Area

In the given figure consider ABD and BCD

Area of ABD = x base x height = x AB x BD

= x 5 x 7 = -------- 1

Area of BCD = x base x height = x DC x DB

= x 5 x 7 = ----------2

From 1 and 2 we can tell that area of two triangle that is ABD and BCD are equal

Since the diagonal BD divides ABCD into two triangles of equal area and opp sides AB = DC

ABCD is a parallelogram

Area of parallelogram ABCD = Area of ABD + Area of BCD

= = cm² = 35 cm²

Area of parallelogram ABCD = 35cm²

Given

AB = 10 cm

DL = 6 cm

BM = 8 cm

AD = ? (To find)

Here, Area of parallelogram = base x height

In the given figure if we consider AB as base Area = AB x DL

If we consider DM as base Area = AD x BM

Area = AB x DL = AD x BM

⇒ 10 x 6 = AD x 8

⇒ 60 = 8 x AD

Here, Let ABCD be Rhombus with diagonals AC and BD

Here let AC = 24 and BD = 16

We know that, in a Rhombus, diagonals are perpendicular bisectors to each other

if we consider ABC AC is base and OB is height

Similarly, in ADC AC is base and OD is height

Now, Area of Rhombus = Area of ABC + Area of ADC

= x AC x OB + x AC x OD

= x 24 x + x 24 x (Since AC and BC are perpendicular bisectors OB = OD = )

= x 24 x + x 24 x = 96 + 96 = 192 cm²

Area of Rhombus ABCD is 192cm²

Given

AB = a = 9 cm

DC = b = 6 cm

Height (h) = 8 cm

We know that area of trapezium is x (sum of parallel sides) x height

Therefore, Area of trapezium ABCD = x (AB + DC) x h = x (9 + 6) x 8 = 60 cm²

Area of Trapezium ABCD = 60cm²

Given

AD = 9 cm

BC = 8 cm

DC = 17 cm

Here Area of Quad ABCD = Area of ABD + Area of BCD

= x AB x AD + x BC x BD

By Pythagoras theorem in BCD

DC² = BD² + BC²

17² = BD² + 8²

BD² = 17² - 8² = 289 – 64 = 225

BD = 15 cm

Similarly in ABD using Pythagoras theorem

BD² = AD² + AB²

15² = 9² + AB²

AB² = 15 ² - 9² = 225 – 81 = 144

AB = 12 cm

Now, Area of Quad ABCD = Area of ABD + Area of BCD

= x AB x AD + x BC x BD

= x 12 x 9 + x 8 x 15 = 54 + 60 = 114 cm²

Area of Quadrilateral ABCD = 114 cm²

Given :- Right trapezium

RS = 8 cm

PT = 8cm

TQ = 8 cm

RQ = 17 cm

Here PQ = PT + TQ = 8 + 8 = 16

We know that area of trapezium is x (sum of parallel sides) x height

That is x (AB + DC) x RT

Consider TQR

By Pythagoras theorem

RQ² = TQ² + RT²

17² = 8² + RT²

RT² = 17 ² - 8² = 289 – 64 = 225

RT = 15 cm

Area of trapezium = x (RS + PQ) x RT

= x (8 + 16) x 15 = 180 cm²

Area of trapezium PQRS = 180cm²

Given

AB = 7 cm

AD = BC 5 cm

AL = BM = 4cm (height)

DC = ?

Here in the given figure AB = LM

LM = 7 cm ------------1

Now Consider ALD

By Pythagoras theorem

AD² = AL² + DL²

5² = 4² + DL²

DL² = 5 ² - 4² = 25 – 16 = 9

DL = 3 cm --------------2

Similarly in BMC

By Pythagoras theorem

BC² = BM² + MC²

5² = 4² + MC²

MC² = 5 ² - 4² = 25 – 16 = 9

MC = 3 cm --3

from 1 2 and 3

DC = DL + LM + MC = 3 + 7 + 3 = 13 cm

We know that area of trapezium is x (sum of parallel sides) x height

Area of trapezium = x (AB + DC) x AL

= x (7 + 13) x 4 = 40 cm²

Area of trapezium ABCD = 180cm²

Given :

AL ⊥ BD and CM ⊥ BD

To prove : ar (quad. ABCD) = x BD x (AL + CM)

Proof:

Area of ABD = x BD x AM

Area of ABD = x BD x CM

Now area of Quad ABCD = Area of ABD + Area of BCD

= x BD x AL + x BD x CM

= x BD x (AL + CM)

Hence proved

Given

AL ⊥ BD and CM ⊥ BD

BD = 14 cm

AL = 8 cm

CM = 6 cm

Here,

Area of ABD = x BD x AM

Area of ABD = x BD x CM

Now area of Quad ABCD = Area of ABD + Area of BCD

= x BD x AL + x BD x CM

= x BD x (AL + CM)

Area of quad ABCD = x BD x (AL + CM) = x 14 x (8 + 6) = 98cm²

Area of quad ABCD = 98cm²

Given

AB ‖ DC

To prove that: area(∆AOD) = area(∆BOC)

Here in the given figure Consider ABD and ABC,

we find that they have same base AB and lie between two parallel lines AB and CD

According to the theorem: triangles on the same base and between same parallel lines have equal areas.

Area of ABD = Area of BCA

Now,

Area of AOD = Area of ABD - Area of AOB ---1

Area of COB = Area of BCA - Area of AOB ---2

From 1 and 2

We can conclude that area(∆AOD) = area(∆BOC) (Since Area of AOB is common)

Hence proved

Given

AB ‖ DC

To prove that : (i) area(∆ACD) = area(∆ABE)

(ii) area(∆OCE) = area(∆OBD)

(i)

Here in the given figure Consider BDE and ECD,

we find that they have same base DE and lie between two parallel lines BC and DE

According to the theorem: triangles on the same base and between same parallel lines have equal

areas.

Area of BDE = Area of ECD

Now,

Area of ACD = Area of ECD + Area of ADE ---1

Area of ABE = Area of BDE + Area of ADE ---2

From 1 and 2

We can conclude that area(∆AOD) = area(∆BOC) (Since Area of ADE is common)

Hence proved

(ii)

Here in the given figure Consider BCD and BCE,

we find that they have same base BC and lie between two parallel lines BC and DE

According to the theorem : triangles on the same base and between same parallel lines have equal

areas.

Area of BCD = Area of BCE

Now,

Area of OBD = Area of BCD - Area of BOC ---1

Area of OCE = Area of BCE - Area of BOC ---2

From 1 and 2

We can conclude that area(∆OCE) = area(∆OBD) (Since Area of BOC is common)

Hence proved

Given

A triangle ABC in which points D and E lie on AB and AC of ∆ABC such that ar(∆BCE) = ar(∆BCD).

To prove: DE ‖ BC

Proof:

Here, from the figure we know that BCE and BCD lie on same base BC and

It is given that area(∆BCE) = area(∆BCD)

Since two triangle have same base and same area they should equal altitude(height)

That means they lie between two parallel lines

That is DE ‖ BC

DE ‖ BC

Hence proved

Given : A parallelogram ABCD with a point ‘O’ inside it.

To prove : (i) area(∆OAB) + area(∆OCD) = area(‖gm ABCD),

(ii)area(∆OAD) + area(∆OBC) = area(‖gm ABCD).

Construction : Draw PQ ‖ AB and RS ‖ AD

Proof:

(i)

∆AOB and parallelogram ABQP have same base AB and lie between parallel lines AB and PQ.

According to theorem: If a triangle and parallelogram are on the same base and between the same

parallel lines, then the area of the triangle is equal to half of the area of the parallelogram.

area(∆AOB) = area(‖gm ABQP) –--1

Similarly, we can prove that area(∆COD) = area(‖gm PQCD) –--2

Adding –1 and –2 we get,

area(∆AOB) + area(∆COD) = area(‖gm ABQP) + area(‖gm PQCD)

area(∆AOB) + area(∆COD) = = area(‖gm ABCD)

area(∆AOB) + area(∆COD) = area(‖gm ABCD)

Hence proved

(ii)

∆OAD and parallelogram ASRD have same base AD and lie between parallel lines AD and RS.

According to theorem: If a triangle and parallelogram are on the same base and between the same

parallel lines, then the area of the triangle is equal to half of the area of the parallelogram.

area(∆OAD) = area(‖gm ASRD) –--1

Similarly, we can prove that area(∆OBC) = area(‖gm BCRS) –--2

Adding –1 and –2 we get,

area(∆OAD) + area(∆OBC) = area(‖gm ASRD) + area(‖gm BCRS)

area(∆OAD) + area(∆OBC) = = area(‖gm ABCD)

area(∆OAD) + area(∆OBC) = area(‖gm ABCD)

Hence proved

Given : ABCD is a quadrilateral in which a line through D drawn parallel to AC which meets BC produced in P.

To prove: area of (∆ABP) = area of (quad ABCD)

Proof:

Here, in the given figure

∆ACD and ∆ACP have same base and lie between same parallel line AC and DP.

According to the theorem : triangles on the same base and between same parallel lines have equal

areas.

area of (∆ACD) = area of (∆ACP) -------------1

Now, add area of (∆ABC) on both side of (1)

area of (∆ACD) + (∆ABC) = area of (∆ACP) + (∆ABC)

Area of (quad ABCD) = area of (∆ABP)

area of (∆ABP) = Area of (quad ABCD)

Hence proved

Given : ∆ABC and ∆DBC having same base BC and area(∆ABC) = area(∆DBC).

To prove: OA = OD

Construction : Draw AP ⊥ BC and DQ ⊥ BC

Proof :

Here area of ∆ABC = x BC x AP and area of ∆ABC = x BC x DQ

since, area(∆ABC) = area(∆DBC)

x BC x AP = x BC x DQ

AP = DQ -------------- 1

Now in ∆AOP and ∆QOD, we have

APO = DQO = 90 and

AOP = DOQ [Vertically opposite angles]

AP = DQ [from 1]

Thus by AAS congruency

∆AOP ∆QOD [AAS]

Thus By corresponding parts of congruent triangles law [C.P.C.T]

OA = OD [C.P.C.T]

Hence BC bisects AD

Hence proved

Given : A ∆ABC in which AD is the median and P is a point on AD

To prove: (i) ar(∆BDP) = ar(∆CDP),

(ii) ar(∆ABP) = ar(∆ACP).

(i)

In ∆BPC, PD is the median. Since median of a triangle divides the triangles into two equal areas

So, area(∆BDP) = area(∆CDP)----1

Hence proved

(ii)

In ∆ABC AD is the median

So, area(∆ABD) = area(∆ADC) ----2 and

area(∆BDP) = area(∆CDP) [from 1]

Now subtracting area(∆BDP) from ---2 , we have

area(∆ABD) - area(∆BDP) = area(∆ADC) - area(∆BDP)

area(∆ABD) - area(∆BDP) = area(∆ADC) - area(∆CDP) [since area(∆BDP) = area(∆CDP) from –1]

area(∆ABP) = area(∆ACP)

Hence proved.

Given : A quadrilateral ABCD with diagonals AC and BD and BO = OD

To prove: Area of (∆ABC) = area of (∆ADC)

Proof : BO = OD [given]

Here AO is the median of ∆ABD

Area of (∆AOD) = Area of (∆AOB) ---------------- 1

And OC is the median of ∆BCD

Area of (∆COD) = Area of (∆BOC) ---------------- 2

Now by adding –1 and –2 we get

Area of (∆AOD) + Area of (∆COD) = Area of (∆AOB) + Area of (∆BOC)

Area of (∆ABC) = Area of (∆ADC)

Hence proved

Given : A ∆ABC in which AD is the median and E is the midpoint on line AD

To prove: area(∆BED) = area(∆ABC)

Proof : here in ∆ABC AD is the midpoint

Area of (∆ABD) = Area of (∆ADE)

Hence Area of (∆ABD) = [Area of (∆ABC)] ------------------ 1

No in ∆ABD E is the midpoint of AD and BE is the median

Area of (∆BDE) = Area of (∆ABE)

Hence Area of (∆BED) = [Area of (∆ABD)] -------------- 2

Substituting (1) in (2), we get

Hence Area of (∆BED) = [ Area of (∆ABC)]

area(∆BED) = area(∆ABC)

Hence proved

Given : A ∆ABC in which AD is a line where D is a point on BC and E is the midpoint of AD

To prove: ar(∆BEC) = ar(∆ABC)

Proof: In ∆ABD E is the midpoint of side AD

Area of (∆BDE) = Area of (∆ABE)

Hence Area of (∆BDE) = [Area of (∆ABD)] –1

Now, consider ∆ACD in which E is the midpoint of side AD

Area of (∆ECD) = Area of (∆AEC)

Hence Area of (∆ECD) = [Area of (∆ACD)] –2

Now, adding –1 and –2, we get

Area of (∆BDE) + Area of (∆ECD) = [Area of (∆ABD)] + [Area of (∆ACD)]

area(∆BEC) = [area(∆ABD) + area(∆ACD)]

Area(∆BEC) = Area(∆ABC)

Hence proved

Given : D is the midpoint of side BC of ∆ABC and E is the midpoint of BD and O is the midpoint of AE

To prove : ar(∆BOE) = ar(∆ABC)

Proof : Consider ∆ABC here D is the midpoint of BC

Area of (∆ABD) = Area of (∆ACD)

Area(∆ABD) = Area(∆ABC)—1

Now, consider ∆ABD here E is the midpoint of BD

Area of (∆ABE) = Area of (∆AED)

Area(∆ABE) = Area(∆ABD)—2

Substituting –1 in –2 , we get

Area(∆ABE) = ( Area(∆ABC))

Area(∆ABE) = Area(∆ABC)—3

Now consider ∆ABE here O is the midpoint of AE

Area of (∆BOE) = Area of (∆AOB)

Area(∆BOE) = Area(∆ABE)—4

Now, substitute –3 in –4 , we get

Area(∆BOE) = ( Area(∆ABC))

area(∆BOE) = area(∆ABC)

Hence proved

Given : A parallelogram ABCD in which AC is the diagonal and O is some point on the diagonal AC

To prove: area(∆AOB) = area(∆AOD)

Construction : Draw a diagonal BD and mark the intersection as P

Proof:

We know that in a parallelogram diagonals bisect each other, hence P is the midpoint of ∆ABD

Area of (∆APB) = Area of (∆APD)—1

Now consider ∆BOD here OP is the median, since P is the midpoint of BD

Area of (∆OPB) = Area of (∆OPD)—2

Adding –1 and –2 we get

Area of (∆APB) + Area of (∆OPB) = Area of (∆APD) + Area of (∆OPD)

Area of (∆AOB) = Area of (∆AOD)

Hence proved

Given : ABCD is a parallelogram and P,Q,R,S are the midpoints of AB,BC,CD,AD respectively

To prove: (i) PQRS is a parallelogram

(ii) Area(‖gm PQRS) = x area(‖gm ABCD)

Construction : Join AC ,BD,SQ

Proof:

(i)

As S and R are midpoints of AD and CD respectively, in ∆ACD

SR || AC [By midpoint theorem] ------------------- (1)

Similarly in ∆ABC , P and Q are midpoints of AB and BC respectively

PQ || AC [By midpoint theorem] ------------------ (2)

From (1) and (2)

SR || AC || PQ

SR || PQ ------------------- (3)

Again in ∆ACD as S and P are midpoints of AD and CB respectively

SP || BD [By midpoint theorem] ------------------ (4)

Similarly in ∆ABC , R and Q are midpoints of CD and BC respectively

RQ || BD [By midpoint theorem] -------------------- (5)

From (4) and (5)

SP || BD || RQ

SP || RQ ----------- (6)

From (3) and (6)

We can say that opposite sides are Parallel in PQRS

Hence we can conclude that PQRS is a parallelogram.

(ii)

Here ABCD is a parallelogram

S and Q are midpoints of AD and BC respectively

SQ || AB

SQAB is a parallelogram

Now, area(∆SQP) = x area of (SQAB) -------------- 1

[Since ∆SQP and ||gm SQAB have same base and lie between same parallel lines]

Similarly

S and Q are midpoints of AD and BC respectively

SQ || CD

SQCD is a parallelogram

Now, area(∆SQR) = x area of (SQCD) ------------------- 2

[Since ∆SQR and ||^gm SQCD have same base and lie between same parallel lines]

Adding (1) and (2) we get

area(∆SQP) + area(∆SQR) = x area of (SQAB) + x area of (SQCD)

area(PQRS) = (area of (SQAB) + area of (SQCD))

Area(‖gm PQRS) = x area(‖gm ABCD)

Hence proved

Given : ABCDE is a pentagon EG is drawn parallel to DA which meets BA produced at G and CF is drawn parallel to DB which meets AB produced at F

To prove: area(pentagon ABCDE) = area(∆DGF)

Proof:

Consider quadrilateral ADEG. Here,

area(∆AED) = area(∆ADG) ------------- (1)

[since two triangles are on same base AD and lie between parallel line i.e, AD||EG]

Similarly now, Consider quadrilateral BDCF. Here,

area(∆BCD) = area(∆BDF) ---------------- (2)

[since two triangles are on same base AD and lie between parallel line i.e, AD||EG]

Adding Eq (1) and (2) we get

area(∆AED) + area(∆BCD) = area(∆ADG) + area(∆BDF) ------------------ (3)

Now add area(∆ABD) on both sides of Eq (3), we get

area(∆AED) + area(∆BCD) + area(∆ABD) = area(∆ADG) + area(∆BDF) + area(∆ABD)

area(pentagon ABCDE) = area(∆DGF)

Hence proved

Given : A ∆ABC with D as median

To prove : Median D divides a triangle into two triangles of equal areas.

Constructions: Drop a perpendicular AE onto BC

Proof: Consider ∆ABD

area(∆ABD) = x BD x AE

Now , Consider ∆ACD

area(∆ACD) = x CD x AE

since D is the median

BD = CD

x BD x AE = x CD x AE

Hence , area(∆ABD) = area(∆ACD)

we can say that Median D divides a triangle into two triangles of equal areas.

Hence proved

Given: A parallelogram ABCD with a diagonal BD

To prove: area(∆ABD) = area(∆BCD)

Proof:

We know that in a parallelogram opposite sides are equal, that is

AD = BC and AB = CD

Now, consider ∆ABD and ∆BCD

Here AD = BC

AB = CD

BD = BD (common)

Hence by SSS congruency

∆ABD ∆BCD

By this we can conclude that both the triangles are equal

area(∆ABD) = area(∆BCD)

Hence proved

Given: A ∆ABC with a point D on BC such that BD = DC

To prove: area(∆ABD) = x area(∆ABC)

Construction: Drop a perpendicular onto BC

Proof: area(∆ABC) = x BC x AE ---------------(1)

and, area(∆ABD) = x BD x AE ----------------- (2)

given that BD = DC ------------------ (3)

so, BC = BD + DC = BD + 2BD = 3BD [from 2]

∴ BD = (BC)

Sub BD in (1), we get

area(∆ABD) = x ((BC) X AE)

area(∆ABD) = x ( BC X AE)

area(∆ABD) = x area(∆ABC) [from 1]

Hence proved

Given : A ∆ABC in which a point D divides the Side BC in the ratio m:n.

To prove: area(∆ABD): area(∆ABC) = m:n

Construction : Drop a perpendicular AL on BC

Proof:

area(∆ABD) = x BD x AL ---------------- (1)

and, area(∆ADC) = x DC x AL ------------------ (2)

BD:DC = m:n

--------------(3)

sub Eq (3) in eq (1)

area(∆ABD) = x ( x DC) x AL

area(∆ABD) = x ( x DC x AL)

area(∆ABD) = x area(∆ADC)

=

Area(∆ABD): Area(∆ABC) = m:n

Hence proved

Here, ΔPQR and ΔSQR are on the same base QR but there is no parallel line to QR.

∴ Here, Figure in option B is on the same base but not between the same parallels.

Here parallelogram ABCD and parallelogram ABQP lie on the same base AB and lie between the parallel line AB and DP.

∴ Here, Figure in option C is on the same base and between the same parallels.

In ΔABC, AD is the median
Hence BD = DC
Draw AE ⊥ BC
Area of ΔABD = Area of ΔADC
Thus median of a triangle divides it into two triangles of equal area.

Given:

∠ABC = 90°

∠ACD = 90°

CD = 8cm

AB = 9cm

AD = 17cm

Consider ΔACD

Here, By Pythagoras theorem : AD² = CD² + AC²

17² = 8² + AC²

⇒ AC² = 17²—8²

⇒ AC² = 289 – 64 = 225

⇒ AC = 15

Now, Consider ΔABC

Here, By Pythagoras theorem : AC² = AB² + BC²

15² = 9² + AC²

⇒ BC² = 15²—9²

⇒ BC² = 225 – 81 = 144

⇒ BC = 12

Here,

Area (quad.ABCD) = Area (ΔABC) + Area (ΔACD)

Area (quad.ABCD) = 1/2×AB×BC + 1/2×AC×CD

Area (quad.ABCD) = 1/2×9×12 + 1/2×15×8 = 54 + 60 = 104cm²

∴ Area (quad.ABCD) = 114cm²

Given:

∠BEC = 90°

∠DAE = 90°

CD = AE = 8cm

BE = 15cm

BC = 17cm

Consider ΔCEB

Here, By Pythagoras theorem

BC² = CE² + EB²

17² = CE² + 15²

CE² = 17² – 15²

CE² = 289 – 225 = 64

CE = 8

Here,

∠AEC = 90°

CD = CE = 8cm

∴ AECD is a Square.

∴ Area (Trap. ABCD) = Area (Sq. AECD) + Area (ΔCEB)

Area (Trap. ABCD) = AE×EC + 1/2×CE×EB

Area (Trap. ABCD) = 8×8 + 1/2×8×15 = 64 + 60 = 104cm²

∴ Area (Trap. ABCD) = 124cm²

Given:

AB = CD = 5cm

BD⊥DC

BD = 6.8cm

Now, consider the parallelogram ABCD

Here, let DC be the base of the parallelogram then BD becomes its altitude (height).

Area of the parallelogram is given by: Base × Height

∴ area of ‖gm ABCD = CD×BD = 5×6.8 = 34cm²

∴area of ‖gm ABCD = 34cm².

Given: ABCD is a ‖gm in which diagonals AC and BD intersect at O and ar(‖gm ABCD) is 52cm².

Here,

Ar (∆ABD) = ar(∆ABC)

(∵ ΔABD and ΔABC on same base AB and between same parallel lines AB and CD)

Here,

ar(∆ABD) = ar(∆ABC) = 1/2 × ar(||gm ABCD)

(∵ ΔABD and ΔABC on same base AB and between same parallel lines AB and CD are half the area of the parallelogram)

∴ ar(∆ABD) = ar(∆ABC) = 1/2 × 52 = 26cm²

Now, consider ΔABC

Here OB is the median of AC

(∵ diagonals bisect each other in parallelogram)

∴ ar(∆AOB) = ar(∆BOC)

(∵median of a triangle divides it into two triangles of equal area)

ar(∆AOB) = 1/2 × ar(ΔABC)

ar(∆AOB) = 1/2 × 26 = 13cm²

∴ ar(∆AOB) = 13cm²

Area of parallelogram is: base × height

Here,

Base = AB = 10cm

Height = DL = 4cm

∴ ar(‖gm ABCD) = AB ×DL = 10×4 = 40cm²

∴ ar(‖gm ABCD) = 40cm²

Given:

AB = 10cm

DL ⊥ AB

BM ⊥ AD

DL = 6cm

BM = 8cm

Now, consider the parallelogram ABCD

Here, let AB be the base of the parallelogram then DL becomes its altitude (height).

Area of the parallelogram is given by: Base × Height

∴ area of ‖gm ABCD = AB×DL = 10×6 = 60cm²

Now,

Consider AD as base of the parallelogram then BM becomes its altitude (height)

∴ area of ‖gm ABCD = AD × BM = 60cm²

AD × 8 = 60cm²

AD = 60/8 = 7.5cm

∴length of AD = 7.5cm.

Given:

Length of diagonals of rhombus: 12cm and 16cm.

Area of the rhombus is given by:

∴ Area of the rhombus = = 96cm²

Given:

Lengths of parallel sides of trapezium: 12cm and 8cm

Distance between two parallel lines (height): 6.5cm

Area of the trapezium is given by:

∴ Area of the trapezium = = 65cm²

Given:

AL ⊥ DC

BM ⊥ DC

AB = 7cm

BC = AD = 5cm

AL = BM = 4cm

Here,

MC = DL and AB = LM = 7 cm

Consider the ΔBMC

Here, by Pythagoras theorem

BC² = BM² + MC²

5² = 4² + MC²

MC² = 25—16

MC² = 9

MC = 3cm

∴ MC = DL = 3cm

CD = DL + LM + MC = 3 + 7 + 3 = 13cm

Now,

Area of the trapezium is given by:

∴ Area of the rhombus = = 40cm²

Given:

BD = 16cm

AL ⊥ BD

CM ⊥ BD

AL = 9cm

CM = 7cm

Here,

Area of quadrilateral ABCD = area(ΔABD) + area(ΔBCD)

Area of triangle = 1/2 × base × height

area(ΔABD) = 1/2 × base × height = 1/2 × BD × CM = 1/2 × 16 × 7 = 56cm²

area(ΔBCD) = 1/2 × base × height = 1/2 × BD × AL = 1/2 × 16 × 9 = 64cm²

∴ Area of quadrilateral ABCD = area(ΔABD) + area(ΔBCD) = 56 + 64 = 120cm²

Given:∠DCB = 60°

Let the length of the side be x

Here, consider ΔBCD

BC = DC (all sides of rhombus are equal)

∴ ∠CDB = ∠CBD (angles opposite to equal sides are equal)

Now, by angle sum property

∠CDB + ∠CBD + ∠BCD = 180°

2× ∠CBD = 180° –60°

2 × ∠CBD = 180° – 60°

∴ 2× ∠CBD = 120°

∠ CBD = = 60°

∴ ∠CDB = ∠CBD = 60°

∴ Δ ADC is equilateral triangle

∴ BC = CD = BD = x cm

In Rhombus diagonals bisect each other.

Consider Δ COD

By Pythagoras theorem

CD² = OD² + OC²

x² = ² + OC²

OC² = x² – ²

OC =

OC = cm

∴ AC = 2× OC = 2 × = x

AC: BD = x : x = : 1

∴ AC: BD = : 1

Given: ar(quad. EABC) = 17cm² and ar(‖gm ABCD) = 25cm²

We know that any two or parallelogram having the same base and lying between the same parallel lines are equal in area.

∴ Area (‖gm ABCD) = Area (||gm ABFE) = 25cm²

Here,

Area (||gm ABFE) = Area (quad. EABC) + Area (∆BCF)

25cm² = 17cm² + Area (∆BCF)

Area (∆BCF) = 25 – 17 = 8cm²

∴ Area (∆BCF) = 8cm²

Given: ∆ABC and ∆BDE are two equilateral triangles, D is the midpoint of BC.

Consider ΔABC

Here, let AB = BC = AC = x cm (equilateral triangle)

Now, consider ΔBED

Here,

BD = 1/2 BC

∴ BD = ED = EB = 1/2 BC = x/2 (equilateral triangle)

Area of the equilateral triangle is given by: (a is side length)

∴ ar(∆BDE): ar(∆ABC) = := :1 = 1:4

Given:

P and Q are midpoints of AB and CD respectively

ar(‖gm ABCD) = 16cm²

Now, consider the (‖gm ABCD)

Here,

Q is the midpoint of DC and P is the midpoint of AB.

∴ By joining P and Q (‖gm ABCD) is divided into two equal parallelograms.

That is, ar(‖gm ABCD) = ar(‖gmAPQD) + ar(‖gmPQCB)

ar(‖gm ABCD) = 2×ar(‖gmAPQD) (∵ar(‖gmAPQD) = ar(‖gmPQCB) )

2×ar(‖gmAPQD) = 16cm² (∵ar(‖gm ABCD) = 16cm²)

ar(‖gmAPQD) = 16/2 = 8cm²

∴ ar(‖gmAPQD) = 8cm²

Given: D is the midpoint of BC and E is the midpoint of AD

Here,

D is the midpoint of BC and AD is the median of ΔABC

Area (Δ ABD) = Area (Δ ADC) (∵ median divides the triangle into two triangles of equal areas)

∴ Area (Δ ABD) = Area (Δ ADC) = Area (∆ABC)

Now, consider Δ ABD

Here, BE is the median

Area (Δ ABE) = Area (Δ BED)

∴ Area (Δ ABE) = Area (Δ BED) = Area (∆ABD)

Area (Δ BED) = Area (∆ABD)

Area (Δ BED) = × (∵Area (Δ ABD) = Area (∆ABC) )

Area (Δ BED) = Area (∆ABC)

∴ Area (Δ BED) = Area (∆ABC)

Given:

Here,

D is the midpoint of BC and AD is the median of ΔABC

Area (Δ ABD) = Area (Δ ADC) (∵ median divides the triangle into two triangles of equal areas)

∴ Area (Δ ABD) = Area (Δ ADC) = Area (∆ABC)

Now, consider Δ ABD

Here, BE is the median

Area (Δ ABE) = Area (Δ BED)

∴ Area (Δ ABE) = Area (Δ BED) = Area (∆ABD)

Area (Δ BED) = Area (∆ABD)

Area (Δ BED) = × (∵Area (Δ ABD) = Area (∆ABC) ) –1

Area (Δ BED) = Area (∆ABC)

Similarly,

Area (Δ EDC) = Area (∆ABC) –2

Add –1 and –2

Area (Δ BED) + Area (Δ EDC) = Area (∆ABC) + Area (∆ABC) = Area (∆ABC)

∴ Area (Δ BEC) = Area (∆ABC)

Given: D is the midpoint of BC; E is the midpoint of BD and O is the midpoint of AE.

Here,

D is the midpoint of BC and AD is the median of ΔABC

Area (Δ ABD) = Area (Δ ADC) (∵ median divides the triangle into two triangles of equal areas)

∴ Area (Δ ABD) = Area (Δ ADC) = Area (∆ABC)

Now, consider Δ ABD

Here, AE is the median

Area (Δ ABE) = Area (Δ BED)

∴ Area (Δ ABE) = Area (Δ BED) = Area (∆ABD)

Area (Δ ABE) = Area (∆ABD)

Area (Δ ABE) = × (∵Area (Δ ABD) = Area (∆ABC) ) –1

Area (Δ ABE) = Area (∆ABC)

Consider Δ ABE

Here, BO is the median

Area (Δ BOE) = Area (Δ BOA)

∴ Area (Δ BOE) = Area (Δ BOA) = Area (∆ABE)

Area (Δ BOE) = × (∵Area (Δ ABE) = Area (∆ABC) )

Area (Δ BOE) = Area (∆ABC)

∴ Area (Δ BOE) = Area (∆ABC)

Given:

We know that when a parallelogram and a triangle lie on same base and between same parallel lines then, area of the triangle is half the area of the parallelogram.

Area (ΔABF) = 1/2 Area(||gm ABCD) –1

Area (ΔABF) : Area (||gm ABCD) = 1/2 Area(||gm ABCD) : Area(||gm ABCD) (from –1 )

Area (ΔABF) : Area (||gm ABCD) = 1/2 : 1 = 1:2

∴ Area (ΔABF) : Area (||gm ABCD) = 1:2

Given: ABCD is a trapezium, AB‖DC, AB = a cm and DC = b cm, E and F are the midpoints of AD and BC.

Since E and F are midpoints of AD and BC, EF will be parallel to both AB and CD.

EF =

Height between EF and DC and height between EF and AB are equal, because E and F are midpoints OF AD and BC and EF||AB||DC.

Let height between EF and DC and height between EF and AB be h cm.

Area of trapezium = 1/2 × (sum of parallel lines) × height

Now,

Area (Trap.ABFE) = 1/2 × (a + ) × h.

and

Area (Trap.ABFE) = 1/2 × (b + ) × h.

Area (Trap.ABFE) : Area (Trap.ABFE) = 1/2 × (a + ) × h : 1/2 × (b + ) × h

Area (Trap.ABFE) : Area (Trap.ABFE) = : = 3a + b : a + 3b

∴ Area (Trap.ABFE) : Area (Trap.ABFE) = 3a + b : a + 3b

Given: a quadrilateral whose diagonal AC divides it into two parts, equal in area.

Here,

A quadrilateral is any shape having four sides, it is given that diagonal AC of the quadrilateral

divides it into two equal parts.

We know that the rectangle, parallelogram and rhombus are all quadrilaterals, in these

quadrilaterals if a diagonal is drawn say AC it divides it into equal areas.

∵ This diagonal divide the quadrilateral into two equal or congruent triangles.

Given: Area (‖gm ABCD) = Area (rectangle ABEF)

Consider ΔAFD

Clearly AD is the hypotenuse

∴ AD > AF

Perimeter of Rectangle ABEF = 2× (AB + AF) –1

Perimeter of Parallelogram ABCD = 2× (AB + AD) –2

On comparing –1 and –2, we can see that

Perimeter of ABCD > perimeter of ABEF (∵AD > AF)

Given: ABCD is a rectangle inscribed in a quadrant of a circle of radius 10cm and AD = 25cm

Consider Δ ADC

By Pythagoras theorem

AC² = AD² + DC²

10² = (25)² + AC²

AC² = 10²—(25)²

AC² = 100 – 20 = 80

AC = 45

Area of rectangle = length × breadth = DC × AD

Area of rectangle = 45 × 25 = 40cm²

∴ Area of rectangle = 40cm²

Consider Statement (I) :

Two or more parallelograms on the same base and between the same parallels are equal in area. Rectangle is also a parallelogram.

∴ It is true.

Consider Statement (II) :

Here, let AB be the base of the parallelogram then DE becomes its altitude (height).

Area of the parallelogram is given by: Base × Height

∴ Area of ‖gm ABCD = AB×DE = 10×6 = 60cm²

Now,

Consider AD as base of the parallelogram then BF becomes its altitude (height)

∴ area of ‖gm ABCD = AD × BF = 60cm²

AD × 8 = 60cm²

AD = = 7.5cm

∴length of AD = 7.5cm.

∴ Statement (II) is correct.

Consider Statement (III)

Area of parallelogram is base× height

∴ Statement (III) is false

∴ Statement (I) and (II) are true and statement (III) is false

Assertion:

Here, Area (ΔABD) = Area(ΔABC) (∵ Triangles on same base and between same parallel lines) –1

Subtract Area (Δ AOB) on both sides of –1

Area (ΔABD) – Area (Δ AOB) = Area (ΔABC) – Area (Δ AOB)

Area (∆AOD) = Area (∆BOC)

∴ Both Assertion and Reason are true and Reason is a correct explanation of Assertion.

Given: ∠DCB = 60°

Let the length of the side be x

Here, consider ΔBCD

BC = DC (all sides of rhombus are equal)

∴ ∠CDB = ∠CBD (angles opposite to equal sides are equal)

Now, by angle sum property

∠CDB + ∠CBD + ∠BCD = 180°

2× ∠CBD = 180° –60°

2 × ∠CBD = 180° – 60°

∴ 2× ∠CBD = 120°

∠ CBD = = 60°

∴ ∠CDB = ∠CBD = 60°

∴ Δ ADC is equilateral triangle

∴ BC = CD = BD = x cm

In Rhombus diagonals bisect each other.

Consider Δ COD

By Pythagoras theorem

CD² = OD² + OC²

x² = ² + OC²

OC² = x² – ²

OC =

OC = cm

∴ AC = 2× OC = 2 × = x

AC: BD = x : x = : 1

∴ AC: BD = : 1

∴ Both Assertion but Reason are true and Reason is not a correct explanation of Assertion.

Consider Δ ABD

We know that diagonals in a parallelogram bisect each other

∴ E is the midpoint of BD, AE is median of Δ ABD

∴ Area (Δ ADE) = Area (Δ AEB) (∵ Median divides the triangle into two triangles of equal areas)

Similarly we can prove

Area (Δ ADE) = Area (Δ DEC)

Area (Δ DEC) = Area (Δ CEB)

Area (Δ CEB) = Area (Δ AEB)

∴ Diagonals of a ‖ gm divide into four triangles of equal area.

∴ Both Assertion and Reason are true and Reason is a correct explanation of Assertion.

Area of trapezium = 1/2 × (sum of parallel sides) × height = 1/2 × (25 + 15)×6 = 120cm²

∴ Area of trapezium = 120cm²

∴ Assertion is correct.

Area of an equilateral triangle is given by: × a² (here ‘a’ is length of the side)

∴ Area of an equilateral triangle with side length 8 cm = × 8² = 16

∴ Reason is correct

∴ Both Assertion but Reason are true and Reason is not a correct explanation of Assertion.

Here, let AB be the base of the parallelogram then DE becomes its altitude (height).

Area of the parallelogram is given by: Base × Height

∴ Area of ‖gm ABCD = AB×DE = 16×8 = 128cm²

Now,

Consider AD as base of the parallelogram then BF becomes its altitude (height)

∴ area of ‖gm ABCD = AD × BF = 128cm²

AD × 10 = 128cm²

AD = = 12.8cm

∴length of AD = 12.8cm

∴Assertion is false and Reason is true

The correct answer is Option (D)

Δ ABC and Δ BCD does not lie between parallel lines and also Δ AOB and Δ COD are not congruent.

The correct answer is Option (B)

Area of parallelogram = base × corresponding height.

Area of the ||gm is Base×Height

Here, height is distance between the Base and its corresponding parallel side.

∴ Area (||gm ABCD) = Base × Height = DC × DL

(∵ Here DC is taken as length and DL is the distance between DC and its corresponding parallel side AB).

We know that any two or parallelogram having the same base and lying between the same parallel lines are equal in area.

Consider two ||gms ABCD and PQRS which are on same base and lie between same parallel lines.

∴ ar(||gm ABCD) = ar(||gm PQRS) –1

∴ ar(||gm ABCD ) : ar(||gm PQRS) = 1:1 (∵ eq –1)

Quadrilateral is any closed figure which has four sides.

Rhombus, Rectangle, Parallelograms are few Quadrilaterals.

When a Diagonal AC of a quadrilateral divides it into two parts of equal areas, it is not necessary for the figure to be a Rhombus or a Rectangle or a Parallelogram, it can be any Quadrilateral.

We know that any two or parallelogram having the same base and lying between the same parallel lines are equal in area.

∴ ar(||gm ABCD) = ar(||gm ABPQ) –1

We also know that when a parallelogram and a triangle lie on same base and between same parallel lines then, area of the triangle is half the area of the parallelogram.

∴ ar(∆BMP) = ar(‖gm ABPQ)

But, from –1

ar(||gm ABCD) = ar(||gm ABPQ)

∴ ar(∆BMP) = ar(‖gm ABCD)

Join EF

Here Area (ΔAEF) = Area (ΔBDF) = Area (ΔDEF) = Area (ΔDEC) = Area (ΔABC) – 1

Consider any vertex of the triangle.

Let us consider Vertex B

Here, BDEF form a parallelogram.

Area (||gm BDEF) = Area (ΔBDF) + Area (ΔDEF)

Area (||gm BDEF) = Area (ΔABC) + Area (ΔABC) = Area (ΔABC) (from –1)

∴ Area (||gm BDEF) = Area (ΔABC)

Similarly, we can prove for other vertices.

Given:

AD = 6cm

DL ⊥AB

BM ⊥ AD

DL = 8cm

BM = 10cm

Now, consider the parallelogram ABCD

Here, let AD be the base of the parallelogram then BM becomes its altitude (height).

Area of the parallelogram is given by: Base × Height

∴ area of ‖gm ABCD = AD×BM = 6×10 = 60cm²

Now,

Consider AB as base of the parallelogram then DL becomes its altitude (height)

∴ area of ‖gm ABCD = AB × DL = 60cm²

AB × 8 = 60cm²

AB = = 7.5cm

∴length of AB = 7.5cm.

Given: Length of parallel sides 14 cm and 10 cm, height is 6cm

We know that area of trapezium is given by: 1/2 (sum of parallel sides)×height

∴ Area of trapezium = 1/2 (14 + 10)×6 = 72cm²

∴ Area of trapezium = 72cm²

Consider the Figure

Here,

In ΔABC, AD is the median
Hence BD = DC
Draw AE ⊥ BC
Area of ΔABD = Area of ΔADC
Thus median of a triangle divides it into two triangles of equal area.

We know that when a parallelogram and a triangle lie on same base and between same parallel lines then, area of the triangle is half the area of the parallelogram.

Consider the figure,

Here,

Area(ΔABF) = 1/2 Area(||gm ABCD) (From above statement) –1

Area(||gm ABCD) = Base× Height –2

Sub –2 in –1

Area(ΔABF) = 1/2 × Base× Height

Given: BD = 14cm, AL = 8 cm, CM = 6 cm and also, AL ⊥ BD and CM ⊥ BD.

Here,

Area (Quad.ABCD) = Area (ΔABD) + Area (ΔABC)

Area (ΔABD) = 1/2 base × height = 1/2 × BD×AL = 1/2 × 14 × 8 = 56cm²

Area (ΔABC) = 1/2 base × height = 1/2 × BD×CM = 1/2 × 14 × 6 = 42cm²

∴ Area (Quad.ABCD) = Area (ΔABD) + Area (ΔABC) = 56 + 42 = 98 cm²

∴ Area (Quad.ABCD) = 98 cm²

Given: AC ||DP

We know that any two or Triangles having the same base and lying between the same parallel lines are equal in area.

∴ Area (Δ ACD) = Area (Δ ACP) –1

Add Area (Δ ABC) on both sides of eq –1

We get,

Area (Δ ACD) + Area (Δ ABC) = Area (Δ ACP) + Area (Δ ABC)

That is,

Area (quad.ABCD) = Area (Δ ABP)

Given: BE ||AC

We know that any two or more Triangles having the same base and lying between the same parallel lines are equal in area.

∴ Area (Δ ACE) = Area (Δ ACB) –1

Add Area (Δ ADC) on both sides of eq –1

We get,

Area (Δ ACE) + Area (Δ ADC) = Area (Δ ACB) + Area (Δ ADC)

That is,

Area (Δ ADE) = Area (quad. ABCD)

Given: area of ‖ gm ABCD is 80 cm²

We know that any two or parallelogram having the same base and lying between the same parallel lines are equal in area.

∴ ar(||gm ABCD) = ar(||gm ABEF) –1

We also know that when a parallelogram and a triangle lie on same base and between same parallel lines then, area of the triangle is half the area of the parallelogram.

∴ ar(∆ABD) = 1/2 × ar(||gm ABCD) and,

ar(∆BEF) = 1/2 × ar(||gm ABEF)

(i)

ar(||gm ABCD) = ar(||gm ABEF)

∴ ar(||gm ABEF) = 80cm² (∵ar(||gm ABCD) = 80cm²)

(ii)

ar(∆ABD) = 1/2 × ar(||gm ABCD)

ar(∆ABD) = 1/2 × 80 = 40cm² (∵ar(||gm ABCD) = 80cm²)

∴ ar(∆ABD) = 40cm²

(iii)

ar(∆BEF) = 1/2 × ar(||gm ABEF)

ar(∆BEF) = 1/2 × 80 = 40cm² (∵ar(||gm ABEF) = 80cm²)

∴ ar(∆BEF) = 40cm²

Given: AB‖DC and L is the midpoint of BC, PQ ‖ AD

Construction: Drop a perpendicular DM from D onto AP

Consider ΔPBL and ΔCQL

Here,

∠LPB = ∠LQC (Alternate interior angles, AB|| DQ)

BL = LC (L is midpoint of BC)

∠PLB = ∠QLC (vertically opposite angles)

∴ By AAS congruency

ΔPBL ΔCQL

∴ PB = CQ (C.P.C.T)

Area (||gm APQD) = base× height = AP × DM –1

Area (Trap.ABCD) = 1/2 × (sum of parallel sides) × height = 1/2 × (AB + DC) × DM

Area (Trap.ABCD) = 1/2 × (AB + DC) × DM = 1/2 × (AP + PB + DC) × DM (∵ AB = AP + PB)

Area (Trap.ABCD) = 1/2 × (AP + CQ + DC) × DM (∵ PB = CQ)

Area (Trap.ABCD) = 1/2 × (AP + DQ) × DM (∵ DC + CQ = DQ)

Area (Trap.ABCD) = 1/2 × (2× AP) × DM (∵ AP = DQ)

Area (Trap.ABCD) = AP × DM –2

From –1 and –2

Area (Trap.ABCD) = Area (||gm APQD)

Given: ABCD is a ‖ gm and O is a point on the diagonal AC.

Construction: Drop perpendiculars DM and BN onto diagonal AC.

Here,

DM = BN (perpendiculars drawn from opposite vertices of a ||gm to the diagonal are equal)

Now,

Area (ΔAOB) = 1/2 × base × height = 1/2 × AO × BN –1

Area (ΔAOD) = 1/2 × base × height = 1/2 × AO × DM –2

From –1 and –2

Area (∆AOB) = Area (∆AOD) (∵ BN = DM)

Given:∆ABC and ∆BDE are two equilateral triangles, D is the midpoint of BC.

Consider ΔABC

Here, let AB = BC = AC = x cm (equilateral triangle)

Now, consider ΔBED

Here,

BD = 1/2 BC

∴ BD = ED = EB = 1/2 BC = x/2 (equilateral triangle)

Area of the equilateral triangle is given by: (a is side length)

∴ ar(∆BDE): ar(∆ABC) = := :1 = 1:4

That is =

∴ ar(∆BDE) = ar(∆ABC)

Hence Proved

Given: D is the midpoint of AB and P Point is any point on BC, CQ‖ PD

In Quadrilateral DPQC

Area (Δ DPQ) = Area (Δ DPC)

Add Area (Δ BDP) on both sides

We get,

Area (Δ DPQ) + Area (Δ BDP) = Area (Δ DPC) + Area (Δ BDP)

Area (Δ BPQ) = Area (Δ BCD) –1

D is the midpoint BC, and CD is the median

∴ Area (Δ BCD) = Area (Δ ACD) = 1/2 × Area (Δ ABC) –2

Sub –2 in –1

Area (Δ BPQ) = 1/2 × Area (Δ ABC) (∵Area (Δ BCD) = 1/2 × Area (Δ ABC))

Consider Δ ABD

We know that diagonals in a parallelogram bisect each other

∴ E is the midpoint of BD, AE is median of Δ ABD

∴ Area (Δ ADE) = Area (Δ AEB) (∵ Median divides the triangle into two triangles of equal areas)

Similarly we can prove

Area (Δ ADE) = Area (Δ DEC)

Area (Δ DEC) = Area (Δ CEB)

Area (Δ CEB) = Area (Δ AEB)

∴ Diagonals of a ‖ gm divide into four triangles of equal area.

Given: BD ‖ CA, E is the midpoint of CA and BD = CA

Consider Δ BCD and Δ DEC

Here,

BD = EC (∵ E is the midpoint of AC that is CE = CA, BD = CA)

CD = CD (common)

∠BDC = ∠ECD (alternate interior angles, DB||AC)

∴ By SAS congruency

Δ BCD Δ DEC

∴ Area (Δ BCD) = Area (Δ DEC) –1

Here,

Area (Δ BCE) = Area (Δ DEC) (triangles on same base CE and between same parallel lines) –2

E is the midpoint of AC, BE is the median of ΔABC

∴ Area (Δ BCE) = Area (Δ ABE) = 1/2 × Area (Δ ABC)

∴ Area (Δ DEC) = 1/2 × Area (Δ ABC) (∵Area (Δ BCE) = Area (Δ DEC))

∴ Area (Δ BCD) = 1/2 × Area (Δ ABC) (∵Area (Δ DEC) = Area (Δ BCD))

Given: EG||DA, CF||DB

Here, in Quadrilateral ADEG

Area (Δ AED) = Area (Δ ADG) –1

In Quadrilateral CFBD

Area (Δ CBD) = Area (Δ BCF) –2

Add –1 and –2

Area (Δ AED) + Area (Δ CBD) = Area (Δ ADG) + Area (Δ BCF) –3

Add Area (Δ ABD) to –3

Area (Δ AED) + Area (Δ CBD) + Area (Δ ABD) = Area (Δ ADG) + Area (Δ BCF) + Area (Δ ABD)

Area (pentagon ABCDE) = Area (∆DGF)

Given: D divides the side BC of ∆ABC in the ratio m:n

Area (Δ ABD) = 1/2 × BD × AL

Area (Δ ADC) = 1/2 × CD × AL

Area (∆ABD): Area (∆ADC) = 1/2 × BD × AL: 1/2 × CD × AL

Area (∆ABD): Area (∆ADC) = BD: CD

Area (∆ABD): Area (∆ADC) = m: n (∵ BD:CD = m:n)

Given: X and Y are the midpoints of AC and AB respectively, QP ‖ BC and CYQ and BXP are straight lines.

Construction: Join QB and PC

In Quadrilateral BCQP

Area (Δ QBC) = Area (Δ BCP) (Triangles on same base BC and between same parallel lines are equal in area) –1 and,

Area (||gm ACBQ) = Area (||gm ABCP) (parallelograms on same base BC and between same parallel lines are equal in area) –2

Subtract –1 from –2

Area (||gm ACBQ) – Area (Δ QBC) = Area (||gm ABCP) – Area (Δ BCP)

Area (∆ACQ) = Area (∆ABP)

∴ Area(∆ABP) = Area(∆ACQ)