**NCERT Solutions for Class 9
Maths Chapter 9 Areas of Parallelograms **and Triangles** Ex 9.4**

NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.4 are the part of NCERT Solutions for Class 9 Maths. Here you can find the NCERT Solutions for Chapter 9 Areas of Parallelograms and Triangles Ex 9.4.

**Ex 9.4 Class 9 Maths 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.

**Solution:****
**We have a parallelogram ABCD and
rectangle ABEF such that

ar (||

^{gm}ABCD) = ar (rect. ABEF)

AB = EF [Opposite sides of a rectangle]

⇒ CD = EF

⇒ AB + CD = AB + EF …..…(1)

BE < BC and AF < AD

[In
a right-angled triangle, hypotenuse is the longest side]

⇒ (BC + AD) > (BE + AF) …..…(2)

From eqs. (1) and (2), we have

(AB + CD) + (BC + AD) > (AB + EF) + (BE + AF)

⇒ (AB + BC + CD + DA) > (AB + BE
+ EF + FA)

⇒ Perimeter of parallelogram ABCD
> Perimeter of rectangle ABEF

**Ex 9.4 Class 9 Maths Question 2.****
**In figure, D and E are two points
on BC such that BD = DE = EC. Show that ar (ABD) = ar (ADE) = ar (AEC).

**Solution:**

Let us draw AF, perpendicular to BC such that AF is the height of ∆ABD, ∆ADE and ∆AEC.

**Ex 9.4 Class 9 Maths Question 3.
**In figure, ABCD, DCFE and ABFE are parallelograms.
Show that ar (ADE) = ar (BCF).

**Solution:**

Since, ABCD is a parallelogram. [Given]

i.e., AD = BC …..…(1)

Now, ∆ADE and ∆BCF are on equal bases AD = BC [from (1)] and between the same parallels AB and EF.

So, ar (∆ADE) = ar (∆BCF).

**Ex 9.4 Class 9 Maths Question 4.
**In figure, ABCD is a 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.]

**Solution:****
**We have a parallelogram ABCD and AD
= CQ. Let us join AC.

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

Since, ∆QAC and ∆QDC are on the same base QC and between the same parallels AD and BQ.

∴ ar (∆QAC) = ar (∆QDC)

Subtracting ar (∆QPC) from both sides, we have

ar (∆QAC) – ar (∆QPC) = ar (∆QDC) – ar (∆QPC)

⇒ ar (∆PAC) = ar (∆QDP) …..…(1)

Since, ∆PAC and ∆PBC are on the same base PC and between the same parallels AB and CD.

∴ ar (∆PAC) = ar (∆PBC) …..…(2)

From eq. (1) and (2), we get

ar (∆PBC) = ar (∆QDP)

Or ar
(BPC) = ar (DPQ) Hence proved.

**Ex 9.4 Class 9 Maths Question 5.****
**In 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

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

**Solution:****
**Let us join EC and AD. Draw EP ⊥ BC.

Let AB = BC = CA = a, then

BD = a/2 = DE = BE

**(ii)** Since, ∆ABC and ∆BED are equilateral triangles.

⇒ ∠ACB = ∠DBE = 60°

⇒ BE || AC

∆BAE and ∆BEC are on the same base BE and between the same parallels BE and AC.

ar (∆BAE) = ar (∆BEC)

⇒ ar (∆BAE) = 2 ar (∆BDE) [DE is median of ∆EBC. ∴ ar (∆BEC) = 2 ar (∆BDE)]

⇒ ar (∆BDE) = ½ ar (∆BAE)

**(iii)**
ar (∆ABC) = 4 ar (∆BDE)
…….(1) [Proved in
part (i)]

ar (∆BEC) = 2 ar (∆BDE)
……..(2) [∵ DE is median of
∆BEC]

⇒ ar (∆ABC) = 2 ar (∆BEC) [From (1) and (2)]

**(iv)**
Since, ∆ABC and ∆BDE are equilateral triangles.

⇒ ∠ABC = ∠BDE = 60°

⇒ AB || DE

∆BED and ∆AED are on the same base ED and between the same parallels AB and DE.

∴ ar (∆BED) = ar (∆AED)

Subtracting ar (∆EFD) from both sides, we get

⇒ ar (∆BED) – ar (∆EFD) = ar (∆AED) – ar (∆EFD)

⇒ ar (∆BEE) = ar (∆AFD)

**(v)**
In right-angled ∆ABD, we get

From (1) and (2), we get

ar (∆AFD) = 2 ar (∆EFD)

Ar (∆AFD) = ar (∆BEF) [From part (iv)]

⇒ ar (∆BFE) = 2 ar (∆EFD)

**(vi)** ar (∆AFC) = ar (∆AFD) + ar (∆ADC)

= ar (∆BFE)
+ ½ ar (∆ABC) [From part (iv)]

= ar (∆BFE)
+ ½ × 4 × ar (∆BDE) [From part (i)]

= ar (∆BFE) + 2 ar
(∆BDE)

= 2 ar (∆FED) +
2[ar (∆BFE) + ar (∆FED)]

= 2 ar (∆FED) +
2[2 ar (∆FED) + ar (∆FED)] [From
part (v)]

= 2 ar (∆FED) +
2[3 ar (∆FED)]

= 2 ar (∆FED) + 6
ar (∆FED)

= 8 ar (∆FED)

∴ ar (∆FED) = 1/8 ar (∆AFC)

**Ex 9.4 Class 9 Maths Question 6.
**Diagonals AC and BD of a quadrilateral ABCD
intersect each other at P. Show that

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

[

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

**Solution:****
**We have a quadrilateral ABCD such
that its diagonals AC and BD intersect at P.

Let us draw AM ⊥ BD and CN ⊥ BD.

**Ex 9.4 Class 9 Maths 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

We have a ∆ABC such that P is the mid-point of AB and Q is the mid-point of BC.

Also, R is the mid-point of AP. Let us join AQ, RQ, PC and RC.

**(i)** In ∆APQ, R is the mid-point of AP. [Given]

⇒ ar (∆PRQ) = ½ ar (∆APQ) …..…(1)

In ∆ABQ, P is the mid-point of AB.

∴ QP is a median of ∆ABQ.

**Ex 9.4 Class 9 Maths Question 8.
**In 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)**∆MBC = ∆ABD

**(ii)**ar(BYXD) = 2 ar(MBC)

**(iii)**ar(BYXD) = ax(ABMN)

**(iv)**∆FCB ≅ ∆ACE

**(v)**ar(CYXE) = 2 ar(FCB)

**(vi)**ar(CYXE) = ax(ACFG)

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

**Solution:****
**We have a right ∆ABC such that
BCED, ACFG and ABMN are squares on its sides BC, CA and AB respectively. Line
segment AX 1 DE is also drawn such that it meets BC at Y.

**(i)**
∠CBD = ∠MBA [Each
90°]

∴ ∠CBD + ∠ABC = ∠MBA + ∠ABC

(By adding ∠ABC on both sides)

or ∠ABD = ∠MBC

In ∆ABD and ∆MBC, we have

AB = MB [Sides of
a square]

BD = BC

∠ABD = ∠MBC [Proved above]

∴ ∆ABD ≅ ∆MBC [By SAS congruency rule]

**(ii)** Since parallelogram BYXD and ∆ABD are on the same base BD and
between the same parallels BD and AX.

∴ ar (∆ABD) = ½ ar (||^{gm} BYXD)

But ∆ABD ≅ ∆MBC [From part (i)]

Since, congruent triangles have equal areas.

∴ ar (∆MBC) = ½ ar (||^{gm} BYXD)

⇒ ar (||^{gm} BYXD) = 2 ar (∆MBC)

**(iii)**
Since, ar (||^{gm} BYXD) = 2 ar (∆MBC) …..…(1) [From part (ii)]

and ar (square ABMN) = 2 ar (∆MBC) …..…(2)

[ABMN and AMBC are on the same base MB and between the same parallels MB and
NC]

From (1) and (2), we have

ar (BYXD) = ar (ABMN) .

**(iv)**
∠FCA = ∠BCE [Each 90°]

or ∠FCA + ∠ACB = ∠BCE + ∠ACB

[By adding ∠ACB on both sides]

⇒ ∠FCB = ∠ACE

In ∆FCB and ∆ACE, we have

FC = AC [Sides of a square]

CB = CE [Sides
of a square]

∠FCB = ∠ACE [Proved above]

⇒ ∆FCB ≅ ∆ACE [By SAS congruency
rule]

**(v) **Since,
||^{gm} CYXE and ∆ACE are on the same base CE and between the same
parallels CE and AX.

∴ ar (||^{gm} CYXE) = 2 ar (∆ACE)

But ∆ACE ≅ ∆FCB [From part (iv)]

Since, congruent triangles are equal in areas.

∴ ar (||^{gm} CYXE) = 2 ar (∆FCB)

**(vi)**
Since, ar (||^{gm} CYXE) = 2 a r(∆FCB) ………(3) [From part (v)]

Also (quad. ACFG) and ∆FCB are on the same base FC and between the same
parallels FC and BG.

⇒ ar (quad. ACFG) = 2 ar (∆FCB) ………(4)

From (3) and (4), we get

ar (quad. CYXE) = ar (quad. ACFG) …..…(5)

**(vii)** We have ar (quad. BCED) = ar (quad. CYXE) + ar (quad. BYXD)

= ar (quad. CYXE) + ar (quad. ABMN)

[From part (iii)]

Thus, ar (quad. BCED) = ar (quad. ABMN) + ar (quad. ACFG)

[From
part (vi)]

**NCERT Solutions for Maths Class 10**