myhelper: RD Sharma 2020 solution class 9 chapter 14 Area of Parallelograms and Triangles Exercise 14.3

Exercise 14.3

Page-14.44

Question 1:

In the given figure, compute the area of quadrilateral ABCD.

Answer 1:

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 =

Question 2:

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

Answer 2:

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

Question 3:

Compute the area of trapezium PQRS in the given figure.

Answer 3:

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

Question 4:

In the given figure, ∠AOB = 90°, AC = BC, OA = 12 cm and OC = 6.5 cm. Find the area of Δ AOB.

Answer 4:

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

Question 5:

In the given figure, 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.

Answer 5:

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

Page-14.45

Question 6:

In the given figure, OCDE is a rectangle inscribed in a quadrant of a circle of radius 10 cm. If OE = 2 $\sqrt{5}$ , find the area of the rectangle.

Answer 6:

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 =

Question 7:

In the given figure, ABCD is a trapezium in which AB || DC. Prove that
ar(Δ AOD) = ar(Δ BOC).

Answer 7:

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

Question 8:

In the given figure, ABCD, ABFE and CDEF are parallelograms. Prove that
ar(Δ ADE) = ar(Δ BCF)

Answer 8:

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

Question 9:

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

Answer 9:

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

Question 10:

If AD is a median of a triangle ABC, then prove that triangles ADB and ADC are equal in area. If G is the mid-point of median AD, prove that ar(Δ BGC)= 2 ar (Δ AGC).

Answer 10:

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)

Question 11:

A point D is taken on the side BC of a ΔABC such that BD = 2DC. Prove that ar (Δ ABD) = 2 ar (Δ ADC).

Answer 11:

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

Page-14.46

Question 12:

ABCD is a parallelogram whose diagonals intersect at O. If P is any point on BO, prove that:

(i) ar (Δ ADO) = ar(Δ CDO)

(ii) ar (Δ ABP) = ar (Δ CBP).

Answer 12:

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)

Question 13:

ABCD is a parallelogram in which BC is produced to E such that CE = BC. AE intersects CD at F.

(i) Prove that ar (Δ ADF) = ar (Δ ECF)

(ii) If the area of Δ DFB = 3 cm², find the area of ||^gm ABCD.

Answer 13:

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)

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)

Question 14:

ABCD is a parallelogram whose diagonals AC and BD intersect at O. A line through O intersects AB at P and DC at Q. Prove that ar (Δ POA) = ar (Δ QOC).

Answer 14:

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

Area (ΔPOA) = Area (ΔQOC)

Hence it is proved that

Question 15:

In the given figure, D and E are two points on BC such that BD = DE = EC. Show that a (Δ ABD) = ar (Δ ADE) = ar (Δ AEC).

Answer 15:

Given: In ΔABCD, D and E are two points on BC such that BD = DE = EC

To prove:

Proof: The ΔABD, ΔADE, and ΔAEC, are on the equal bases and their heights are equal

Therefore their areas are equal

Hence we get the result as

Question 16:

Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that:
ar(Δ APB) ✕ ar (Δ CPD) = ar (Δ APD) ✕ ar (Δ BPC)

Answer 16:

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

Question 17:

If P is any point in the interior of a parallelogram ABCD, then prove that area of the triangle APB is less than half the area of parallelogram.

Answer 17:

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

Question 18:

ABCD is a parallelogram. E is a point on BA such that BE = 2 EA and F is a point on DC such that DF = 2 FC. Prove that AE CF is a parallelogram whose area is one third of the area of parallelogram AB CD.

Answer 18:

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

Question 19:

In a Δ ABC, P and Q are respectively the mid-points of AB and BC and R is the mid-point of AP. Prove that:

(i) ar (Δ PBQ) = ar (Δ ARC)

(ii) ar (Δ PRQ) = $\frac{1}{2}$ ar (Δ ARC)

(iii) ar (Δ RQC) = $\frac{3}{8}$ ar (Δ ABC).

Answer 19:

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

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)

Question 20:

ABCD is a parallelogram, G is the point on AB such that AG = 2 GB, E is a point of DC such that CE = 2DE and F is the point of BC such that BF = 2FC. Prove that:

(i) ar (ADEG) = ar (GBCE)

(ii) ar (Δ EGB) = $\frac{1}{6}$ ar (ABCD)

(iii) ar (Δ EFC) = $\frac{1}{2}$ ar (Δ EBF)

(iv) ar (Δ EBG) = ar (Δ EFC)

(v) Find what portion of the area of parallelogram is the area of Δ EFG

Answer 20:

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

So, and

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

Since

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

Page-14.47

Question 21:

In the given figure, CD || AE and CY || BA.

(i) Name a triangle equal in area of ΔCBX

(ii) Prove that ar (Δ ZDE) = ar (Δ CZA)

(iii) Prove that ar (BCZY) = ar (Δ EDZ).

Answer 21:

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

Question 22:

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

Answer 22:

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

Question 23:

In the given figure, 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) = $\frac{9}{11}$ ar (trap. (XYBA)

Answer 23:

Given: ABCD IS A trapezium in which

(a) AB||DC

(b) DC = 40 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.

Question 24:

D is the mid-point of side BC of Δ ABC and E is the mid-point of BD. If O is the mid-point of AE, prove that ar (Δ BOE) = $\frac{1}{8}$ ar (Δ ABC).

Answer 24:

Given: In ΔABC

(1) D is the midpoint of the side BC

(2) E is the midpoint of the side BD

(3) O is the midpoint of the side AE

To prove:

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

Since AD and AE are the medians of ΔABC and ΔABD respectively. And OB is the median of ΔABE

…… (1)

…… (2)

…… (3)

Therefore

Hence we have proved that

Question 25:

In the given figure, 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).

Answer 25:

Given:

(1) X and Y are the, midpoints of AC and AB respectively.

(2) QP|| BC

(3) CYQ and BXP are straight lines.

To prove:

Proof: Since X and Y are the, midpoints of AC and AB respectively.

So XY||BC

ΔBYC and ΔBXC are on the same base BC and between the same parallels XY and BC.

Therefore

…… (1)

Similarly the quadrilaterals XYAP and XYQA are on the same base XY and between the same parallels XY and PQ. Therefore

…… (2)

Adding equation 1 and 2 we get

Hence we had proved that

Question 26:

In the given figure, ABCD and AEFD are two parallelograms. Prove that

(i) PE = FQ

(ii) ar (Δ APE): ar (ΔPFA) = ar Δ (QFD) : ar (Δ PFD)

(iii) ar (Δ PEA) = ar (Δ QFD).

Answer 26:

Given: ABCD and AEFD are two parallelograms

To prove:

(i) PE = FQ

(ii)

(iii)

Proof: (i) and (iii)

In ΔAPE and ΔDQF

Therefore

, and

(ii) ΔPFA and ΔPFD are on the same base PF and between the same parallels PQ and AD.

From (1) and (2) we get

Page-14.48

Question 27:

In the given figure, ABCD is a ||^gm. O is any point on AC. PQ || AB and LM || AD. Prove that ar (||^gm DLOP) = ar (||^gm BMOQ)

Answer 27:

Given:

(1) ABCD is a parallelogram

(2) O is any point of AC.

(3) PQ||AB and LM||AD

To prove:

Calculation:

We know that the diagonal of a parallelogram divides it into two triangles of equal area

Therefore we have

Since OC and AO are diagonals of parallelogram OQCL and AMOP respectively. Therefore

Subtracting (2) and (3) from (1) we get

Hence we get the result

Question 28:

In a Δ ABC, if L and M are points on AB and AC respectively such that LM || BC. Prove that:

(i) ar (Δ LCM) = ar (Δ LBM)

(ii) ar (Δ LBC) = (Δ MBC)

(iii) ar (Δ ABM) = ar (Δ ACL)

(iv) ar (Δ LOB) = ar (Δ MOC)

Answer 28:

Given:

In ΔABC, if L and M are points on AB and AC such that LM||BC

To prove:

(i)

(ii)

(iii)

(iv)

Proof: We know that triangles between the same base and between the same parallels are equal in area.

(i) Here we can see that ΔLMB and ΔLMC are on the same base BC and between the same parallels LM and BC

Therefore

…… (1)

(ii) Here we can see that ΔLBC and ΔLMC are on the same base BC and between the same parallels LM and BC

Therefore

…… (2)

(iii) From equation (1) we have,

(iv) From (2) we have,

Question 29:

In the given figure, 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 (Δ BDE) = $\frac{1}{4}$ ar (Δ ABC)

(ii) ar (Δ BDE) = $\frac{1}{2}$ ar (Δ BAE)

(iii) ar (Δ BFE) =ar (Δ AFD)

(iv) ar (ΔABC) = 2 ar (Δ BEC)

(v) ar (Δ FED) = $\frac{1}{8}$ ar (Δ AFC)

(vi) ar (Δ BFE) = 2 ar (EFD)

Answer 29:

Given:

(a) ΔABC and Δ BDE are two equilateral triangles

(b) D is the midpoint of BC

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

…… (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

Question 30:

If the given figure, ABC is a right triangle right angled at A, BCED, ACFGand ABMN are squares on the sides BC, CA and AB respectively. Line segment AX ⊥ DE meets BC at Y.

Show that
(i) Δ MBC $≅$ Δ ABD

(ii) ar (BYXD) = 2 ar (Δ MBC)

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

(iv) Δ FCB $≅$ Δ ACE

(v) ar (CYXE) = 2 ar (ΔFCB)

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

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