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
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.
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
(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)]
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