Areas of Parallelograms and Triangles, Class 9 Maths R.D Sharma Solutions

Page. 15.3 Ex.15.1

Q1.

GIVEN: Here in the question figure 1 to 6 are shown.

To Find :

(1) The figures which lie on the same base and between the same parallels.

(2) Write the common base and parallels.

As we know that ‘Two geometric figures are said to be on the same base and between the same parallels, if they have a common side(base) and the vertices (or vertex) opposite to the common base of each figure lie on a line parallel to the base.’

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

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

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

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

(5) ΔQRT and Parallelogram PQRS are on the same base QR and between the same parallels PS and QR.

(6) Parallelogram PQRS, AQRD, and BCQR are between the same parallels. Also Parallelogram PQRS, BPSC and APSD are between the same parallels.

Page. 15.14 Ex.15.2

Q1.

Given: Here in the question it is given

(1) ABCD is a parallelogram,

(2) and

(3) , AB = 16 cm

(4) AE = 8cm

(5) CF = 10cm

To Find : AD =?

Calculation: We know that formula for calculating the

Therefore,

Area of paralleogram ABCD = DC × AE (Taking base as DC and Height as AE )

Area of paralleogram ABCD = AB × AE (AB = DC as opposite side of the parallelogram are equal)

Therefore,

Area of paralleogram ABCD = 16 × 8 ……(1)

Taking the base of Parallelogram ABCD as AD we get

Area of paralleogram ABCD = AD × CF (taking base as AD and height as CF)

Area of paralleogram ABCD = AD × 10 ……(2)

Since equation 1 and 2 both represent the Area of the same Parallelogram ABCD , both should be equal.

Hence fro equation (1) and (2),

This means that,

Hence we get the result as

Q2.

Given: Here in the question it is given that

(1) ABCD is a parallelogram,

(2) and

(3)

(4) AD = 6 cm

(5) AE = 8cm

(6) CF = 10cm

To Find : AB =?

Calculation: We know that formula for calculating the

_{Area of paralleogram = base × height}

Therefore,

Area of paralleogram ABCD = DC × AE (Taking base as DC and Height as AE )

Area of paralleogram ABCD = AB × AE (AB = DC as opposite side of the parallelogram are equal)

Therefore, Area of paralleogram ABCd = 16 × 8

Area of Parallelogram ABCD = AB× 8 ……(1)

Taking the base of Parallelogram ABCD as AD we get

Area of paralleogram ABCD = AD × CF (taking base as AD and height as CF)

Area of paralleogram ABCD = 6 × 10 ……(2)

Since equation 1and 2 both represent the Area of the same Parallelogram ABCD , both should be equal.

Hence equation 1 is equal to equation 2

Which means that,

Hence we got the measure of AB equal to

Q3.

Given: Here in the question it is given that

(1) Area of paralleogram ABCD = 124 cm²

(2) E is the midpoint of AB, which means

(3) F is the midpoint of CD, which means

To Find : Area of Parallelogram AEFD

Calculation: We know that formula for calculating the

Area of Parallelogram = base × height

Therefore,

Area of paralleogram ABCD = AB × AD (Taking base as AB and Height as AD ) ……(1)

Therefore,

Area of paralleogram AEFD = AE × AD (Taking base as AB and Height as AD ) ……(2)

()

= Area of Parallelogram ABCD (from equation1)

Hence we got the result Area of Parallelogram AEFD

Q4.

Given: Here in the question it is given that

(1) ABCD is a Parallelogram

To Prove :

(1)

(2)

(3)

(4)

Construction: Draw

Calculation: We know that formula for calculating the

Area of Parallelogram = base × height

Area of paralleogram ABCD = BC × AE (Taking base as BC and Height as AE ……(1)

We know that formula for calculating the

Area of ΔADC = Base × Height

(AD is the base of ΔADC and AE is the height of ΔADC)

= Area of Parallelogram ABCD (from equation1)

Hence we get the result

Similarly we can show that 

(2)

(3)

(4)

Page. 15.40 Ex.15.3

Q1.

Given: Here from the given figure we get

(1) ABCD is a quadrilateral with base AB,

(2) ΔABD is a right angled triangle

(3) ΔBCD is a right angled triangle with base BC right angled at B

To Find: Area of quadrilateral ABCD

Calculation:

In right triangle ΔBCD, by using Pythagoreans theorem

.So

In right triangle ABD

Hence we get Area of quadrilateral ABCD =

Q2.

Given: Here from the given figure we get

(1) PQRS is a square,

(2) T is the midpoint of PS which means

(3) U is the midpoint of PS which means

(4) QU = 8 cm

To find: Area of ΔOTS

Calculation:

Since it is given that PQ = 8 cm. So

Since T and U are the mid points of PS and QR respectively. So

Therefore area of triangle OTS is equals to

Hence we get the result that Area of triangle OTS is

Q3.

Given:

(1) PQRS is a trapezium in which SR||PQ..

(2) PT = 5 cm.

(3) QT = 8 cm.

(4) RQ = 17 cm.

To Calculate: Area of trapezium PQRS.

Calculation:

In triangle

.So

No area of rectangle PTRS

Therefore area of trapezium PQRS is

Hence the answer is

Q4.

Given: In figure:

(1) ∠AOB = 90°

(2) AC = BC,

(3) OA = 012 cm,

(4) OC = 6.5 cm.

To find: Area of ΔAOB

Calculation:

It is given that AC = BC where C is the mid point of AB

We know that the mid point of hypotenuse of right triangle is equidistant from the vertices

Therefore

CA = BC = OC

⇒ CA = BC = 6.5

⇒ AB = 2 × 6.5 = 13 cm

Now inn triangle OAB use Pythagoras Theorem

So area of triangle OAB

Hence area of triangle is

Q5.

Given: Here from the given figure we get

(1) ABCD is a trapezium

(2) AB = 7 cm,

(3) AD = BC = 5 cm,

(4) DC = x cm

(5) Distance between AB and DC is 4 cm

To find:

(a) The value of x

(b) Area of trapezium

Construction: Draw AL⊥ CD, and BM ⊥ CD

Calculation:

Since AL ⊥ CD, and BM ⊥ CD

Since distance between AB and CD is 4 cm. So

AL = BM = 4 cm, and LM = 7 cm

In triangle ADL use Pythagoras Theorem

Similarly in right triangle BMC use Pythagoras Theorem

Now

We know that,

We get the result as

Area of trapezium is

Q6.

Given: Here from the given figure we get

(1) OCDE is a rectangle inscribed in a quadrant of a circle with radius 10cm,

(2) OE = 2√5cm

To find: Area of rectangle OCDE.

Calculation:

In right triangle ΔODE use Pythagoras Theorem

We know that,

Hence we get the result as area of Rectangle OCDE =

Page. 15.41 Ex.15.3

Q7.

Given:

ABCD is a trapezium with AB||DC

To prove: Area of ΔAOD = Area of ΔBOC

Proof:

We know that ‘triangles between the same base and between the same parallels have equal area’

Here ΔABC and ΔABD are between the same base and between the same parallels AB and DC.

Therefore

Hence it is proved that

Q8.

Given:

(1) ABCD is a parallelogram,

(2) ABFE is a parallelogram

(3) CDEF is a parallelogram

To prove: Area of ΔADE = Area of ΔBCF

Proof:

We know that,” opposite sides of a parallelogram are equal”

Therefore for

Parallelogram ABCD, AD = BC

Parallelogram ABFE, AE = BF

Parallelogram CDEF, DE = CF.

Thus, in ΔADE and ΔBCF, we have

So be SSS criterion we have

This means that

Hence it is proved that

Q9.

Given:

(1) ABCD is a quadrilateral,

(2) Diagonals AC and BD of quadrilateral ABCD intersect at P.

To prove: Area ofΔ APB ×Area of ΔCPD = Area of ΔAPD × Area of ΔBPC

Construction: Draw AL perpendicular to BD and CM perpendicular to BD

Proof:

We know that

Area of triangle = × base× height

Area of ΔAPD = × DP × AL …… (1)

Area of ΔBPC = × CM × BP …… (2)

Area of ΔAPB = × BP × AL …… (3)

Area of ΔCPD = × CM × DP …… (4)

Therefore

Hence it is proved that

Q10.

Given:

(1) ABC and ABD are two triangles on the same base AB,

(2) CD bisect AB at O which means AO = OB

To Prove: Area of ΔABC = Area of ΔABD

Proof:

Here it is given that CD bisected by AB at O which means O is the midpoint of CD.

Therefore AO is the median of triangle ACD.

Since the median divides a triangle in two triangles of equal area

Therefore Area of ΔCAO = Area of ΔAOD …… (1)

Similarly for Δ CBD, O is the midpoint of CD

Therefore BO is the median of triangle BCD.

Therefore Area of ΔCOB = Area of ΔBOD …… (2)

Adding equation (1) and (2) we get

Area of ΔCAO + Area of ΔCOB = Area of ΔAOD + Area of ΔBOD

⇒ Area of ΔABC = Area of ΔABD

Hence it is proved that

Q11.

Given: Here from the question we get

(1) ABCD is a parallelogram

(2) P is any point in the interior of parallelogram ABCD

To prove:

Construction: Draw DN perpendicular to AB and PM perpendicular AB

Proof: Area of triangle = × base× height

Area of ΔAPB = × AB × PM …… (1)

Also we know that: Area of parallelogram = base× height

Area of parallelogram ABCD = AB × DN …… (2)

Now PM < DN (Since P is a point inside the parallelogram ABCD)

Hence it is proved that

Q12.

Given:

(1) ABC is a triangle

(2) AD is the median of ΔABC

(3) G is the midpoint of the median AD

To prove:

(a) Area of Δ ADB = Area of Δ ADC

(b) Area of Δ BGC = 2 Area of Δ AGC

Construction: Draw a line AM perpendicular to AC

Proof: Since AD is the median of ΔABC.

Therefore BD = DC

So multiplying by AM on both sides we get

In ΔBGC, GD is the median

Since the median divides a triangle in to two triangles of equal area. So

Area of ΔBDG = Area of ΔGCD

⇒ Area of ΔBGC = 2(Area of ΔBGD)

Similarly In ΔACD, CG is the median

⇒ Area of ΔAGC = Area of ΔGCD

From the above calculation we have

Area of ΔBGD = Area of ΔAGC

But Area of ΔBGC = 2(Area of ΔBGD)

So we have

Area of ΔBGC = 2(Area of ΔAGC)

Hence it is proved that

(1)

(2)

Q13.

Given:

(1) ABC is a triangle

(2) D is a point on BC such that BD = 2DC

To prove: Area of ΔABD = 2 Area of ΔAGC

Proof:

In ΔABC, BD = 2DC

Let E is the midpoint of BD. Then,

BE = ED = DC

Since AE and AD are the medians of ΔABD and ΔAEC respectively

and

The median divides a triangle in to two triangles of equal area. So

Hence it is proved that

Q14.

Given: Here from the given figure we get

(1) ABCD is a parallelogram

(2) BD and CA are the diagonals intersecting at O.

(3) P is any point on BO

To prove:

(a) Area of ΔADO = Area ofΔ CDO

(b) Area of ΔAPB = Area ofΔ CBP

Proof: We know that diagonals of a parallelogram bisect each other.

O is the midpoint of AC and BD.

Since medians divide the triangle into two equal areas

In ΔACD, DO is the median

Area of ΔADO = Area ofΔ CDO

Again O is the midpoint of AC.

In ΔAPC, OP is the median

⇒ Area of ΔAOP = Area of ΔCOP …… (1)

Similarly O is the midpoint of AC.

In ΔABC, OB is the median

⇒ Area of ΔAOB = Area of ΔCOB …… (2)

Subtracting (1) from (2) we get,

Area of ΔAOB − Area of ΔAOP = Area of ΔCOB − Area of ΔCOP

⇒ Area of ΔABP = Area of ΔCBP

Hence it is proved that

(a)

(b)

Q15.

Given: Here from the given figure we get

(1) ABCD is a parallelogram with base AB,

(2) BC is produced to E such that CE = BC

(3) AE intersects CD at F

(4) Area of ΔDFB = 3 cm

To find:

(a) Area of ΔADF = Area of ΔECF

(b) Area of parallelogram ABCD

Proof: Δ ADF and ΔECF, we can see that

∠ADF = ∠ECF (Alternate angles formed by parallel sides AD and CE)

AD = EC

∠DFA = ∠CFA (Vertically opposite angles)

(ASA condition of congruence)

As

DF = CF

Since DF = CF. So BF is a median in ΔBCD

Since median divides the triangle in to two equal triangles. So

Since .So

Hence Area of parallelogram ABCD

Hence we get the result

(a)

(b)

Page. 15.42 Ex.15.3

Q16.

Given:

(1) Diagonals AC and BD of a parallelogram ABCD intersect at point O.

(2) A line through O intersects AB at P point.

(3) A line through O intersects DC at Q point.

To find: Area of (ΔPOA) = Area of (ΔQOC)

Proof:

From ΔPOA and ΔQOC we get that

=

OA = OC

=

So, by ASA congruence criterion, we have

So

Area (ΔPOA) = Area (ΔQOC)

Hence it is proved that

Q17.

Given:

(1) ABCD is a parallelogram.

(2) E is a point on BA such that BE = 2EA

(3) F is a point on DC such that DF = 2FC.

To find:

Area of parallelogram

Proof: We have,

BE = 2EA and DF = 2FC

AB − AE = 2AE and DC − FC = 2FC

AB = 3AE and DC = 3FC

AE = AB and FC = DC

AE = FC [since AB = DC]

Thus, AE || FC such that AE = FC

Therefore AECF is a parallelogram.

Clearly, parallelograms ABCD and AECF have the same altitude and

AE = AB.

Therefore

Hence proved that

Q18.

Given:

(1) In a triangle ABC, P is the mid-point of AB.

(2) Q is mid-point of BC.

(3) R is mid-point of AP.

To prove:

(a) Area of ΔPBQ = Area of ΔARC

(b) Area of ΔPRQ = Area of ΔARC

(c) Area of ΔRQC = Area of ΔABC

Proof: We know that each median of a triangle divides it into two triangles of equal area.

(a) Since CR is a median of ΔCAP

Therefore …… (1)

Also, CP is a median of ΔCAB.

Therefore …… (2)

From equation (1) and (2), we get

Therefore …… (3)

PQ is a median of ΔABQ

Therefore

Since

Put this value in the above equation we get

…… (4)

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

Therefore …… (5)

(b)

…… (6)

…… (7)

From equation (6) and (7)

…… (8)

From equation (7) and (8)

(c)

= …… (9)

Q19.

Given:

ABCD is a parallelogram

G is a point such that AG = 2GB

E is a point such that CE = 2DE

F is a point such that BF = 2FC

To prove:

(i)

(ii)

(iii)

(iv)

What portion of the area of parallelogram ABCD is the area of ΔEFG

Construction: draw a parallel line to AB through point F and a perpendicular line to AB through

PROOF:

(i) Since ABCD is a parallelogram,

So AB = CD and AD = BC

Consider the two trapeziums ADEG and GBCE:

Since AB = DC, EC = 2DE, AG = 2GB

, and

, and

So, and

Since the two trapeziums ADEG and GBCE have same height and their sum of two parallel sides are equal

Since

So

Hence

(ii) Since we know from above that

. So

Hence

(iii) Since height of triangle EFC and triangle EBF are equal. So

Hence

(iv) Consider the trapezium in which

(From (iii))

Now from (ii) part we have

(v) In the figure it is given that FB = 2CF. Let CF = x and FB = 2x

Now consider the tow triangles CFI and CBH which are similar triangles

So by the property of similar triangle CI = k and IH = 2k

Now consider the triangle EGF in which

Now

(Multiply both sides by 2)

…… (2)

From (1) and (2) we have

Q20.

Given:

(1) CD||AE.

(2) CY||BA.

To find:

(i) Name a triangle equal in area of ΔCBX.

(ii) .

(iii) .

Proof:

(i) Since triangle BCY and triangle YCA are on the same base and between same parallel, so their area should be equal. Therefore

Therefore area of triangle CBX is equal to area of triangle AXY

(ii) Triangle ADE and triangle ACE are on the same base AE and between the same parallels AE and CD.

(iii) Triangle ACY and BCY are on the same base CY and between same parallels CY and BA. So we have

Now we know that

Q21.

Given:

(i) PSDA is a parallelogram.

(ii) .

(iii)

To find:

Proof:

Since AP||BQ||CR||DS and AD||PS

So PQ = CD …… (1)

In ΔBED, C is the mid point of BD and CF||BE

This implies that F is the mid point of ED. So

EF = FD …… (2)

In ΔPQE and ΔCFD, we have

PE = FD

, and [Alternate angles]

PQ = CD.

So, by SAS congruence criterion, we have

ΔPQE = ΔDCF

Hence proved that

Q22.

Given: ABCD IS A trapezium in which

(a) AB||DC

(b) DC = 40 cm

(c) AB = 60 cm

(d) X is the midpoint of AD

(e) Y is the midpoint of BC

To prove:

(i) XY = 50 cm

(ii) DCYX is a trapezium

(iii)

Construction: Join DY and produce it to meet AB produced at P.

Proof:

(i) In ΔBYP and ΔCYD

Y is the midpoint of BC also X is the midpoint of AD

Therefore XY||AP and

(ii) We have proved above that XY||AP

XY|| AP and AB||DC (Given in question)

XY|| DC

(iii) Since X and Y are the midpoints of AD and BC respectively.

Therefore DCYX and ABYX are of the same height say h cm.

Q23.

Given:

(a) ΔABC and Δ BDE are two equilateral triangles

(b) D is the midpoint of BC

(c) AE intersect BC in F.

To prove:

(i)

(ii)

(iii)

(iv)

(v)

(vi)

Proof: Let AB = BC = CA = x cm.

Then BD = = DE = BE

(i) We have

(ii) We Know that ΔABC and ΔBED are equilateral triangles

BE||AC

(iii) We Know that ΔABC and ΔBED are equilateral triangles

AB || DE

(iv) Since ED is a median of Δ BEC

(v) We basically want to find out FD. Let FD = y

Since triangle BED and triangle DEA are on the same base and between same parallels ED and BE respectively. So

Since altitude of altitude of any equilateral triangle having side x is

So

…… (1)

Now

…… (2)

From (1) and (2) we get

(vi) Now we know y in terms of x. So

……. (3)

…… (4)

From (3) and (4) we get