Chapters
Chapter 2: Profit , Loss and Discount
Chapter 3: Compound Interest
Chapter 4: Expansions
Chapter 5: Factorisation
Chapter 6: Changing the subject of a formula
Chapter 7: Linear Equations
Chapter 8: Simultaneous Linear Equations
Chapter 9: Indices
Chapter 10: Logarithms
Chapter 11: Triangles and their congruency
Chapter 12: Isosceles Triangle
Chapter 13: Inequalities in Triangles
Chapter 14: Constructions of Triangles
Chapter 15: Mid-point and Intercept Theorems
Chapter 16: Similarity
Chapter 17: Pythagoras Theorem
Chapter 18: Rectilinear Figures
Chapter 19: Quadrilaterals
Chapter 20: Constructions of Quadrilaterals
Chapter 21: Areas Theorems on Parallelograms
Chapter 22: Statistics
Chapter 23: Graphical Representation of Statistical Data
Chapter 24: Perimeter and Area
Chapter 25: Surface Areas and Volume of Solids
Chapter 26: Trigonometrical Ratios
Chapter 27: Trigonometrical Ratios of Standard Angles
Chapter 28: Coordinate Geometry

Chapter 21: Areas Theorems on Parallelograms
Frank solutions for Class 9 Maths ICSE Chapter 21 Areas Theorems on Parallelograms Exercise 21.1 [Page 25]
ABCD is a parallelogram having an area of 60cm2. P is a point on CD. Calculate the area of ΔAPB.
PQRS is a rectangle in which PQ = 12cm and PS = 8cm. Calculate the area of ΔPRS.
In the figure, PT is parallel to SR. QTSR is a parallelogram and PQSR is a rectangle. If the area of ΔQTS is 60cm2, find:
(i) the area o || gm QTSR
(ii) the area of the rectangle PQRS
(iii) the area of the triangle PQS.
In the given figure area of ∥ gm PQRS is 30 cm2. Find the height of ∥ gm PQFE if PQ = 6 cm.
In the given figure, PQRS is a ∥ gm. A straight line through P cuts SR at point T and QR produced at N. Prove that area of triangle QTR is equal to the area of triangle STN.
In the given figure, ST ∥ PR. Prove that: area of quadrilateral PQRS = area of ΔPQT.
In the figure, ABCD is a parallelogram and APD is an equilateral triangle of side 80cm, Calculate the area of parallelogram ABCD.
In the figure, if the area of ||gm PQRS is 84cm2; find the area of
(i) || gm PQMN
(ii) ΔPQS
(iii) ΔPQN
In the figure, PQR is a straight line. SQ is parallel to Tp. Prove that the quadrilateral PQST is equal in area to the ΔPSR.
In the given figure, if AB ∥ DC ∥ FG and AE is a straight line. Also, AD ∥ FC. Prove that: area of ∥ gm ABCD = area of ∥ gm BFGE.
In the given figure, the perimeter of parallelogram PQRS is 42 cm. Find the lengths of PQ and PS.
In the given figure, PT ∥ QR and QT ∥ RS. Show that: area of ΔPQR = area of ΔTQS.
*Question modified
In the given figure, ΔPQR is right-angled at P. PABQ and QRST are squares on the side PQ and hypotenuse QR. If PN ⊥ TS, show that:
(a) ΔQRB ≅ ΔPQT
(b) Area of square PABQ = area of rectangle QTNM.
In the figure, AE = BE. Prove that the area of triangle ACE is equal in area to the parallelogram ABCD.
The diagonals of a parallelogram ABCD intersect at O. A line through O meets AB in P and CD in Q. Show that
(a) Area of APQD = `(1)/(2)` area of || gm ABCD
(b) Area of APQD = Area of BPQC
Prove that the median of a triangle divides it into two triangles of equal area.
AD is a median of a ΔABC.P is any point on AD. Show that the area of ΔABP is equal to the area of ΔACP.
In the given figure AF = BF and DCBF is a parallelogram. If the area of ΔABC is 30 square units, find the area of the parallelogram DCBF.
Prove that the diagonals of a parallelogram divide it into four triangles of equal area.
The diagonals AC and BC of a quadrilateral ABCD intersect at O. Prove that if BO = OD, then areas of ΔABC an ΔADC area equal.
Prove that the area of a rhombus is equal to half the rectangle contained by its diagonals.
PQRS is a parallelogram and O is any point in its interior. Prove that: area(ΔPOQ) + area(ΔROS) - area(ΔQOR) + area(ΔSOP) = `(1)/(2)`area(|| gm PQRS)
A quadrilateral ABCD is such that diagonals BD divides its area into two equal parts. Prove that BD bisects AC.
In the given figure, BC ∥ DE.
(a) If area of ΔADC is 20 sq. units, find the area of ΔAEB.
(b) If the area of ΔBFD is 8 square units, find the area of ΔCEF
ΔPQR and ΔSQR are on the same base QR with P and S on opposite sides of line QR, such that area of ΔPQR is equal to the area of ΔSQR. Show that QR bisects PS.
If the medians of a ΔABBC intersect at G, show that ar(ΔAGB) = ar(ΔAGC) = ar(ΔBGC) = `(1)/(3)"ar(ΔABC)"`.
In ΔABC, the mid-points of AB, BC and AC are P, Q and R respectively. Prove that BQRP is a parallelogram and that its area is half of ΔABC.
In the given figure, PQ ∥ SR ∥ MN, PS ∥ QM and SM ∥ PN. Prove that: ar. (SMNT) = ar. (PQRS).
In ΔPQR, PS is a median. T is the mid-point of SR and M is the mid-point of PT. Prove that: ΔPMR = `(1)/(8)Δ"PQR"`.
In the figure, ABCD is a parallelogram and CP is parallel to DB. Prove that: Area of OBPC = `(3)/(4)"area of ABCD"`
The medians QM and RN of ΔPQR intersect at O. Prove that: area of ΔROQ = area of quadrilateral PMON.
In the given figure, ABC is a triangle and AD is the median.
If E is any point on the median AD. Show that: Area of ΔABE = Area of ΔACE.
In the given figure, ABC is a triangle and AD is the median.
If E is the midpoint of the median AD, prove that: Area of ΔABC = 4 × Area of ΔABE
In a parallelogram PQRS, M and N are the midpoints of the sides PQ and PS respectively. If area of ΔPMN is 20 square units, find the area of the parallelogram PQRS.
In a parallelogram PQRS, T is any point on the diagonal PR. If the area of ΔPTQ is 18 square units find the area of ΔPTS.
ABCD is a quadrilateral in which diagonals AC and BD intersect at a point O. Prove that: areaΔAOD + areaΔBOC + areaΔABO + areaΔCDO.
In the given figure, AB ∥ SQ ∥ DC and AD ∥ PR ∥ BC. If the area of quadrilateral ABCD is 24 square units, find the area of quadrilateral PQRS.
Chapter 21: Areas Theorems on Parallelograms

Frank solutions for Class 9 Maths ICSE chapter 21 - Areas Theorems on Parallelograms
Frank solutions for Class 9 Maths ICSE chapter 21 (Areas Theorems on Parallelograms) include all questions with solution and detail explanation. This will clear students doubts about any question and improve application skills while preparing for board exams. The detailed, step-by-step solutions will help you understand the concepts better and clear your confusions, if any. Shaalaa.com has the CISCE Class 9 Maths ICSE solutions in a manner that help students grasp basic concepts better and faster.
Further, we at Shaalaa.com provide such solutions so that students can prepare for written exams. Frank textbook solutions can be a core help for self-study and acts as a perfect self-help guidance for students.
Concepts covered in Class 9 Maths ICSE chapter 21 Areas Theorems on Parallelograms are Introduction of Rectilinear Figures, Names of Polygons, Concept of Quadrilaterals - Sides, Adjacent Sides, Opposite Sides, Angle, Adjacent Angles and Opposite Angles, Diagonal Properties of Different Kinds of Parallelograms, Property: The Diagonals of a Rectangle Are of Equal Length., Property: The diagonals of a square are perpendicular bisectors of each other., Types of Quadrilaterals, Property: The Opposite Sides of a Parallelogram Are of Equal Length., Property: The Opposite Angles of a Parallelogram Are of Equal Measure., Property: The diagonals of a rhombus are perpendicular bisectors of one another..
Using Frank Class 9 solutions Areas Theorems on Parallelograms exercise by students are an easy way to prepare for the exams, as they involve solutions arranged chapter-wise also page wise. The questions involved in Frank Solutions are important questions that can be asked in the final exam. Maximum students of CISCE Class 9 prefer Frank Textbook Solutions to score more in exam.
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