSRT
- ΔMNT

ABCD is a rectangle, AB = 15 cm, AC = 17 cm. Find the area of parallelogram PQCA.

ABCD is a square of side 10 cm. Find the area of parallelogram BDPQ.

The perimeter of parallelogram ABCD is 42 cm. If AP = 3 cm and AQ = 4 cm, calculate the sides of parallelogram ABCD.

P is any point on the side AB of parallelogram ABCD. Prove that area (ΔAPD) + area (ΔPBC) = `1/2` area (|| gm ABCD).

PQRS is a parallelogram. A is any point on SR. PA is produced to meet QR produced at B. Prove that area (ΔQAR) = area (ΔSAB).

ABCD and PBCQ are parallelograms. Area of ΔPBC = 8 cm2. Calculate the area of parallelograms ABCD, PBCQ and ΔPCQ.

In the given figure, AC || BH and AD || EF. Show area (ΔAHF) = area (pentagon ABCDE). [Hint: Join BC and DE.]

[Hint: Join BC and DE.]
ABCD is a rectangle, AB = 4 cm, BC = 10 cm. Find the area of ΔRDC and || gm PQCD.

ABCD and ABPQ are two parallelograms. Side AQ and BC are produced to meet at R. Show that area (ΔBCD) = area (ΔBPR).

In the given figure, TQ || SR and PQ || TS. Prove that area (ΔPTS) = area (ΔTRQ).

[Hint: Join QS.]
ABCD and APDR are parallelograms. P is the mid-point of BC and Q is on CD such that DQ : QC = 2 : 1. Area (ΔPQC) = 20 cm2. Find the areas of
- ΔPDC
- ΔABP
- || gm APDR

ABCD and BEFG are parallelograms and CE || AG. Prove that Area (|| gm ABCD) = Area (|| gm BEFG)

[Hint: Prove Area (ΔABC) = Area (ΔGBE)]
ABCD is a parallelogram, BC is produced to Q so that ∠DQC = 90°. AP ⊥ DC. If AP = 2.5 cm, BC = 4 cm and DQ = 5 cm, find the length of AB.

[Hint: Area of || gm = CD × AP = AD × DQ]
ABCD is a parallelogram and AF || DE. Prove that area (|| gm DEFH) = area (|| gm ABCD).

[Hint: Each || gm is equal in area to || gm ADEG]
In ΔABC, M is the mid-point of BC, L is a point on AB such that AL = 2LB. Find the area of ΔALM if area of ΔABC = 72 cm2.

ABCD is a parallelogram, prove that
- Area (ΔABC) = Area (ΔPAD)
- Area (ΔPAB) = Area (quadrilateral ACPD)

D is the mid-point of AC. E is on BC such that BE = `1/3` BC. Area of ΔABC = 48 cm2. Find the area of ΔADB and ΔBDE.

Area of ΔPCD = 10 cm2. Find the area of ΔADQ and parallelogram ABCD.

B Nirmala Shastry solutions for मॅथेमॅटिक्स [इंग्रजी] इयत्ता ९ आयसीएसई 13 Theorems on Area MULTIPLE CHOICE QUESTIONS [Pages 163 - 165]
P is a point on side AB of parallelogram ABCD. Which of the following is true?

- Area of ΔAPD = `1/4` area || gm ABCD
- Area of ΔPDC = `1/2` area || gm ABCD
- Area of ΔAPD + area of ΔPBC = `1/2` parallelogram ABCD
only A
only B
A and B
B and C
In || gm ABCD, M is a point on AB, ∠DMC = 90°. DM = 15 cm, DC = 17 cm. ∴ Area of || gm ABCD is:

60 cm2
68 cm2
120 cm2
136 cm2
PQRS is a rectangle and AQRB is a || gm, SR = 9 cm, PS = 12 cm

Area of ΔCQR =
108 cm2
72 cm2
54 cm2
42 cm2
PQRS is a rectangle and AQRB is a || gm, SR = 9 cm, PS = 12 cm

Area of || gm AQRB =
2 area of ΔBSR
2 area of ΔPAQ
108 cm2
144 cm2
Area of ΔPMQ = 72 cm2, Area of || gm LMNO =

`1/2` area LMNQ
108 cm2
144 cm2
72 cm2
ABCD is a rectangle. PACQ is a || gm. AB = 12 cm, AC = 15 cm. Area of || gm PACQ =

180 cm2
108 cm2
54 cm2
90 cm2
Area of || gm ABCD =

AP × PC
AD × RC
AD × AP
AB × AP
AM is the median of ΔABC. Area of ΔABC = 140 cm2. AN : NC = 3 : 2. Area of ΔMNC =

70 cm2
35 cm2
28 cm2
42 cm2
P and Q are mid points of AB and AC of ΔABC. Which of the following is not true?

Area of ΔBQC = Area of ΔBPC
Area of ΔPBQ = Area of ΔPCQ
Area of ΔAPQ = Area of ΔBOC
Area of ΔABQ = Area of ΔBQC
In ΔABC, P, M and N are mid points of sides AB, BC and AC. If area of ΔABC = 120 cm2, Area of || gm APMN =

60 cm2
72 cm2
75 cm2
80 cm2
ABCD is a rectangle, DBCE is a || gm. Area of || gm DBCE =

320 cm2
192 cm2
160 cm2
240 cm2
Direction for Questions 11 to 14: In each of the following questions, a statement of assertion (A) is given and a statement of Reason (R) given below it choose the correct option for each question.
Assertion: If area of ΔAOD = 18 cm2 then area of parallelogram ABCD = 72 cm2.
Reason: The diagonals of a parallelogram bisect each other.

Both A and R are true and R is the correct reason for A.
Both A and R are true but R is the incorrect reason for A.
A is true but R is false.
A is false but R is true.
Assertion: Area of ΔPDC = 160 cm2 and area of parallelogram ABCD = 320 m2.

Reason: Area of a right angled Δ = `1/2` product of sides containing right angle.
Both A and R are true and R is the correct reason for A.
Both A and R are true but R is the incorrect reason for A.
A is true but R is false.
A is false but R is true.
Assertion: A median of a triangle divides it into 2 triangles of equal area.
Reason: The two triangles so formed are congruent.
Both A and R are true and R is the correct reason for A.
Both A and R are true but R is the incorrect reason for A.
A is true but R is false.
A is false but R is true.
Assertion: AB || DC in the figure. Area of ΔAOD = Area of ΔBOC.

Reason: Two Δs on the same base and between the same parallel lines have equal area.
Both A and R are true and R is the correct reason for A.
Both A and R are true but R is the incorrect reason for A.
A is true but R is false.
A is false but R is true.
B Nirmala Shastry solutions for मॅथेमॅटिक्स [इंग्रजी] इयत्ता ९ आयसीएसई 13 Theorems on Area MISCELLANEOUS EXERCISE [Pages 165 - 166]
ABCD is a trapezium in which AD || BC. M is the mid-point of CD. Through M, PQ is drawn || to AB with P on AD produced and Q on BC.
- Show that ΔDPM ≅ ΔCQM
- Prove that area of ΔABM = `1/2` × area of trapezium ABCD.

The area of ΔABC = 18 cm2. Find the distance between the || lines AD and BC if BC = 8 cm. Calculate the area of || gm BCDE.

ABCD is a rectangle and BCED is a parallelogram. If BC = 12 cm and CE = 15 cm, find the area of
- Parallelogram BCED
- ΔCDE

In ΔPQR, PM and QN are medians. Area of ΔGQM = 15 cm2. Find the area of ΔPQR.

ABCD is a trapezium with AB with AB || DC and diagonals AC and BD intersect at O. Prove that area of ΔAOD = area of ΔBOC.

Area of ΔABD = 30 cm2. Find the area of ΔADC, if BD = `5/8` BC.

AM is the median of ΔABC. N is on AC such that AN : NC = 3 : 2. If area of ΔMNC = 12 cm2, find the area of ΔABC.

In ΔABC, PQ || BC. Prove that

- Area (ΔPBC) = Area (ΔQBC)
- Area (ΔPOB) = Area (ΔQOC)
In adjoining figure, ABCD is a parallelogram, AQ = QB. CB and DQ are produced to meet at P. Prove that

- Area (ΔAQD) = Area (ΔAQP)
- Area (ΔDCP) = Area (|| gm ABCD)
In the given figure, PQRS and PXYZ are parallelograms. Prove that they are of equal area.

[Hint: Join XQ. Area (ΔXPQ) = `1/2` of each || gm]
In the given figure, PQRS and PXYZ are two parallelograms of equal area. Prove that SX || YR.

Solutions for 13: Theorems on Area
![B Nirmala Shastry solutions for मॅथेमॅटिक्स [इंग्रजी] इयत्ता ९ आयसीएसई chapter 13 - Theorems on Area B Nirmala Shastry solutions for मॅथेमॅटिक्स [इंग्रजी] इयत्ता ९ आयसीएसई chapter 13 - Theorems on Area - Shaalaa.com](/images/mathematics-english-class-9-icse_6:a927b361d63845f4b2afea4ec6bbe35a.jpg)
B Nirmala Shastry solutions for मॅथेमॅटिक्स [इंग्रजी] इयत्ता ९ आयसीएसई chapter 13 - Theorems on Area
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Concepts covered in मॅथेमॅटिक्स [इंग्रजी] इयत्ता ९ आयसीएसई chapter 13 Theorems on Area are Figures Between the Same Parallels, Triangles with the Same Vertex and Bases Along the Same Line, Concept of Area.
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