NCERT Solutions Class 9th Maths Chapter 9 Areas of Parallelograms and Triangles Exercise 9.4

NCERT Solutions Class 9th Maths Chapter 9 Areas of Parallelograms and Triangles Exercise 9.4

NCERT Solutions Class 9th Maths Chapter 9 Areas of Parallelograms and Triangles Exercise 9.4

?Chapter – 9?

✍Areas of Parallelograms and Triangles✍

?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.
Solution: We have a parallelogram ABCD and rectangle ABEF such that
ar(||gm ABCD) = ar( rect. ABEF)

AB = CD [Opposite sides of parallelogram]
and AB = EF [Opposite sides of a rectangle]
⇒ CD = EF
⇒ AB + CD = AB + EF … (1)
BE < BC and AF < AD [In a right triangle, hypotenuse is the longest side] ⇒ (BC + AD) > (BE + AF) …(2)
From (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.

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.

Question 3. In figure, ABCD, DCFE and ABFE are parallelograms. Show that ar(ADE) = ax(BCF).

Solution:
Since, ABCD is a parallelogram [Given]
∴ Its opposite sides are parallel and equal.
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).

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) = ax(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(∆QAQ – ar(∆QPC) = ar(∆QDC) – ar(∆QPC)
⇒ ar(∆PAQ = 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 (1) and (2), we get
ar(∆PBC) = ar(∆QDP)

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 = a2 = 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) = || ar(∆BDE)]
⇒ ar(ABDE) = 1/2ar(∆BAE)

(iii) ar(∆ABC) = 4 ar(∆BDE)[Proved in (i) part]
ar(∆BEC) = 2 ar(∆BDE)
[ ∵ DE is median of ∆BEC]
⇒ ar(∆ABC) = 2 ar(∆BEC)

(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(AEFD) 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 (iv) part]
⇒ ar(∆BFE) = 2 ar(∆EFD)

(vi) ar(∆AFC) = ar(∆AFD) + ar(∆ADC)
= ar(∆BFE) + 1/2 ar(∆ABC) [From (iv) part]
= ar(∆BFE) + 1/2 x 4 x ar(∆BDE) [From (i) part]
= ar(∆BFE) + 2ar(∆BDE)
= 2ar(∆FED) + 2[ar(∆BFE) + ar(∆FED)]
= 2ar(∆FED) + 2[2ar(∆FED) + ar(∆FED)] [From (v) part]
= 2ar(∆FED) + 2[3ar(∆FED)]
= 2ar(∆FED) + 6ar(∆FED)
= 8ar(∆FED)
∴ ar(∆FED) = 1/8 ar(∆AFC)

Question 6. Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that
ar(APB) x ar(CPD) = ar(APD) x 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.

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

Solution:
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 PC.

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

∴RQ is a median of ∆APQ.
⇒ ar(∆PRQ) = 1/2ar(∆APQ) …(1)
In ∆ABQ, P is the mid-point of AB.
∴ QP is a median of ∆ABQ.
∴ ar(∆APQ) = 1/2ar(∆ABQ) …(2)

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 [Each90°]
∴ ∠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]

(ii) Since parallelogram BYXD and ∆ABD are on the same base BD and between the same parallels BD and AX.
∴ ar(∆ABD) = 1/2ar(||^gm BYXD)
But ∆ABD ≅ ∆MBC [From (i) part]
Since, congruent triangles have equal
areas.
∴ ar(∆MBC) = 1/2ar(||^gm BYXD)
⇒ ar(||^gm BYXD) = 2ar(∆MBC)

(iii) Since, ar(||^gm BYXD) = 2ar(∆MBC) …(1) [From (ii) part]
and or(square ABMN) = 2or(∆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]

(v) Since, ||^gm CYXE and ∆ACE are on the same base CE and between the same parallels CE and AX.
∴ ar(||^gm CYXE) = 2ar(∆ACE)
But ∆ACE ≅ ∆FCB [From (iv) part]
Since, congruent triangles are equal in areas.
∴ ar (||<^gm CYXE) = 2ar(∆FCB)

(vi) Since, ar(||^gm CYXE) = 2ar(∆FCB) …(3)
[From (v) part]
Also (quad. ACFG) and ∆FCB are on the same base FC and between the same parallels FC and BG.
⇒ ar(quad. ACFG) = 2ar(∆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 (iii) part]
Thus, ar (quad. BCED)
= ar(quad. ABMN) + ar(quad. ACFG)
[From (vi) part]