# NCERT solution class 9 chapter 9 Areas of Parallelograms and Triangles exercise 9.4 mathematics

## EXERCISE 9.4

#### Question 1:

Parallelogram ABCD and rectangle ABEF are on the same base AB and have equal areas. Show that the perimeter of the parallelogram is greater than that of the rectangle.

As the parallelogram and the rectangle have the same base and equal area, therefore, these will also lie between the same parallels.

Consider the parallelogram ABCD and rectangle ABEF as follows.

Here, it can be observed that parallelogram ABCD and rectangle ABEF are between the same parallels AB and CF.

We know that opposite sides of a parallelogram or a rectangle are of equal lengths. Therefore,

AB = EF (For rectangle)

AB = CD (For parallelogram)

∴ CD = EF

⇒ AB + CD = AB + EF … (1)

Of all the line segments that can be drawn to a given line from a point not lying on it, the perpendicular line segment is the shortest.

And similarly, BE < BC

∴ AF + BE < AD + BC … (2)

From equations (1) and (2), we obtain

AB + EF + AF + BE < AD + BC + AB + CD

Perimeter of rectangle ABEF < Perimeter of parallelogram ABCD

#### Question 2:

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

Can you answer the question that you have left in the ’Introduction’ of this chapter, whether the field of Budhia has been actually divided into three parts of equal area?

[Remark: Note that by taking BD = DE = EC, the triangle ABC is divided into three triangles ABD, ADE and AEC of equal areas. In the same way, by dividing BC into n equal parts and joining the points of division so obtained to the opposite vertex of BC, you can divide ΔABC into n triangles of equal areas.]

Let us draw a line segment AM ⊥ BC.

We know that,

Area of a triangle × Base × Altitude

It is given that DE = BD = EC

⇒

⇒ Area (ΔADE) = Area (ΔABD) = Area (ΔAEC)

It can be observed that Budhia has divided her field into 3 equal parts.

#### Question 3:

In the following figure, ABCD, DCFE and ABFE are parallelograms. Show that ar (ADE) = ar (BCF).

It is given that ABCD is a parallelogram. We know that opposite sides of a parallelogram are equal.

∴ AD = BC … (1)

Similarly, for parallelograms DCEF and ABFE, it can be proved that

DE = CF … (2)

And, EA = FB … (3)

AD = BC [Using equation (1)]

DE = CF [Using equation (2)]

EA = FB [Using equation (3)]

∴ ΔADE ≅ BCF (SSS congruence rule)

⇒ Area (ΔADE) = Area (ΔBCF)

#### Question 4:

In the following figure, ABCD is parallelogram and BC is produced to a point Q such that AD = CQ. If AQ intersect DC at P, show that

ar (BPC) = ar (DPQ).

[Hint: Join AC.]

It is given that ABCD is a parallelogram.

AD || BC and AB || DC(Opposite sides of a parallelogram are parallel to each other)

Join point A to point C.

Consider ΔAPC and ΔBPC

ΔAPC and ΔBPC are lying on the same base PC and between the same parallels PC and AB. Therefore,

Area (ΔAPC) = Area (ΔBPC) … (1)

In quadrilateral ACDQ, it is given that

Since ABCD is a parallelogram,

AD || BC (Opposite sides of a parallelogram are parallel)

CQ is a line segment which is obtained when line segment BC is produced.

We have,

AC = DQ and AC || DQ

Hence, ACQD is a parallelogram.

Consider ΔDCQ and ΔACQ

These are on the same base CQ and between the same parallels CQ and AD. Therefore,

Area (ΔDCQ) = Area (ΔACQ)

⇒ Area (ΔDCQ) − Area (ΔPQC) = Area (ΔACQ) − Area (ΔPQC)

⇒ Area (ΔDPQ) = Area (ΔAPC) … (2)

From equations (1) and (2), we obtain

Area (ΔBPC) = Area (ΔDPQ)

#### Question 5:

In the following figure, ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, show that

(i)

(ii)

(iii)

(iv)

(v)

(vi)

[Hint: Join EC and AD. Show that BE || AC and DE || AB, etc.]

(i) Let G and H be the mid-points of side AB and AC respectively.

Line segment GH is joining the mid-points. Therefore, it will be parallel to third side BC and also its length will be half of the length of BC (mid-point theorem).

⇒ GH = BC and GH || BD

⇒ GH = BD = DC and GH || BD (D is the mid-point of BC)

GH ||BD and GH = BD

Two line segments joining two parallel line segments of equal length will also be equal and parallel to each other.

Therefore, BG = DH and BG || DH

Hence, quadrilateral GHDB is a parallelogram.

We know that in a parallelogram, the diagonal bisects it into two triangles of equal area.

Hence, Area (ΔBDG) = Area (ΔHGD)

Similarly, it can be proved that quadrilaterals DCHG, GDHA, and BEDG are parallelograms and their respective diagonals are dividing them into two triangles of equal area.

ar (ΔGDH) = ar (ΔCHD) (For parallelogram DCHG)

ar (ΔGDH) = ar (ΔHAG) (For parallelogram GDHA)

ar (ΔBDE) = ar (ΔDBG) (For parallelogram BEDG)

ar (ΔABC) = ar(ΔBDG) + ar(ΔGDH) + ar(ΔDCH) + ar(ΔAGH)

ar (ΔABC) = 4 × ar(ΔBDE)

Hence,

(ii)Area (ΔBDE) = Area (ΔAED) (Common base DE and DE||AB)

Area (ΔBDE) − Area (ΔFED) = Area (ΔAED) − Area (ΔFED)

Area (ΔBEF) = Area (ΔAFD) (1)

Area (ΔABD) = Area (ΔABF) + Area (ΔAFD)

Area (ΔABD) = Area (ΔABF) + Area (ΔBEF) [From equation (1)]

Area (ΔABD) = Area (ΔABE) (2)

AD is the median in ΔABC.

From (2) and (3), we obtain

2 ar (ΔBDE) = ar (ΔABE)

Or,

(iii)

ar (ΔABE) = ar (ΔBEC) (Common base BE and BE||AC)

ar (ΔABF) + ar (ΔBEF) = ar (ΔBEC)

Using equation (1), we obtain

ar (ΔABF) + ar (ΔAFD) = ar (ΔBEC)

ar (ΔABD) = ar (ΔBEC)

ar (ΔABC) = 2 ar (ΔBEC)

(iv)It is seen that ΔBDE and ar ΔAED lie on the same base (DE) and between the parallels DE and AB.

∴ar (ΔBDE) = ar (ΔAED)

⇒ ar (ΔBDE) − ar (ΔFED) = ar (ΔAED) − ar (ΔFED)

∴ar (ΔBFE) = ar (ΔAFD)

(v)Let h be the height of vertex E, corresponding to the side BD in ΔBDE.

Let H be the height of vertex A, corresponding to the side BC in ΔABC.

In (i), it was shown that

In (iv), it was shown that ar (ΔBFE) = ar (ΔAFD).

∴ ar (ΔBFE) = ar (ΔAFD)

= 2 ar (ΔFED)

Hence,

(vi) Area (AFC) = area (AFD) + area (ADC)

Now, by (v), … (6)

Therefore, from equations (5), (6), and (7), we get:

#### Question 6:

Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that

[Hint: From A and C, draw perpendiculars to BD]

Let us draw AM ⊥ BD and CN ⊥ BD

Area of a triangle

∴ ar (APB) × ar (CPD) = ar (APD) × ar (BPC)

#### Question 7:

P and Q are respectively the mid-points of sides AB and BC of a triangle ABC and R is the mid-point of AP, show that

(i)  (ii)

(iii)

Take a point S on AC such that S is the mid-point of AC.

Extend PQ to T such that PQ = QT.

Join TC, QS, PS, and AQ.

In ΔABC, P and Q are the mid-points of AB and BC respectively. Hence, by using mid-point theorem, we obtain

PQ || AC and PQ

⇒ PQ || AS and PQ = AS (As S is the mid-point of AC)

∴ PQSA is a parallelogram. We know that diagonals of a parallelogram bisect it into equal areas of triangles.

∴ ar (ΔPAS) = ar (ΔSQP) = ar (ΔPAQ) = ar (ΔSQA)

Similarly, it can also be proved that quadrilaterals PSCQ, QSCT, and PSQB are also parallelograms and therefore,

ar (ΔPSQ) = ar (ΔCQS) (For parallelogram PSCQ)

ar (ΔQSC) = ar (ΔCTQ) (For parallelogram QSCT)

ar (ΔPSQ) = ar (ΔQBP) (For parallelogram PSQB)

Thus,

ar (ΔPAS) = ar (ΔSQP) = ar (ΔPAQ) = ar (ΔSQA) = ar (ΔQSC) = ar (ΔCTQ) = ar (ΔQBP) … (1)

Also, ar (ΔABC) = ar (ΔPBQ) + ar (ΔPAS) + ar (ΔPQS) + ar (ΔQSC)

ar (ΔABC) = ar (ΔPBQ) + ar (ΔPBQ) + ar (ΔPBQ) + ar (ΔPBQ)

= ar (ΔPBQ) + ar (ΔPBQ) + ar (ΔPBQ) + ar (ΔPBQ)

= 4 ar (ΔPBQ)

⇒ ar (ΔPBQ) =  ar (ΔABC) … (2)

(i)Join point P to C.

In ΔPAQ, QR is the median.

… (3)

In ΔABC, P and Q are the mid-points of AB and BC respectively. Hence, by using mid-point theorem, we obtain

PQ

Also, PQ || AC  PT || AC

Hence, PACT is a parallelogram.

ar (PACT) = ar (PACQ) + ar (ΔQTC)

= ar (PACQ) + ar (ΔPBQ [Using equation (1)]

∴ ar (PACT) = ar (ΔABC) … (4)

(ii)

(iii)In parallelogram PACT,

#### Question 8:

In the following figure, ABC is a right triangle right angled at A. BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively. Line segment AX ⊥ DE meets BC at Y. Show that:

(i) ΔMBC ≅ ΔABD

(ii)

(iii)

(iv) ΔFCB ≅ ΔACE

(v)

(vi)

(vii)

Note: Result (vii) is the famous Theorem of Pythagoras. You shall learn a simpler proof of this theorem in class X.

(i) We know that each angle of a square is 90°.

Hence, ∠ABM = ∠DBC = 90º

⇒ ∠ABM + ∠ABC = ∠DBC + ∠ABC

⇒ ∠MBC = ∠ABD

In ΔMBC and ΔABD,

∠MBC = ∠ABD (Proved above)

MB = AB (Sides of square ABMN)

BC = BD (Sides of square BCED)

∴ ΔMBC ≅ ΔABD (SAS congruence rule)

(ii) We have

ΔMBC ≅ ΔABD

⇒ ar (ΔMBC) = ar (ΔABD) … (1)

It is given that AX ⊥ DE and BD ⊥ DE (Adjacent sides of square

BDEC)

⇒ BD || AX (Two lines perpendicular to same line are parallel to each other)

ΔABD and parallelogram BYXD are on the same base BD and between the same parallels BD and AX.

Area (BYXD) = 2 area (ΔMBC) [Using equation (1)] … (2)

(iii) ΔMBC and parallelogram ABMN are lying on the same base MB and between same parallels MB and NC.

2 ar (ΔMBC) = ar (ABMN)

ar (BYXD) = ar (ABMN) [Using equation (2)] … (3)

(iv) We know that each angle of a square is 90°.

∴ ∠FCA = ∠BCE = 90º

⇒ ∠FCA + ∠ACB = ∠BCE + ∠ACB

⇒ ∠FCB = ∠ACE

In ΔFCB and ΔACE,

∠FCB = ∠ACE

FC = AC (Sides of square ACFG)

CB = CE (Sides of square BCED)

ΔFCB ≅ ΔACE (SAS congruence rule)

(v) It is given that AX ⊥ DE and CE ⊥ DE (Adjacent sides of square BDEC)

Hence, CE || AX (Two lines perpendicular to the same line are parallel to each other)

Consider ΔACE and parallelogram CYXE

ΔACE and parallelogram CYXE are on the same base CE and between the same parallels CE and AX.

⇒ ar (CYXE) = 2 ar (ΔACE) … (4)

∴ ΔFCB ≅ ΔACE

ar (ΔFCB) ≅ ar (ΔACE) … (5)

On comparing equations (4) and (5), we obtain

ar (CYXE) = 2 ar (ΔFCB) … (6)

(vi) Consider ΔFCB and parallelogram ACFG

ΔFCB and parallelogram ACFG are lying on the same base CF and between the same parallels CF and BG.

⇒ ar (ACFG) = 2 ar (ΔFCB)

⇒ ar (ACFG) = ar (CYXE) [Using equation (6)] … (7)

(vii) From the figure, it is evident that

ar (BCED) = ar (BYXD) + ar (CYXE)

⇒ ar (BCED) = ar (ABMN) + ar (ACFG) [Using equations (3) and (7)]