# Frank Solutions for Chapter 21 Areas Theorem on Parallelograms Class 9 Mathematics ICSE

### Exercise 21.1

1. ABCD is a parallelogram having an area of 60 cm2. P is a point on CD. Calculate the area of APB.

Area (△APB) = (1/2) ×area (parallelogram ABCD)

(The area of a triangle is half that of a parallelogram on the same base and between the same parallels)

Area (△APB) = (1/2)×60 cm2

We get,

Area (△APB) = 30 cm2

2. PQRS is a rectangle in which PQ = 12 cm and PS = 8 cm. Calculate the area of PRS.

Given PQRS is a rectangle

Hence,

PQ = SR

SR = 12 cm

PS = 8 cm

Area (△PRS) = (1/2)×base×height

Area (△PRS) = (1/2)×SR×PS

Area (△PRS) = (1/2)×12×8

We get,

Area (△PRS) = 48 cm2

3. In the given figure area of || gm PQRS is 30 cm2. Find the height of || gm PQFE if PQ = 6 cm.

Area (|| gm PQRS) = Area (|| gm PQFE)

(|| gm on same base PQ and between same parallel lines)

Therefore,

Area (|| gm PQFE) = 30 cm2

Base×Height = 30

⇒ 6×Height = 30

⇒ Height = 30/6

We get,

Height = 5 cm

Hence, the height of a parallelogram PQFE is 5 cm

4. In the given figure, ST || PR. Prove that: area of quadrilateral PQRS = area of PQT

Area (△PSR) = Area (△PTR)

(Triangles on the same base PR and between the same parallel lines PR and ST)

Adding Area (△PQR) on both sides

We get,

Area (△PSR) + Area (△PQR) = Area (△PTR) + Area (△PQR)

⇒ Area (Quadrilateral PQRS) = Area (△PQT)

Hence, proved

5. In the figure, ABCD is a parallelogram and APD is an equilateral triangle of side 8 cm. Calculate the area of parallelogram ABCD.

Area (△APD) = (√3s2)/4

Area (△APD) = (√3×82)/4

Area (△APD) = (√3×64)/4

Area (△APD) = (√3×16)

On further calculation, we get,

Area (△APD) = 16√3 cm2

⇒ Area (△APD) = (1/2) ×area (parallelogram ABCD)

The area of a triangle is half that of a parallelogram on the same base and between the same parallels.

Area (parallelogram ABCD) = 2×area (△APD)

⇒ Area (parallelogram ABCD) = 2×16√3 cm2

We get,

Area (parallelogram ABCD) = 32√3 cm2

6. In the figure, if the area of ||gm PQRS is 84 cm2. Find the area of

(i) || gm PQMN

(ii) PQS

(iii) PQN

(i) Area of a rectangle and area of a parallelogram on the same base is equal

Here,

For rectangle PQMN, base is PQ

For parallelogram PQRS, base is PQ

Hence,

Area of rectangle PQMN = Area of parallelogram PQRS

Area of rectangle PQMN = 84 cm2

(ii) Area (△PQS) = (1/2)× area (parallelogram PQRS)

Area (△PQS) = (1/2)×84 cm2

We get,

Area (△PQS) = 42 cm2

(iii) Area (△PQN) = (1/2)× area (rectangle PQMN)

Area (△PQN) = (1/2)×84 cm2

We get,

Area (△PQN) = 42 cm2

7.In the figure, PQR is a straight line. SQ is parallel to TP. Prove that the quadrilateral PQST is equal in area to the △PSR.

Area (△PQS) = (1/2) x area (quadrilateral PQST)

Area (quadrilateral PQST) = 2×area (△PQS) …(i)

In △PSR,

Area (△PSR) = area (△PQS) + area (△QSR)

Since QS is median as QS || TP

Hence,

Area (△PQS) = Area (△QSR)

Area (△PSR) = 2×area (△PQS) …(ii)

From equations (i) and(ii)

Area (quadrilateral PQST) = Area (△PSR)

Hence, proved

8. 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

By joining AC and FE

We get,

△AFC and △AFE are on the same base AF and between the same parallels AF and CE

Area (△AFC) = Area (△AFE)

⇒ Area (△ABF) + Area (△ABC) = Area (△ABF) + Area (△BFE)

We get,

Area (△ABC) = Area (△BFE)

⇒ (1/2) Area (parallelogram ABCD) = (1/2) Area (parallelogram BFGE)

⇒ Area (parallelogram ABCD) = Area (parallelogram BFGE)

Hence, proved

9. In the given figure, the perimeter of parallelogram PQRS is 42 cm. Find the lengths of PQ and PS.

Area of || gm PQRS = PQ×6

Also,

Area of || gm PQRS = PS×8

Therefore,

PQ ×6 = PS×8

⇒ PQ = (8PS)/6

We get,

PQ = (4PS)/3 …(i)

Perimeter of || gm PQRS = PQ + QR + RS + PS

42 = 2PQ + 2PS (Opposite sides of a parallelogram are equal)

⇒ 21 = PQ + PS

Substituting the value of PQ from equation (i), we get,

(4PS/3) + PS = 21

⇒ (4PS + 3PS)/3 = 21

⇒ 7PS = 63

We get,

PS = 9 cm

Now,

Substituting the value of PS in equation (i), we get,

PQ = (4PS)/3

⇒ PQ = (4×9)/3

We get,

PQ = 12 cm

Therefore, PQ = 12 cm and PS = 9 cm

10. In the given figure, PT || QR and QT || RS. Show that: area of PQR = area of TQS.

By joining TR, we get,

△PQR and △QTR are on the same base QR and between the same parallel lines QR and PT

Therefore,

Area (△PQR) = Area (△QTR) …(i)

△QTR and △TQS are on the same base QT and between the same parallel lines QT and RS

Therefore,

Area (△QTR) = Area (△TQS) …(ii)

From equations (i) and (ii), we get,

Area (△PQR) = Area (△TQS)

Hence, proved

11. In the given figure, PQR is right angled at P. PABQ and QRST are squares on the side PQ and hypotenuse QR. If PN TS, show that:

(a) QRB PQT

(b) Area of square PABQ = area of rectangle QTNM.

∠BQR = ∠BQP + ∠PQR

⇒ ∠BQR = 90° + ∠PQR

⇒ ∠PQT = ∠TQR + ∠PQR

⇒ ∠PQT = 90° + ∠PQR

Hence,

∠BQR = ∠PQT …(i)

(a) In △QRB and △PQT,

BQ = PQ …(sides of a square PABQ)

QR = QT …(sides of a square QRST)

∠BQR = ∠PQT …{From (i)}

Therefore,

△QRB ≅ △PQT (By SAS congruence criterion)

Area (△BQR) = Area (△PQT) …(ii)

(b) △PQT and rectangle QTNM are on the same base QT and between the same parallel lines QT and PN

Hence,

Area (△PQT) = (1/2) Area (rectangle QTNM)

Area (rectangle QTNM) = 2×Area (△PQT)

Area (rectangle QTNM) = 2×Area (△BQR) {from (ii)} …(iii)

△BQR and square PABQ are on the same base BQ and between the same parallel lines BQ and AR

Therefore,

2 ×Area (△BQR) = Area (square PABQ) …(iv)

From equations (iii) and (iv), we get,

Area (square PABQ) = Area (rectangle QTNM)

Hence, proved

12. In the figure, AB = BE. Prove that the area of triangle ACE is equal in area to the parallelogram ABCD.

In parallelogram ABCD,

Area (△ABC) = (1/2)×Area (parallelogram ABCD)

(The area of a triangle is half that of a parallelogram on the same base and between the same parallels)

Area (parallelogram ABCD) = 2×Area (△ABC) …(i)

In △ACE,

Area (△ACE) = Area (△ABC) + Area (△BCE)

Since BC is median,

Hence,

Area (△ABC) = Area (△BCE)

Area (ACE) = 2×Area (△ABC) …(ii)

From equation

D is the midpoint

Area (△ABD) = Area (△ADC)

Hence, proved

13. Prove that the median of a triangle divides it into two triangles of equal area.

Draw AL perpendicular to BC

Since AD is median of △ABC

Hence,

D is the midpoint of BC

BD = DC

Multiplying by AL, we get,

BD ×AL = DC×AL

⇒ (1/2) (BD×AL) = (1/2) (DC×AL)

Therefore,

Area (△ABD) = Area (△ADC)

Hence, proved

14. AD is a median of a ABC. P is any point on AD. Show that the area of ABP is equal to the area of ACP.

AD is the median of △ABC.

So, it will divide △ABC into two triangles of equal areas

Hence,

Area (△ABD) = Area (△ACD) …(i)

Now,

PD is the median of △PBC

Hence,

Area (△PBD) = Area (△PCD) …(ii)

On subtracting equation (ii) from equation (i), we get,

Area (△ABD) – Area (△PBD) = Area (△ACD) – Area (△PCD)

⇒ Area (△ABP) = Area (△ACP)

Hence, proved

15. In the given figure AF = BF and DCBF is a parallelogram. If the area of ABC is 30 square units, find the area of the parallelogram DCBF.

In △ABC,

AF = FB and EF || BC …(given)

Hence, AE = EC …(converse of mid-point theorem) …(i)

In △AEF and △CED,

∠FEA = ∠DEC …(vertically opposite angles)

CE = AE …{From (i)}

∠FAE = ∠DCE …(Alternate angles)

Therefore,

△AEF ≅ △CED (By ASA test of congruency)

Area (△AEF) = Area (△CED) …(ii)

⇒ Area (△ABC) = Area (△AEF) + Area (EFBC)

⇒ Area (△ABC) = Area (△CED) + Area (EFBC) ...{from (ii)}

Therefore,

Area (△ABC) = Area (parallelogram DCBF)

Hence, area of parallelogram DCBF is 30 square units

16. In the figure, PT is parallel to SR. QTSR is a parallelogram and PQSR is a rectangle. If the area of QTS is 60 cm2, find:

(i) the area of ||gm QTSR

(ii) the area of the rectangle PQRS

(iii) the area of the triangle PQS.

(i)

(ii)

(iii)

17. 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.

18. The diagonals of a parallelogram ABCD intersect at O. A line through O meets AB in P and CD in Q. Show that

20. Prove that the area of a rhombus is equal to half the rectangle contained by its diagonals.

21. PQRS is a parallelogram and O is any point in its interior. Prove that:

Area(POQ) + area(ROS) = area(QOR) + area(SOP) = 1/2area(||gm PQRS)

24. In the given figure, AB || SQ || DC and AD || PR || BC. If the area of quadrilateral ABCD is 24 square units, find the area quadrilateral PQRS.

25. PQR and SQR are on the same base QR with P and S on opposite sides of line QR, such that area of PQR is equal to the area of SQR. Show that QR bisects PS.

26. If the medians of a ABC intersect G, show that ar(AGB) = ar(AGC) = ar(BGC) = 1/3ar(ABC).

27. In ABC, 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 ABC.

28. In the given figure, PQ || SR || MN, PS || QM and SM || PN. Prove that :

ar.(SMNT) = ar. (PQRS).

29. In PQR, PS is a median. T is the mid-point of SR and M is the mid-point of PT. Prove that : PMR = 1/8 PQR.

30. In the figure, ABCD is a parallelogram and CP is parallel to DB. Prove that :

Area of OBPC = ¾ area of ABCD

31. The medians QM and RN of PQR intersect at O. Prove that: area of ROQ = area of quadrilateral PMON.

32. (a) 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 ABE = Area of ACE.

(b) 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 ABC = 4 × Area of ABE

(a)

(b)

33. In a parallelogram PQRS, M and N are the midpoints of the sides PQ and PS respectively. If area of PMN is 20 square units, find the area of the parallelogram PQRS.

Construction: Join SM and SQ.

34. In a parallelogram PQRS, T is any point on the diagonal PR. If the area of PTQ is 18 square units find the area of PTS.