AREAS

11.1 – 1

D is midpoint of AC, let’s draw a perpendicular on AB from D to E,

It will be parallel to BC and half of BC, so DE = ½ x 6.5 =3.25 cm

Since angle AED = 90⁰, AB is base and ED is height of ΔABD

base = AB = 12 cm, height = DE = 3.25 cm

area(ΔABD) = ½ x 3.25 x 12 = 19.5 cm²

11.1 – 2

In ΔPQS, angle P = 90⁰, so, base = 12 cm, height = 9 cm

area(ΔPQS) = ½ x 12 x 9 = 54 cm²

In ΔQRS, QS² + 8² = 17², so, QS = 15 cm  

angle SQR = 90⁰, so, base = 8 cm, height = 15 cm

area(ΔQRS) = ½ x 8 x 15 = 60 cm²

 area(PQRS) = 54+60 = 114 cm²

11.1 – 3

In rectangle ADCE, length = 8 cm, breadth = 3 cm

area(ADCE) = 8*3 = 24 cm²

In ΔBEC, angle BEC = 90⁰, so, base = 8 cm, height = 3 cm

area(ΔBEC) = ½ x 8 x 3 = 12 cm²

Hence, area(ABCD) = 24+12 = 36 cm²

11.1 – 4

AB || CD, AB = CD, AD = BC

Since diagonals bisect, AO = OC and DO = OB

Therefore ΔAOD and ΔBOC are congruent (SSS congruency)

Hence, area(ΔAOD) = area(ΔBOC)

11.3 – 1

AD is median, it means D is midpoint, BD=DC

E is midpoint, so AE=ED

ΔABD and ΔADC have equal base (BD=DC) and they are between same parallels,

So area(ΔABD) = area(ΔADC)

 ΔBDE and ΔDCE have equal base (BD=DC) and they are between same parallels,

So area(ΔBDE) = area(ΔDCE)

area(ΔABD) – area(ΔBDE) = area(ΔADC) – area(ΔDCE)

Therefore, area(ΔABE) = area(ΔACE)

Similarly, area(ΔAEB)=area(ΔEDB) because base=AE=ED and they are between same parallels

And, area(ΔAEC)=area(ΔEDC) because base=AE=ED and they are between same parallels

That is, area(ΔABE) = area(ΔBED) = area(ΔEDC) = area(ΔAEC)

Or, area(ΔABE) = ¼ area(ΔABC)

11.3 – 2

ABCD is a parallelogram. AB || CD, AD || BC

AC and BD are diagonals, they intersect at O

In a parallelogram, diagonals bisect each other

ΔABC and ΔABD are on the same base AB and are between parallel lines

So, area(ΔABC) = area(ΔABD)

Or, area(ΔABO) + area(AOD) = area(ΔABO) + area(ΔBOC)

So, area(ΔAOD) = area(ΔBOC)

In ΔABO and ΔCDO, AB=CD, AO=OB, CO=OD, they are congruent

So area(ΔABO) = area(ΔCDO)

Thus, area(ΔABO) = area(ΔBCO) = area(ΔCDO) = area(ΔDAO)

Hence, diagonals divide a parallelogram into four triangles of equal area

11.3 – 3

Given AB is bisected at O, CD is bisected at O

AB, CD are diagonals of ABCD

Since diagonals bisect each other, ABCD is a parallelogram

area(ΔABC) = ½ area(ABCD) and area(ΔABD) = ½ area(ABCD)

Therefore, area(ΔABC) = area(ΔABD)

11.3 – 4

Given, BD = DC, CE = EA, AF = FB

FE joins midpoints, so FE || BC or FE || BD

ED joins midpoints, so ED || AB or ED || FB

So, BDEF is a parallelogram

area(ΔBDF) = area(ΔFDE) = half of area(BDEF)

area(ΔBDF) = area(ΔDCE) since BD = DC & they are between parallel lines

area(ΔCDE) = area(ΔEFA) since CE = EA & they are between parallel lines

area(ΔAFE) = area(ΔFBD) since AF = FB & they are between parallel lines

Therefore, area(ΔBDF) = area(ΔFDE) = area(ΔDCE) = area(ΔAFE)

Or, area(ΔFDE) = ¼ area(ΔABC)

BDEF has 2 out of the 4 triangles, so area(BDEF) = ½ area(ΔABC)

11.3 – 5

Given, area(ΔDBC) = area(ΔEBC)

½ x BC x height(ΔDBC) = ½ x BC x height(ΔEBC)

height(ΔDBC) = height(ΔEBC)

It means, D and E are at the same distance from BC

So, DE || BC

11.3 – 6

Given, EA || BC, EB || AC, so EABC is a parallelogram, EA = BC

Given, AF || BC, AB || FC, so AFBC is a parallelogram, AF = BC

Therefore, EA = AF

EA is the base of ΔEAB and AF is the base of ΔAFC

They are between parallel lines, so their heights are same

So, area(ΔEAB) = area(ΔAFC)

11.3 – 7

AB || DC, that means D and C are at the same distance from AB

area(ABD) = area(ABC) since both have base AB and same height

area(ABO) + area(AOD) = area(ABO) + area(BOC)

Therefore, area(AOD) = area(BOC)

11.3 – 8

AC || BF ..... (1)

Angles BAF, AFC are equal (alternate angles)

Angles CAF, AFB are equal (alternate angles)

Therefore, angle BAF+CAF = angle AFC+AFB ..... (2)

Also, angles ACB, CBF are equal (alternate angles)

Angles ABC, BCF are equal (alternate angles)

Therefore, angle ACB+ABC = CBF+BCF ..... (3)

From 1, 2 & 3, ABCF is a parallelogram, so AC = BF, AB = CF

In triangles ABO, CFO,

AB=CF, angle BAO = angle OFC, angle ABO = angle OCF

So, triangles ABO, CFO are congruent, their area is equal.

area (ABC) = area (ACO) + area (ABO)

area (ACF) = area (ACO) + area (CFO)

Since area (ABO) = area (CFO), AREA (ABC) = AREA (ACF)

Area (ABCDE) = Area (AEDC) + Area (ABC)

Area (AEDF) = Area (AEDC) + Area (ACF)

Since area(ABC) = area(ACF), AREA (ABCDE) = AREA (AEDF)

11.3 – 9

Given, area(RSA) = area(RSB)

½ x RS x height of RSA = ½ x RS x height of RSB

So, height of RSA = height of RSB

That is, distance of A from RS and distance of B from RS are same

So, RS || AB

Since two sides are parallel, RSAB is a trapezium.

Given, area(PAS) = area(QBR)

That is, area(RSA) + area(RSP) = area(RSB) + area (RSQ)

But area(RSA) = area(RSB)

That is, area(RSP) = area(RSQ)

½ x RS x height(RSP) = ½ x RS x height (RSQ)

So, height(RSP) = height(RSQ)

Distance of P from RS and distance of Q from RS are same

So,  PQ || RS

Since two sides are parallel, PQRS is a trapezium.