# 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 cm ^{2}. P is a point on CD. Calculate the area of **

**△**

**APB.**

**Answer**

(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 cm^{2}

We get,

Area (△APB) = 30 cm^{2}

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

**Answer**

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 cm^{2}

**3. ****In the given figure area of || gm PQRS is 30 cm ^{2}. Find the height of || gm PQFE if PQ = 6 cm.**

**Answer:**

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

Therefore,

Area (|| gm PQFE) = 30 cm^{2}

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**

**Answer**

(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.**

**Answer:**

Area (△APD) = (√3s^{2})/4

Area (△APD) = (√3×8^{2})/4

Area (△APD) = (√3×64)/4

Area (△APD) = (√3×16)

On further calculation, we get,

Area (△APD) = 16√3 cm^{2}

⇒ 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 cm^{2}

We get,

Area (parallelogram ABCD) = 32√3 cm^{2}

**6. In the figure, if the area of ||gm PQRS is 84 cm ^{2}. Find the area of**

**(i) || gm PQMN**

**(ii) ****△****PQS**

**(iii) ****△****PQN**

**Answer**

**(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 cm^{2}

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

Area (△PQS) = (1/2)×84 cm^{2}

We get,

Area (△PQS) = 42 cm^{2}

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

Area (△PQN) = (1/2)×84 cm^{2}

We get,

Area (△PQN) = 42 cm^{2}

**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.**

**Answer**

In quadrilateral PQST,

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**

**Answer**

By joining AC and FE

We get,

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.**

**Answer**

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**.

**Answer**

By joining TR, we get,

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**.

**Answer**

∠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.**

**Answer**

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.**

**Answer**

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.**

**Answer**

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.**

**Answer**

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 cm ^{2}, find: **

**(i) the area of ****||gm QTSR **

**(ii) the area of the rectangle PQRS **

**(iii) the area of the triangle PQS. **

**Answer**

(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. **

**Answer**

**18. 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. **

**(ii) Area of APQD = Area of BPQC**

**Answer**

**19. The diagonals AC and BD of a quadrilateral ABCD intersect at O. Prove that if BO = OD, then areas of ****△****ABC and ****△****ADC are equal. **

**Answer**

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

**Answer**

**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**)

**Answer**

**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. **

**Answer**

**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. **

**Answer**

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

**Answer**

**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. **

**Answer**

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

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

**Answer**

**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.**

**Answer**

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

**Area of OBPC = ¾ area of ABCD**

**Answer**

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

**Answer**

**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**

**Answer**

(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. **

**Answer**

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. **

**Answer**

Construction: Join QR. Let the diagonals PR and QS intersect each other at point O.

**35. ABCD is a quadrilateral in which diagonals AC and BD intersect at a point O. Prove that: area ****△****AOD + area ****△****BOC = area ****△****ABO + area ****△****CDO. **

**Answer**