Math Ncert Exemplar 2019 Solutions for Class 9 Maths Chapter 9 Areas Of Parallelograms And Triangles are provided here with simple step-by-step explanations. These solutions for Areas Of Parallelograms And Triangles are extremely popular among Class 9 students for Maths Areas Of Parallelograms And Triangles Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Math Ncert Exemplar 2019 Book of Class 9 Maths Chapter 9 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Math Ncert Exemplar 2019 Solutions. All Math Ncert Exemplar 2019 Solutions for class Class 9 Maths are prepared by experts and are 100% accurate.

#### Page No 85:

#### Question 1:

Write the correct answer in each of the following :

The median of a triangle divides it into two

(A) triangles of equal area

(B) congruent triangles

(C) right triangles

(D) isosceles triangles

#### Answer:

A triangle's median is a line segment that connects a vertex to the opposing side's mid-point. As a result, a triangle's median divides it into two equal-area triangles.

Hence, the correct answer is option A.

#### Page No 85:

#### Question 2:

Write the correct answer in each of the following :

In which of the following figures, you find two polygons on the same base and between the same parallels?

#### Answer:

There are two polygons on the same base in figures (A), (B), and (C), but they are not between the same parallels. On the same base and between the same parallels are two polygons (PQRA and BQRS) in figure (D).

Hence, the correct answer is option D.

#### Page No 86:

#### Question 3:

Write the correct answer in each of the following :

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 rectangle of area 24 cm^{2}

(B) a square of area 25 cm^{2}

(C) a trapezium of area 24 cm^{2 }

(D) a rhombus of area 24 cm^{2}

#### Answer:

The length of the rectangle ABCD is 8 cm, and the width is 6 cm. If the midpoints of the sides of rectangle ABCD are E, F, G, and H, then EFGH is a rhombus.

As, length of rectangle ABCD = 8 cm

and breath of rectangle ABCD = 6 cm

If E, F, G, H are the mid-points of the sides of rectangle ABCD, then EFGH is a rhombus.

Then, diagonal of rhombus EFGH are EG and HF.

Thus, EG = BC = 8 cm

and HF = AB = 6 cm

$\begin{array}{rcl}\therefore \mathrm{Area}\mathrm{of}\mathrm{rhombus}& =& \frac{\mathrm{Product}\mathrm{of}\mathrm{diagonals}}{2}\\ & =& \frac{8\times 6}{2}=4\times 6=24{\mathrm{cm}}^{2}\end{array}$

Thereby, joining the mid-points of the adjacent sides of a rectangle forms a rhombus of area 24 cm^{2}.

Hence, the correct answer is option D.

#### Page No 86:

#### Question 4:

Write the correct answer in each of the following :

In the Figure, the area of parallelogram ABCD is :

(A) AB × BM

(B) BC × BN

(C) DC × DL

(D) AD × DL

#### Answer:

The area of a parallelogram is equal to the product of its either side and the corresponding altitude, as we know (or height).When AB is the base, then DL is the height. Area of parallelogram = AB × DL and when AD is the base, the height is equal to BM. Area of parallelogram = AD × BM, when DC equals base, height equals DL. Area of parallelogram = DC × DL and when BC is base, then height is not given.

Hence, the correct answer is option C.

#### Page No 86:

#### Question 5:

Write the correct answer in each of the following :

In the given Figure, if parallelogram ABCD and rectangle ABEF are of equal area, then :

(A) Perimeter of ABCD = Perimeter of ABEM

(B) Perimeter of ABCD < Perimeter of ABEM

(C) Perimeter of ABCD > Perimeter of ABEM

(D) Perimeter of ABCD = $\frac{1}{2}$ (Perimeter of ABEM)

#### Answer:

AB = EM [sides of rectangle] in rectangle ABEM, and CD = AB in parallelogram ABCD.

When both are added together, we get AB + CD = EM + AB .....(1)

The perpendicular distance between two parallel sides of a parallelogram is always shorter than the length of the other parallel sides.

BE < BC and AM < AD [because the hypotenuse of a right-angled triangle is longer than the other side]

On adding both above inequalities, we get

BE + AM < BC + AD or BC + AD > BE + AM

On adding AB + CD both sides, we get

AB + CD + BC + AD > AB + CD + BE + AM

⇒ AB + BC + CD + AD > AB + BE + EM + AM [∴ CD = AB = EM]

Perimeter of parallelogram ABCD > perimeter of rectangle ABEM.

Hence, the correct answer is option C.

****Disclaimer: In the given question, rectangle ABEM is incorrectly written as ABEF.**

#### Page No 87:

#### Question 6:

Write the correct answer in each of the following :

The mid-point of the sides of a triangle along with any of the vertices as the fourth point make a parallelogram of area equal to

(A) $\frac{1}{2}$ar (ABC)

(B) $\frac{1}{3}$ ar (ABC)

(C) $\frac{1}{4}$ ar (ABC)

(D) ar (ABC)

#### Answer:

If D, E and F are the mid-points of the sides BC, CA and AB of a ∆ABC, then all four triangles has equal area i.e.,

ar(∆AFE) = ar(∆BFD) = ar(∆EDC) = ar(∆DEF) ...(1)

∴ Area of ∆DEF = $\frac{1}{4}$ Area of ∆ABC ...(2)

Now, taking D as the fourth vertex, the area of the parallelogram AFDE will be

= Area of ∆AFE + Area of ∆DEF

= Area of ∆DEF + Area of ∆DEF

= 2 Area of ∆DEF [using (1)]

= $2\times \frac{1}{4}$ Area of ∆ABC [using (2)]

= $\frac{1}{2}$Area of ∆ABC

Hence, the correct answer is option A.

#### Page No 87:

#### Question 7:

Write the correct answer in each of the following :

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

(A) 1 : 2

(B) 1 : 1

(C) 2 : 1

(D) 3 : 1

#### Answer:

We already know that a parallelogram with the same bases and parallels has the same area. As a result, their area ratio is 1 : 1.

Hence, the correct answer is option B.

#### Page No 87:

#### Question 8:

Write the correct answer in each of the following :

ABCD is a quadrilateral whose diagonal AC divides it into two parts, equal in area, then ABCD

(A) is a rectangle

(B) is always a rhombus

(C) is a parallelogram

(D) need not be any of (A), (B) or (C)

#### Answer:

ABCD does not have to be a rectangle, rhombus, or parallelogram since if ABCD is a square, its diagonal AC divides it into two equal-sized halves.

Hence, the correct answer is option D.

#### Page No 87:

#### Question 9:

Write the correct answer in each of the following :

If a triangle and a parallelogram are on the same base and between same parallels, then the ratio of the area of the triangle to the area of parallelogram is

(A) 1 : 3

(B) 1 : 2

(C) 3 : 1

(D) 1 : 4

#### Answer:

When a parallelogram and a triangle are on the same base and have the same parallels, we know that the triangle's area is half that of the parallelogram.

i.e., Area of triangle = $\frac{1}{2}$ Area of parallelogram

$\Rightarrow \frac{\mathrm{Area}\mathrm{of}\mathrm{triangle}}{\mathrm{Area}\mathrm{of}\mathrm{parallelogram}}=\frac{1}{2}$

∴ Area of triangle : Area of parallelogram = 1 : 2

Hence, the correct answer is option B.

#### Page No 87:

#### Question 10:

Write the correct answer in each of the following :

ABCD is a trapezium with parallel sides AB = *a *cm and DC = *b *cm. E and F are the mid-points of the non-parallel sides. The ratio of ar (ABFE) and ar (EFCD) is

(A) *a *: *b*

(B) (3*a *+ *b*) : (*a *+ 3*b*)

(C) (*a *+ 3*b*) : (3*a *+ *b*)

(D) (2*a *+ *b*) : (3*a *+ *b*)

#### Answer:

Provided, AB = *a* cm, DC = *b* cm, AB||DC.

E and F are mid-points of AD and BC.

Now, distance between CD, EF and AB, EF will be same say *x*.

Join BD which intersecting EF at M.

Now, in ∆ABD, E is the mid-point of AD and EM || AB.

$\begin{array}{rcl}\mathrm{MF}& =& \frac{1}{2}\mathrm{CD}[\mathrm{mid}-\mathrm{point}\mathrm{theorem}]...\left(2\right)\\ & & \end{array}$

$\mathrm{EM}+\mathrm{MF}=\frac{1}{2}\mathrm{AB}+\frac{1}{2}\mathrm{CD}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{EF}=\frac{1}{2}(\mathrm{AB}+\mathrm{CD})=\frac{1}{2}(a+b)$

Now, area of trapezium ABFE = $\frac{1}{2}$(sum of parallel sides) × (distance between parallel sides)

$=\frac{1}{2}\left(a+\frac{1}{2}\left(a+b\right)\right)\times x=\frac{1}{4}\left(3a+b\right)x\phantom{\rule{0ex}{0ex}}$

Then, area of trapezium EFCD $=\frac{1}{2}\left(b+\frac{1}{2}\left(a+b\right)\right)\times x=\frac{1}{4}\left(3b+a\right)x$

∴ Required ratio $=\frac{\mathrm{Area}\mathrm{of}\mathrm{ABFE}}{\mathrm{Area}\mathrm{of}\mathrm{EFCD}}=\frac{{\displaystyle \frac{1}{4}}\left(3a+b\right)x}{\frac{1}{4}\left(3b+a\right)x}$

$=\frac{{\displaystyle \left(3a+b\right)}}{\left(a+3b\right)}\mathrm{or}\left(3a+b\right):\left(a+3b\right)$

Hence, the correct answer is option B.

#### Page No 88:

#### Question 1:

Write True or False and justify your answer :

ABCD is a parallelogram and X is the mid-point of AB. If ar(AXCD) = 24 cm^{2}, then ar(ABC) = 24 cm^{2}.

#### Answer:

The answer is False.

Given that ABCD is a parallelogram and ar(AXCD) = 24 cm^{2}.

Join AC and set the area of the parallelogram ABCD to 2*y* cm^{2}.

we know that, diagonal divides the area of parallelogram in two equal areas.

∴ ar(∆ABC) = ar(∆ACD) = *y*

As, X is the mid-point of AB.

Thus, ar(∆ACX) = ar(BCX) [since, X is median in ∆ABC]

$=\frac{1}{2}\mathrm{ar}\left(\mathrm{ABC}\right)=\frac{1}{2}y$

Now, ar(AXCD) = ar(∆ADC) + ar(ACX)

⇒ $24=\frac{3y}{2}$

⇒ $y=\frac{24\times 2}{3}=16{\mathrm{cm}}^{2}$

∴ ar(∆ABC) = 16 cm

^{2}

Hence, given statement is false.

#### Page No 88:

#### Question 2:

Write True or False and justify your answer :

PQRS is a rectangle inscribed in a quadrant of a circle of radius 13 cm. A is any point on PQ. If PS = 5 cm, then ar (PAS) = 30 cm^{2}.

#### Answer:

The answer is True.

Given, PS = 5 cm radius of circle = SQ = 13 cm

In right angled ∆SPQ,

SQ^{2} = PQ^{2} + PS^{2} [by Pythagoras theorem]

⇒ (13)^{2} = PQ^{2} + (5)^{2}

⇒ PQ^{2 }= 169 – 25 = 144

⇒ PQ = 12 cm [Neglecting negative taking positive square root, because length is always positive]

Then, area of ∆APS = $\frac{1}{2}$ × Base × Height

$=\frac{1}{2}\times \mathrm{PS}\times \mathrm{PQ}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\times 5\times 12=30{\mathrm{cm}}^{2}$

Hence, the statement is true, if A and Q coincide.

#### Page No 88:

#### Question 3:

Write True or False and justify your answer :

PQRS is a parallelogram whose area is 180 cm^{2 }and A is any point on the diagonal QS. The area of ∆ASR = 90 cm^{2}.

#### Answer:

The answer is False

Given that the area of the parallelogram PQRS = 180 cm^{2}, and QS is the diagonal that divides it into two equal-area triangles.

Therefore, area of triangle ASR cannot be 90 cm^{2} as area of triangle QSR is 90 cm^{2}.

#### Page No 88:

#### Question 4:

Write True or False and justify your answer :

ABC and BDE are two equilateral triangles such that D is the mid-point of BC.

Then ar(BDE) = $\frac{1}{4}$ ar(ABC).

#### Answer:

True

As, ∆ABC and ∆BDE are two equilateral triangles.

∴ Area of an equilateral ∆ABC $=\frac{\sqrt{3}}{4}\times {\left(\mathrm{side}\right)}^{2}=\frac{\sqrt{3}}{4}{\left(\mathrm{BC}\right)}^{2}...\left(1\right)$

[∵ in equilateral ∆ABC, AB = BC = AC]

Also, given D is the mid-point of BC.

∴ $\mathrm{BD}=\mathrm{DC}=\frac{1}{2}\mathrm{BC}...\left(2\right)$

Now, area of an equilateral ∆BDE $=\frac{\sqrt{3}}{4}\times {\left(\mathrm{side}\right)}^{2}$

$=\frac{\sqrt{3}}{4}\times {\left(\mathrm{BD}\right)}^{2}\left[\because \mathrm{in}\mathrm{equilateral}\u2206\mathrm{BDE},\mathrm{BD}=\mathrm{DE}=\mathrm{BE}\right]\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{3}}{4}\times {\left(\frac{1}{2}\mathrm{BC}\right)}^{2}\left[\mathrm{from}\left(2\right)\right]\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{3}}{4}\times \frac{1}{4}{\mathrm{BC}}^{2}=\frac{1}{4}{\left(\frac{\sqrt{3}}{4}\mathrm{BC}\right)}^{2}$

∴ Area of ∆BDE $=\frac{1}{4}\mathrm{Area}\mathrm{of}\u2206\mathrm{ABC}$

#### Page No 88:

#### Question 5:

Write True or False and justify your answer :

In the Figure, ABCD and EFGD are two parallelograms and G is the mid-point of CD. Then ar (DPC) = $\frac{1}{2}$ ar (EFGD).

#### Answer:

The answer is False

Join PG in the given figure. Since, G is the mid-point of CD.

Thus, PG is a median of ∆DPC and it divides the triangle into parts of equal areas.

Then, ar(∆DPG) = ar(∆GPC) = $\frac{1}{2}$ ar(∆DPC) ...(1)

If a parallelogram and a triangle lie on the same base and between the same parallels, then area of triangle is equal to half of the area of parallelogram.

Thus, parallelogram EFGD and ∆DPG lie on the same base DG and between the same parallels DG and EF.

So, ar(∆DPG) $=\frac{1}{2}\mathrm{ar}\left(\mathrm{EFGD}\right)...\left(2\right)$

From (1) and (2),

$\frac{1}{2}\mathrm{ar}(\u2206\mathrm{DPC})=\frac{1}{2}\mathrm{ar}\left(\mathrm{EFGD}\right)$

⇒ ar(∆DPC) = ar(EFGD)

#### Page No 89:

#### Question 1:

In the Figure, PSDA is a parallelogram. Points Q and R are taken on PS such that PQ = QR = RS and PA || QB || RC. Prove that ar (PQE) = ar (CFD).

#### Answer:

In a parallelogram PSDA, points Q and R are on PS such that PQ = QR = RS and PA || QB || RC.

To prove: ar(PQE) = ar(CFD)

Proof: In parallelogram PABQ, and PA||QB

So, PABQ is a parallelogram. PQ = AB …(1)

Similarly, QBCR is also a parallelogram.

QR = BC …(2)

and RCDS is a parallelogram.

RS =CD …(3)

Now, PQ=QR = RS …(4)

From. (1), (2) (3) and (4),

PQ || AB [∴ in parallelogram PSDA, PS || AD]

In ΔPQE and ΔDCF, ∠QPE = ∠FDC [As, PS || AD and PD is transversal, then alternate interior angles are equal]

PQ=CD [from (3) and (4)]

and ∠PQE = ∠FCD [∴ ∠PQE = ∠PRC corresponding angles and ∠PRC = ∠FCD alternate interior angles]

ΔPQE = ΔDCF [ASA congruence rule]

∴ ar(ΔPQE) ≅ ar(ΔCFD) [As, congruent figures have equal area]

Hence proved.

#### Page No 90:

#### Question 2:

X and Y are points on the side LN of the triangle LMN such that LX = XY = YN. Through X, a line is drawn parallel to LM to meet MN at Z. Prove that ar (LZY) = ar (MZYX)

#### Answer:

As, X and Y are points on the side LN such that LX = XY = YN and XZ || LM

To prove: ar(ΔLZY) = ar(MZYX)

Proof: Since, ΔXMZ and ΔXLZ are on the same base XZ and between the same parallel lines LM and XZ.

Thus, ar(ΔXMZ) = ar(ΔXLZ) …(1)

On adding ar (ΔXYZ) both sides of (1), we have

ar(ΔXMZ) + ar(ΔXYZ) = ar(ΔXLZ) + ar(ΔXYZ)

⇒ ar(MZYX) = ar(ΔLZY)

Hence proved.

#### Page No 90:

#### Question 3:

The area of the parallelogram ABCD is 90 cm^{2}.^{ } Find

(i) ar (ABEF)

(ii) ar (ABD)

(iii) ar (BEF)

#### Answer:

As, area of parallelogram, ABCD = 90 cm^{2}.

1. We know that parallelograms with the same base and parallel lines have the same area. Parallelograms ABCD and ABEF are on the same base AB and between the same parallels AB and CF in this situation.

Thus, ar(ΔBEF) = ar(ABCD) = 90 cm^{2}

2. We know that if a triangle and a parallelogram are on the same base and have the same parallels, the triangle's area is half of the parallelogram's area. The parallelogram ABCD and the ΔABD are on the same base AB and between the same parallels AB and CD in this case.

So, ar (ΔABD) = $\frac{1}{2}$ar(ABCD)

$=\frac{1}{2}\times 90=45{\mathrm{cm}}^{2}$ [∴ ar(ABCD) = 90 cm^{2}]

Hence, ar(ΔABD) = 45 cm^{2}

3. The parallelograms ABEF and ABEF are on the same base EF and between the same parallels AB and EF in this case.

ar(ΔBEF) = $\frac{1}{2}$ar(ABEF)

$=\frac{1}{2}\times 90=45{\mathrm{cm}}^{2}$ [∴ ar(ABEF) = 90 cm^{2}, from part (1)]

Hence, ar(ΔBEF) = 45 cm^{2}

#### Page No 90:

#### Question 4:

In ∆ABC, D is the mid-point of AB and P is any point on BC. If CQ || PD meets AB in Q (Figure 9.14), then prove that ar (BPQ) = $\frac{1}{2}$ ar (ABC).

#### Answer:

Given: In ∆ABC, D is the mid-point of AB and P is any point on BC.

And CQ || PD meets AB in Q.

To Prove: ar(∆BPQ) = $\frac{1}{2}$ar(∆ABC)

Construction: Join CD.

Proof: As, D is the mid-point of AB. So CD is the median of ∆ABC.

We know, a median of a triangle divides it into two triangles of equal areas.

$\therefore \mathrm{ar}(\u2206\mathrm{BDC})=\frac{1}{2}\mathrm{ar}\left(\mathrm{ABC}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{ar}(\u2206\mathrm{BPD})+\mathrm{ar}(\u2206\mathrm{DPC})=\frac{1}{2}\mathrm{ar}(\u2206\mathrm{ABC})...\left(1\right)\phantom{\rule{0ex}{0ex}}$

Now, ∆DPQ and ∆DPC are on the same base DP and between the same parallel lines DP and CQ.

So, ar(∆DPQ) = ar(∆DPC) ...(2)

On putting the value from (2) in (1), we get

ar(∆BPD) + (∆DPQ) = $\frac{1}{2}$ ar(∆ABC)

⇒ ar(∆BPQ) = $\frac{1}{2}$ar(∆ABC)

Hence proved.

#### Page No 90:

#### Question 5:

ABCD is a square. E and F are respectively the midpoints of BC and CD. If R is the mid-point of EF (Figure 9.15), prove that ar(AER) = ar(AFR).

#### Answer:

Given: square ABCD, E and F are the mid-points of BC and CD and R is the mid-point of EF.

To prove: ar(AER) = ar(AFR).

Construction: Draw AN ⊥ EF.

Proof: ar(∆AER) = $\frac{1}{2}$ × Base × Height

$=\frac{1}{2}\times \mathrm{ER}\times \mathrm{AN}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\times \mathrm{FR}\times \mathrm{AN}$

[∵ R is the mid-point of EF, so ER = FR]

= ar(∆AER)

Hence proved.

#### Page No 91:

#### Question 6:

O is any point on the diagonal PR of a parallelogram PQRS (figure 9.16). Prove that ar(PSO) = ar(PQO).

#### Answer:

Given: In a parallelogram PQRS, O is any point on the diagonal PR.

To prove: ar(PSO) = ar(PQO)

Construction: Join SQ which intersects PR at B.

Proof: we know, diagonals of a parallelogram bisect each other, so B is the mid-point of SQ.

Then, PB is a median of ∆QPS and we know that, a median of a triangle divides it two triangles of equal area.

∴ ar(∆BPQ) = ar(∆BPS) ...(1)

Also, OB is the median of ∆OSQ.

∴ ar(∆OBQ) = ar(∆OBS) ...(2)

On adding (1) and (2), we get

ar(∆BPQ) + ar(∆OBQ) = ar(∆BPS) + ar(∆OBS)

⇒ ar(∆PQO) = ar(∆PSO)

Hence proved.

#### Page No 91:

#### Question 7:

ABCD is a parallelogram in which BC is produced to E such that CE = BC (figure 9.17). AE intersects CD at F. If ar (DFB) = 3 cm^{2}, find the area of the parallelogram ABCD.

#### Answer:

ABCD is a parallelogram and CE = BC i.e., C is the mid-point of BE.

∆ADF and ∆DFB are on the same base DF and between parallels CD and AB. Then,

ar(∆ADF) = ar(DFB) = 3 cm^{2 } ...(1)

In ∆ABE, by the converse of mid-point theorem,

EF = AF [As, C is mid-point of BE] ...(2)

In ∆ADF and ∆ECF,

∠AFD = ∠CFE [vertically opposite angles]

AF = EF [from (2)]

and ∠DAF = ∠CEF [since, BE || AD and AE is transversal, then alternate interior angles are equal]

∴ ∆ADF ≅ ∆ECF [ASA congruence rule]

Then, ar(∆ADF) = ar(∆CFE) [congruent figures have equal area]

∴ ar(∆ADF) = ar(∆CFE) = 3 cm^{2} [from (1)] ...(3)

Now, in ∆BFE, C is the mid-point of BE then CF is median of ∆BFE.

∴ ar(∆CEF) = ar(∆BFC) [median of a triangle divides it into two triangles of equal area]

⇒ ar(∆BFC) = 3 cm^{2} ...(4)

Now, ar(∆BDC) = ar(∆DFB) + ar(∆BFC)

= 3 + 3 = 6 cm^{2} [from (1) and (4)]

We know, diagonal of a parallelogram divides it into two congruent triangles of equal areas.

∴ Area of parallelogram ABCD = 2 × area of ∆BDC

= 2 × 6 = 12 cm^{2}

Hence, the area of parallelogram ABCD is 12 cm^{2}.

#### Page No 91:

#### Question 8:

In trapezium ABCD, AB || DC and L is the mid-point of BC. Through L, a line PQ || AD has been drawn which meets AB in P and DC produced in Q (figure 9.18). Prove that ar (ABCD) = ar (APQD)

#### Answer:

Given: In trapezium ABCD, AB || DC, DC produced in Q and L is the mid-point of BC.

∴ BL = CL

To Prove: ar(ABCD) = ar(APQD)

Proof: Since, DC produced in Q and AB || DC.

So,

In ∆CLO and ∆BLP,

CL = BL [As, L is the mid-point of BC]

∠LCQ = ∠LBP [alternate interior angles as BC is a transversal]

∠CQL = ∠LPB [alternate interior angles as PQ is a transversal]

∴ ∆CLQ ≅ ∆BLP [AAS congruence rule]

Then, ar(∆CLQ = ar(∆BLP) ...(1) [As, congruent triangles have equal area]

Now, ar(ABCD) = ar(APQD) – ar(∆CLQ) + ar(∆BLP)

= ar(APQD) – ar(∆BLP) + ar(∆BLP) [from (1)]

⇒ ar(ABCD) = ar(APQD)

Hence proved.

#### Page No 92:

#### Question 9:

If the mid-points of the sides of a quadrilateral are joined in order, prove that the area of the parallelogram so formed will be half of the area of the given quadrilateral figure.

#### Answer:

Given: ABCD is a quadrilateral. P, F, R and S are the mid-points of the sides BC, CD, AD and AB and PFRS is a parallelogram.

To prove: ar(parallelogram PFRS) = $\frac{1}{2}$ar(quadrilateral ABCD)

Construction: Join BD and BR.

Proof: Median BR divides ∆BDA into two triangles of equal area.

∴ ar(∆BRA) = $\frac{1}{2}$ar(∆BDA) ...(1)

Similarly, median RS divides ∆BRA into two triangles of equal area.

ar(∆ASR) = $\frac{1}{2}$ar(∆BRA) ...(2)

From (1) and (2),

ar(∆ASR) = $\frac{1}{4}$ar(∆BDA) ...(3)

Then, ar(∆CFP) = $\frac{1}{4}$ ar(∆BCD) ...(4)

On adding (3) and (4), we have

ar(∆ASR) + ar(∆CFP) = $\frac{1}{4}$ar(quadrilateral BCDA) ...(5)

Similarly, ar(∆DRF) + ar(∆BSP) = $\frac{1}{4}$ar(quadrilateral BCDA) ...(6)

On adding (5) and (6) we get

ar(∆ASR) + ar(∆CFP) + ar(∆DRF) + ar(∆BSP) = $\frac{1}{2}$ar(quadrilateral BCDA) ...(7)

Also, ar(∆ASR) + ar(∆CFP) + ar(∆DRF) + ar(∆BSP) + ar(parallelogram PFRS) = ar(quadrilateral BCDA) ...(8)

On subtracting (7) from (8) we have

ar(parallelogram PFRS) = $\frac{1}{2}$ar(quadrilateral BCDA)

Hence proved.

#### Page No 94:

#### Question 1:

A point E is taken on the side BC of a parallelogram ABCD. AE and DC are produced to meet at F. Prove that ar (ADF) = ar (ABFC)

#### Answer:

A parallelogram is ABCD, and E is a point on BC. AE and DC are produced to meet at F.

∴ AB || CD and BC || AD ...(1)

To prove: ar(∆ADF) = ar(ABFC)

Construction: Join AC and DE.

Proof: As, AC is a diagonal of parallelogram ABCD.

So, ar(∆ABC) = ar(∆ACD) ...(2)

As, ∆ABF and ∆ABC are on the same base AB and between the same parallels AB and DF,

∴ ar(∆ABF) = ar(∆ABC) ...(3)

From (2) and (3),

ar(∆ABC) = ar(∆ACD) = ar(∆ABF)

On subtracting ar(∆ABE) from both sides of (3), we get

ar(∆ABF) – ar(∆ABE) = ar(∆ABC) – ar(∆ABE)

⇒ ar(∆BEF) = ar(∆AEC) ...(4)

Now, ar(AECD) = ar(∆AEC) + ar(∆ACD) = ar(∆ABC) + ar(∆BEF) [from (2) and (4)]

On adding ar(∆CEF) both sides, we get

ar(AECD) + ar(∆CEF) = ar(∆ABC) + ar(∆BEF) + ar(∆CEF)

⇒ ar(∆ADF) = ar(ABFC)

Hence proved.

#### Page No 94:

#### Question 2:

The diagonals of a parallelogram ABCD intersect at a point O. Through O, a line is drawn to intersect AD at P and BC at Q. Show that PQ divides the parallelogram into two parts of equal area.

#### Answer:

Given: In a parallelogram ABCD, diagonals interect at O and draw a line PQ, which intersects AD and BC.

Tp prove: PQ divides the parallelogram ABCD into two parts of equal of equal area.

⇒ ar(ABQP) = ar(CDPQ)

Proof: As, diagonals of a parallelogram bisect each other.

∴ OA = OC and OB = OD ...(1)

In ∆AOB and ∆COD,

OA = OC

OB = OD [from (1)]

Also, ∠AOB = ∠COD [vertically opposite angles]

∴ ∆AOB ≅ ∆COD [by SAS congruence rule]

⇒ ar(∆AOB) = ar(∆COD) ...(2)

[since, congruent figures have equal area]

Now, in ∆AOP and ∆COQ,

∠PAO = ∠OCQ [alternate interior angles]

⇒ OA = OC [from (1)]

and ∠AOP = ∠COQ [vertically opposite angles]

∴ ∆AOP ≅ ∆COQ [by ASA congruence rule]

⇒ ar(∆AOP = ar(∆COQ) ...(3)

[since, congruent figures have equal area]

Then, ar(∆BOQ) = ar(∆POD) ...(4)

Now, ar(∆COQ) + ar(∆COD) + ar(∆POD) = ar(∆AOP) + ar(∆AOB) + ar(∆BOQ) [Adding (2), (3) and (4)]

⇒ ar(ABQP) = ar(CDPQ)

Hence proved.

#### Page No 94:

#### Question 3:

The medians BE and CF of a triangle ABC intersect at G. Prove that the area of ∆GBC = area of the quadrilateral AFGE.

#### Answer:

Given: In ∆ABC, median BE and CF intersect each other at G.

To prove: ar(∆GBC) = ar(AFGE)

Proof: As BE is the median of ∆ABC and a median of a triangle divides it into two parts of equal area.

So, ar(∆ABE) = ar(∆CBE)

⇒ ar(∆ABE) = $\frac{1}{2}$ar(∆ABC) ...(1)

As, CF is the median of ∆ABC.

Then, ar(∆ACF) = ar(∆BCF)

⇒ ar(∆BCF) = $\frac{1}{2}$ ar(∆ABC) ...(2)

From (1) and (2),

ar(∆ABE) = ar(∆BCF) ...(3)

On substracting ar(∆GBF) from both sides of (3), we have

ar(∆ABE) – ar(∆GBF) = ar(∆BCF) – ar(∆GBF)

⇒ ar(AFGE) = ar(GBC)

Hence proved.

#### Page No 95:

#### Question 4:

In the Figure, CD || AE and CY || BA. Prove that ar (CBX) = ar (AXY)

#### Answer:

In figure, CD||AE and CY || BA

To prove: ar(ΔCBX) = ar(ΔAXY)

Proof: The areas of triangles with the same base and parallels are equal.

As, ΔABY and ΔABC both lie on the same base AB and between the same parallels CY and BA therefore,

ar(ΔABY) = ar(ΔABC)

⇒ ar(ABX) + ar(AXY) = ar(ABX) + ar(CBX)

⇒ ar(AXY) = ar(CBX) [eliminating ar(ABX) from both sides]

Hence proved.

#### Page No 95:

#### Question 5:

ABCD is a trapezium in which AB || DC, DC = 30 cm and AB = 50 cm. If X and Y are, respectively the mid-points of AD and BC, prove that ar (DCYX) = $\frac{7}{9}$ ar (XYBA)

#### Answer:

Given: A trapezium ABCD, where AB || DC, DC = 30 cm and AB = 50 cm.

Also X and Y are the mid-points of AD and BC.

To prove: ar(DCYX) = $\frac{7}{9}$ar(XYBA)

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

Proof: In ∆DCY and ∆PBY,

CY = BY [As Y is the mid-point of BC]

⇒ ∠DCY = ∠PBY [alternate interior angles]

and ∠2 = ∠3 [vertically opposite angles]

∴ ∆DCY ≅ ∆PBY [ASA congruence rule]

Then, DC = BP [CPCT]

Also, DC = 30 cm [given]

∴ DC = BP = 30 cm

Now, AP = AB + BP

AP = 50 + 30 = 80 cm

In ∆ADP, using mid-point theorem,

XY = $\frac{1}{2}$ AP = $\frac{1}{2}$ × 80 = 40 cm

Assume distance between AB, XY and XY, DC is *x* cm.

Now, area of trapezium DCYX = $\frac{1}{2}$*x*(30 + 40)

[∵ area of trapezium = $\frac{1}{2}$sum of parallel sides × distance between them]

= $\frac{1}{2}$(*x*)(70) = 35*x* cm^{2}

Then, area of trapezium XYBA = $\frac{1}{2}$*x*(40 + 50) = $\frac{1}{2}$*x* × 90 = 45*x* cm^{2}

$\therefore \frac{\mathrm{ar}\left(\mathrm{DCYX}\right)}{\mathrm{ar}\left(\mathrm{XYBA}\right)}=\frac{35x}{45x}=\frac{7}{9}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{ar}\left(\mathrm{DCYX}\right)=\frac{7}{9}\mathrm{ar}\left(\mathrm{XYBA}\right)$

Hence proved.

#### Page No 95:

#### Question 6:

In ∆ABC, if L and M are the points on AB and AC, respectively such that LM || BC. Prove that ar (LOB) = ar (MOC)

#### Answer:

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

To prove: ar(ΔLOB) = ar(ΔMOC)

Proof: Triangles on the same base and between the same parallels are equal in area.

As, ΔLBC and ΔMBC lie on the same base BC and between the same the same parallels BC and LM.

So, ar(ΔLBC) = ar(ΔMBC)

⇒ ar(ΔLOB) + ar(ΔBOC) = ar(ΔMOC) + ar(ΔBOC)

On eliminating ar(ΔBOC) from both sides, we have

ar(ΔLOB) = ar(ΔMOC)

Hence proved.

#### Page No 95:

#### Question 7:

In (Figure 9.25), ABCDE is any pentagon. BP drawn parallel to AC meets DC produced at P and EQ drawn parallel to AD meets CD produced at Q. Prove that ar (ABCDE) = ar (APQ)

#### Answer:

ABCDE is a pentagon where BP || AC and EQ|| AD.

To prove: ar (ABCDE) = ar (APQ)

Proof: The area of triangles with the same base and parallels is the same. Here, ΔADQ and ΔADE lie on the same base AD and between the same parallels AD and EQ.

So, ar(ΔADQ) = ar(ΔADE) ...(1)

Then, ΔACP and ΔACB lie on the same base AC and between the same parallels AC and BP.

So, ar(ΔACP) = ar(ΔACB) …(2)

On adding (1) and (2), we have

ar(ΔADQ) + ar(ΔACP) = ar(ΔADE) + ar(ΔACB)

On adding ar(ΔACD) both sides, we get

ar(ΔADQ) + ar(ΔACP) + ar(ΔACD) = ar(ΔADE) + ar(ΔACB) + ar(ΔACD)

⇒ ar(ΔAPQ) = ar(ABCDE)

Hence proved.

#### Page No 96:

#### Question 8:

If the medians of a ∆ABC intersect at G, show that ar (AGB) = ar (AGC) = ar (BGC) = $\frac{1}{3}$ ar (ABC)

#### Answer:

In ΔABC, AD, BE and CF are medians and intersect at G.

To prove: ar(ΔAGB) = ar(ΔAGC) = ar(ΔBGC) = $\frac{1}{3}$ ar(ΔABC)

Proof: A median of a triangle divides it into two triangles of equal area.

In ΔABC, AD is median,

∴ ar(ΔABD) = ar(ΔACD) ...(1)

In ΔBGC, GD is a median,

∴ ar(ΔGBD) = ar(ΔGCD) ...(2)

On subtracting (2) from (1) we have

ar(ΔABD) − ar(ΔGBD) = ar(ΔACD) − ar(ΔGCD)

⇒ ar(ΔAGB) = ar(ΔAGC) ...(3)

Similarly,

ar(ΔAGB) = ar(ΔBGC) ...(4)

From (3) and (4),

ar(ΔAGC) = ar(ΔBGC) ...(5)

Then,

ar(ΔABC) = ar(ΔAGB) + ar(ΔBGC) + ar(ΔAGC)

⇒ ar(ΔABC) = ar(AGB) + ar(ΔAGB) + ar(ΔAGB)

⇒ ar(ΔABC) = 3 ar(ΔAGB)

⇒ ar(ΔAGB) = $\frac{1}{3}$ ar(ΔABC) ...(6)

From (5) and (6),

ar(ΔBGC) = $\frac{1}{3}$ ar(ΔABC)

Also, ar(ΔAGC) = $\frac{1}{3}$ar(ΔABC)

Hence proved.

#### Page No 96:

#### Question 9:

In the 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 (ABP) = ar (ACQ).

#### Answer:

Given: X and Y are the mid-points of AC and AB respectively. Also, QP|| BC and CYQ, BXP are straight lines.

To prove: ar(ΔABP) = ar(ΔACQ)

Proof: As, X and Y are the mid-points of AC and AB thus, XY || BC.

In terms of area, triangles with the same base and parallels have the same area.

Here, ΔBYC and ΔBXC lie on same base BC and between the same parallels BC and XY.

Then, ar(ΔBYC) = ar(ΔBXC)

On subtracting ar (ΔBOC) from both sides, we have

ar(ΔBYC) – ar(ΔBOC) = ar(ΔBXC) – ar(ΔBOC)

⇒ ar(ΔBOY) = ar(ΔCOX)

On adding ar(ΔXOY) both sides we get,

ar(ΔBOY) + ar(ΔXOY) = ar(ΔCOX) + ar(ΔXOY)

⇒ ar(ΔBYX) = ar(ΔCXY) …(1)

As a result, quadrilaterals XYAP and YXAQ are found on the same XY base and between the same XY and PQ parallels.

ar(XYAP) = ar(YXAQ) …(2)

On adding (1) and (2), we get

ar(ΔBYX) + ar(XYAP) = ar(ΔCXY) + ar(YXAQ)

⇒ ar(ΔABP) = ar(ΔACQ)

Hence proved.

#### Page No 96:

#### Question 10:

In the Figure, ABCD and AEFD are two parallelograms. Prove that ar (PEA) = ar (QFD)

#### Answer:

Given: Two parallelograms are ABCD and AEFD.

To prove: ar(PEA) = ar(QFD)

Proof: In quadrilateral PQDA, AP || DQ [As in ABCD parallelogram, AB || CD]

and PQ || AD [As in the AEFD parallelogram, FE || AD]

Then, quadrilateral PQDA is a parallelogram.

In addition, parallelograms PQDA and AEFD are based on the same base AD and have the same parallels AD and EQ.

⇒ ar(parallelogram PQDA) = ar(parallelogram AEFD)

On subtracting ar(quadrilateral APFD) from both sides, we get

ar(parallelogram PQDA) – ar(quadrilateral APFD) = ar(parallelogram AEFD) – ar(quadrilateral APFD)

⇒ ar(QFD) = ar(PEA)

Hence proved.

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