# RD SHARMA Solutions for Class 9 Maths Chapter 14 - Areas of Parallelograms and Triangles

## Chapter 14 - Areas of Parallelograms and Triangles Exercise 14.60

Two parallelograms are on the same base and between the same parallels. The ratio of their areas is

(a) 1 : 2

(b) 2 : 1

(c) 1 : 1

(d) 3 : 1

Area of parallelogram = Base × height

Base = Length of base

Height = distance between Base and Side parallel to it

In figure, there are two Parallelograms.

Base of both is same, and because both lie under same parallels that's why height is also same.

Thus, the Ratio of Areas of both parallelogram = 1 : 1

Hence, correct option is (c).

A triangle and a parallelogram are on the same base and between the same parallels.

The ratio of the areas of triangle and parallelogram is

(a) 1 : 1

(b) 1 : 2

(c) 2 : 1

(d) 1 : 3

Let ABC be a triangle of area 24 sq. units and PQR be the triangle formed by the mid-points of sides of ΔABC.

Then the area of ΔPQR is

(a) 12 sq. units

(b) 6 sq. units

(c) 4 sq. units

(d) 3 sq. units

When a triangle is formed by joining the mid-points of sides of a triangle, the triangle formed is Congruent to triangles formed around that.

i.e. ΔPQR is congruent to ΔRPA, ΔQBP & ΔCQR.

Hence, Area of all four triangles formed inside ΔABC is same.

So (4 × Area of any one Δ) = Area of ΔABC

4 × (Area of ΔPQR) = 24 sq. units

Area of ΔPQR = 6 sq. units

Hence, correct option is (b).

The median of a triangle divides it into two

(a) congruent triangles

(b) isosceles triangles

(c) right triangles

(d) triangles of equal areas

A median divides the base in two equal parts but height of a triangle remains the same.

Now, since bases and heights are equal, areas of both Δs are equal.

Hence, correct option is (d).

In a ΔABC, D, E, F are the mid-points of sides BC, CA and AB respectively. If ar(ΔABC) = 16 cm^{2} , then ar(trapezium FBCE) =

(a) 4 cm^{2}

(b) 8 cm^{2}

(c) 12 cm^{2}

(d) 10 cm^{2}

ABCD is a parallelogram. P is any point on CD. If ar (ΔDPA) = 15 cm^{2} and ar(ΔAPC) = 20 cm^{2}, then ar(ΔAPB) =

(a) 15 cm^{2}

(b) 20 cm^{2}

(c) 35 cm^{2}

(d) 30 cm^{2}

## Chapter 14 - Areas of Parallelograms and Triangles Exercise 14.61

The area of the figure formed by joining the mid-points of the adjacent sides of a rhombus with diagonals 16 cm and 12 cm is

(a) 28 cm^{2}

(b) 48 cm^{2}

(c) 96 cm^{2}

(d) 24 cm^{2}

A, B, C, D are mid-points of sides of parallelogram PQRS. If ar(PQRS) = 36 cm^{2}, then ar(ABCD) =

(a) 24 cm^{2}

(b) 18 cm^{2}

(c) 30 cm^{2}

(d) 36 cm^{2}

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 rhombus of area 24 cm^{2}

(b) a rectangle of area 24 cm^{2}

(c) a square of area 26 cm^{2}

(d) a trapezium of area 14 cm^{2}

If AD is median of ΔABC and P is a point on AC such that

ar(ΔADP) : ar (ΔABC) = 2 : 3, then ar (ΔPDC) : ar(ΔABC) is

(a) 1 : 5

(b) 1 : 5

(c) 1 : 6

(d) 3 : 5

Medians of ΔABC intersect at G. If ar(ΔABC) = 27 cm^{2}, then ar(ΔBGC) =

(a) 6 cm^{2}

(b) 9 cm^{2}

(c) 12 cm^{2}

(d) 18 cm^{2}

In a ΔABC if D and E are mid-points of BC and AD respectively such that ar(ΔAEC) = 4cm^{2}, then ar(ΔBEC) =

(a) 4 cm^{2}

(b) 6 cm^{2}

(c) 8 cm^{2}

(d) 12 cm^{2}

E is the mid-point of AD and CE is median of ΔACD.

Hence Ar(ΔAEC) = Ar(ΔCED) = 4 cm^{2} ....(1)

(Median divides a Δ in two two equal Areas)

Also AD is median of ΔABC and and ED is median of ΔBEC.

So Ar(ΔBED) = Ar(CED) = 4 cm^{2} [From eq (1)]

So Ar(ΔBEC) = Ar(ΔBED) + Ar(ΔCED) = 4 + 4 = 8 cm^{2}

Hence, correct option is (c).

In figure, ABCD is a parallelogram. If AB = 12 cm, AE = 7.5 cm, CF = 15 cm, then AD =

(a) 3 cm

(b) 6 cm

(c) 8 cm

(d) 10.5 cm

Area of parallelogram = AD × FC = AB × AE

Thus,

AD × 15 = 12 × 7.5

AD = 6 cm

Hence, correct option is (b).

In figure, PQRS is a parallelogram. If X and Y are mid-points of PQ and SR respectively and diagonal SQ is joined. The ratio ar(||^{gm} XQRY) : ar (ΔQSR) =

(a) 1 : 4

(b) 2 : 1

(c) 1 : 2

(d) 1 : 1

Diagonal AC and BD of trapezium ABCD, in which AB || DC, intersect each other at O. The triangle which is equal in area of ΔAOD is

(a) ΔAOB

(b) ΔBOC

(c) ΔDOC

(d) ΔADC

ΔABD & ΔABC have same base AB and are between same parallels.

Then,

Ar(ΔABD) = Ar(ΔABC)

But Ar(ΔAOB) is common in both.

Thus, Ar(ΔAOD) = Ar(ΔBOC)

Hence, correct option is (b).

ABCD is a trapezium in which AB || DC. If ar(ΔABD)= 24 cm^{2} and AB = 8 cm, then height of ΔABC is

(a) 3 cm

(b) 4 cm

(c) 6 cm

(d) 8 cm

## Chapter 14 - Areas of Parallelograms and Triangles Exercise 14.62

In figure, ABCD and FECG are parallelograms equal in area. If ar(ΔAQE) = 12 cm^{2}, then ar(||^{gm} FGBQ) =

(a) 12 cm^{2}

(b) 20 cm^{2}

(c) 24 cm^{2}

(d) 36 cm^{2}

ABCD is a trapezium with parallel sides AB = a and DC = b. If E and F are mid-points of non-parallel sides AD and BC respectively, then the ratio of area of quadrilaterals ABFE and EFCD is

(a) a : b

(b) (a + 3b) : (3a + b)

(c) (3a + b) : (a + 3b)

(d) (2a + b) : (3a + b)

ABCD is a rectangle with O as any point in its interior. If ar (ΔAOD) = 3 cm^{2} and ar (ΔBOC) = 6 cm^{2}, then area of rectangle ABCD is

(a) 9 cm^{2}

(b) 12 cm^{2}

(c) 15 cm^{2}

(d) 18 cm^{2}

## Chapter 14 - Areas of Parallelograms and Triangles Exercise Ex. 14.1

Which of the following figures lie on the same base and between the same parallels. In such a case, write the common base and two parallels:

(i) ΔAPB and trapezium ABCD are on the same base AB and between the same parallels AB and CD.

(ii) Parallelograms ABCD and APQD are on the same base AD and between the same parallels AD and BQ.

(iii) Parallelogram ABCD and ΔPQR are between the same parallels AD and BC but they are not on the same base.

(iv) ΔQRT and parallelogram PQRS are on the same base QR and between the same parallels QR and PS.

(v) Parallelogram PQRS and trapezium SMNR are on the same base SR but they are not between the same parallels.

(vi) Parallelograms PQRS, AQRD, BQRC are between the same parallels. Also, parallelograms PQRS, BPSC and APSD are between the same parallels.

## Chapter 14 - Areas of Parallelograms and Triangles Exercise Ex. 14.2

We know that,

Area of parallelogram = Base x corresponding attitude

Area of parallelogram ABCD = CD x AE = AD x CF

16 cm x 8 cm = AD x 10 cm

In Q. No. 1, if AD = 6 cm, CF = 10 cm, and AE = 8, find AB.

## Chapter 14 - Areas of Parallelograms and Triangles Exercise Ex. 14.3

In fig., compute the area of quadrilateral ABCD.

In the fig., PQRS is a square and T and U are, respectively, the mid-points of PS and QR. Find the area of ΔOTS if PQ = 8 cm.

Compute the area of trapezium PQRS in fig.

In fig., ∠AOB = 90, AC = BC, OA = 12 cm and OC = 6.5 cm. find the area of ΔAOB

In fig., ABCD is a trapezium in which AB = 7 cm, AD = BC = 5 cm, DC = x cm, and distance between AB and DC is 4 cm. Find the value of x and area of trapezium ABCD.

In fig., ABCD is a trapezium in which AB ∥ DC. PRove that ar (ΔAOD) = ar (ΔBOC)

In fig., ABC and ABD are two triangles on the base Ab. If the line segment CD is bisected by AB at O, show that ar (Δ ABC) = ar (Δ ABD).

In fig., D and E are two points on BC such that BD = DE = EC. Show that ar (ΔABD) = ar (ΔADE) = ar(ΔAEC).

Draw a line *l* through A parallel to BC.

Given that, BD = DE = EC.

We observe that the triangles ABD, ADE and AEC are on the equal bases and between the same parallels *l* and BC. Therefore, their areas are equal.

Hence, ar (ΔABD) = ar (ΔADE) = ar(ΔAEC).

(i)

In fig., CD ∥ AE and CY ∥ BA.

(i) Name a triangle equal in area of ΔCBX

(ii) Prove that ar (ΔZDE) = ar (ΔCZA)

(iii) Prove that ar (BCZY) = ar (ΔEDZ)

In fig., PSDA is a parallelogram in which PQ = QR = RS and AP ∥ BQ ∥CR. Prove that ar(Δ PQE) = ar (Δ CFD).

In fig., ABCD is a trapezium in which AB DC and DC = 40 cm and AB = 60 cm. If X and Y are, respectively, the mid - points of AD and BC, prove that:

(i) XY = 50 cm

(ii) DCYX is a trapezium

(iii) ar (trap. DCYX) = (9/11)ar (trap.(XYBA)

In fig., X and Y are the mid-points of AC and AB respectively, QP ∥ BC and CYQ and BXP are straight lines. Prove that ar(Δ ABP) = ar (Δ ACQ)

In fig., ABCD and AEFD are two parallelograms. Prove that

(i) PE = FQ

(ii) ar(ΔAPE) : ar(ΔPFA) = ar Δ(QFD) : ar (ΔPFD)

(iii) ar(ΔPEA) = ar (ΔQFD)

In fig. ABCD is a ∥^{gm}. O is any point on AC. PQ ∥ AB and LM ∥ AD. Prove that ar(∥^{gm} DLOP) = ar (∥^{gm} BMOQ).

In fig., ABC and BDE are two equilateral triangles such that D is the mid-point of BC. AE intersects BC in F. Prove that

(i) ar (_{}BDE) = _{}ar _{}(ABC)

(ii) ar(_{}BDE) = _{}ar(_{}BAE)

(iii) ar (_{}BFE) = ar(_{}AFD)

(iv) ar(_{}ABC) = 2 ar(_{}BEC)

(v) ar (_{}FED) = _{}ar(AFC)

(vi) ar(_{}BFE) = 2 ar (_{}EFD)

Given, ABC and BDE are two equilateral triangles.

Let AB = BC = CA = x. Then, BD = = DE = BE

(i) We have,

ar(ABC) = x^{2}

ar (BDE) =

ar(BDE) = ar (ABC)

(ii) It is given that triangles ABC and BED are equilateral triangles.

ACB = DBE = 60^{o}

BE||AC(Since, alternate angles are equal)

Triangles BAE and BEC are on the same base BE and between the same parallels BE and AC.

ar (BAE) = ar(BEC)

ar (BAE) =2 ar (BDE)

[ ED is a median of EBC ar(BEC) = 2ar(BDE)]

ar (BDE) = ar(BAE)

(iii) Since ABC and BDE are equilateral triangles.

ABC = 60^{o} and BDE = 60^{o}

ABC = BDE

AB||DE(Since, alternate angles are equal)

Triangles BED and AED are on the same base ED and between the same parallels AB and DE.

ar (BED) = ar(AED)

ar (BED) ar(EFD) = ar(AED) ar(EFD)

ar(BEF) = ar(AFD)

(iv) Since ED is a median of BEC

ar (BEC) = 2 ar (BDE)

ar (BEC) = ar (ABC)[From (i), ar (BDE) = ar (ABC)]

ar(BEC) = ar (ABC)

ar (ABC) = 2 ar (BEC)

(v) Let h be the height of vertex E, corresponding to the side BD in triangle BDE.

Let H be the height of vertex A, corresponding to the side BC in triangle ABC.

From part (i),

ar(BDE) = ar (ABC)

From part (iii),

ar (BFE) = ar (AFD)

(vi) ar (AFC) = ar (AFD) + ar (ADC)

= ar (BFE) + ar (ABC)

(Using part (iii); and AD is the median of ABC)

= ar (BFE) + 4 ar (BDE)(Using part (i))

= ar (BFE) + 2 ar (BDE) (2)

Now, from part (v),

ar (BFE) = 2ar (FED) (3)

ar (BDE) = ar (BFE) + ar (FED)

= 2 ar (FED) + ar (FED)

= 3 ar (FED) (4)

From (2), (3) and (4), we get,

ar (AFC) = 2ar (FED) + 2 3 ar (FED) = 8 ar (FED)

Hence, ar (FED) = ar(AFC)

If fig., ABC is a right triangle right angled at A, BCED, ACFG and ABMN are square on the sides BC, CA and AB respectively. Line segment AX DE meets BC at Y. Show that

(i) MBC ABD

(ii) ar (BYXD) = 2ar(MBC)

(iii) ar(BYXD) = ar(ABMN)

(iv) FCB ACE

(v) ar(CYXE) = 2ar (FCB)

(vi) ar(CYXE) = ar (ACFG)

(vii) ar(BCED) = ar (ABMN) + ar (ACFG)

(i) In MBC and ABD, we have

MB = AB

BC = BD

And MBC = ABD

[MBC and ABD are obtained by adding ABC to a right angle]

So, by SAS congruence criterion, we have

MBC ABD

ar (MBC) = ar(ABD) (1)

(ii) Clearly, triangle ABD and rectangle BYXD are on the same base BD and between the same parallels AX and BD.

ar(ABD) = ar (rect. BYXD)

ar (rect. BYXD) = 2 ar(ABD)

ar (rect. BYXD) = 2 ar (MBC)...(2)

[ ar (ABD) = ar (MBC), from (1)]

(iii) Since triangle MBC and square MBAN are on the same base MB and between the same parallels MB and NC.

2 ar (MBC) = ar (MBAN) (3)

From (2) and (3), we have

ar (sq. MBAN) = ar(rec. BYXD)

(iv) In triangles FCB and ACE, we have

FC = AC

CB = CE

And, FCB = ACE

[FCB and ACE are obtained by adding ACB to a right angle]

So, by SAS congruence criterion, we have

FCB ACE

(v) We have,

FCB ACE

ar (FCB) = ar (ACE)

Clearly, ACE and rectangle CYXE are on the same base CE ad between the same parallels CE and AX.

2 ar (ACE) = ar (CYXE)

2 ar (FCB) = ar (CYXE) (4)

(vi) Clearly, FCB and rectangle FCAG are on the same base FC and between the same parallels FC and BG.

2ar (FCB) = ar(FCAG) (5)

From (4) and (5), we get

ar(CYXE) = ar (ACFG)

(vii) Applying Pythagoras theorem in ACB, we have

BC^{2} = AB^{2} + AC^{2}

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