NCERT Solutions for Exercise 9.4 Class 9 Maths Chapter 9 - Areas Of Parallelograms And Triangles

Q1 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.

Answer:

We have ||gm ABCD and a rectangle ABEF both lie on the same base AB such that, ar(||gm ABCD) = ar(ABEF)
for rectangle, AB = EF
and for ||gm AB = CD
$\Rightarrow$ CD = EF
$\Rightarrow$ AB + CD = AB + EF ...........(i)

SInce $\Delta$ BEC and $\Delta$ AFD are right angled triangle
Therefore, AD > AF and BC > BE
$\Rightarrow$ (BC + AD ) > (AF + BE)...........(ii)

Adding equation (i) and (ii), we get

(AB + CD)+(BC + AD) > (AB + BE) + (AF + BE)
$\Rightarrow$ (AB + BC + CD + DA) > (AB + BE + EF + FA)
Hence proved, perimeter of ||gm ABCD is greater than perimeter of rectangle ABEF.

Q2 In Fig. $\small 9.30$ , D and E are two points on BC such that $\small BD=DE=EC$ . Show that $\small ar(ABD)=ar(ADE)=ar(AEC)$ .

Answer:

In $\Delta$ ABC, D and E are two points on BC such that BD = DE = EC

AD is the median of ABE, therefore,

area of ABD= area of AED...................(1)

AE is the median of ACD, therefore,

area of AEC= area of AED...................(2)

From (1) and (2)

area of ABD=area of AED= area of AEC
Hence proved.

Q3 In Fig. $\small 9.31$ , ABCD, DCFE and ABFE are parallelograms. Show that $\small ar(ADE)=ar(BCF)$ .

Answer:

Given,

ABCD is a parallelogram. Therefore, AB = CD & AD = BC & AB || CD .......(i)

Now, $\Delta$ ADE and $\Delta$ BCF are on the same base AD = BC and between same parallels AB and EF.
Therefore, ar ( $\Delta$ ADE) = ar( $\Delta$ BCF)

Hence proved.

Q4 In Fig. $\small 9.32$ , ABCD is a parallelogram and BC is produced to a point Q such that $\small AD=CQ$ . If AQ intersects DC at P, show that $\small ar (BPC) = ar (DPQ)$ . [ Hint : Join AC.]

Answer:

Given,
ABCD is a ||gm and AD = CQ. Join AC.
Since $\Delta$ DQC and $\Delta$ ACD lie on the same base QC and between same parallels AD and QC.
Therefore, ar( $\Delta$ DQC) = ar( $\Delta$ ACD).......(i)

Subtracting ar( $\Delta$ PQC) from both sides in eq (i), we get
ar( $\Delta$ DPQ) = ar( $\Delta$ PAC).............(i)

Since $\Delta$ PAC and $\Delta$ PBC are on the same base PC and between same parallel PC and AB.
Therefore, ar( $\Delta$ PAC) = ar( $\Delta$ PBC)..............(iii)

From equation (ii) and eq (ii), we get

$\small ar (BPC) = ar (DPQ)$

Hence proved.

Q5 (i) In Fig. $\small 9.33$ , ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, show that

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

[ Hint: Join EC and AD. Show that $\small BE\parallel AC$ and $\small DE\parallel AB$ , etc.]

Answer:

Let join the CE and AD and draw $EP \perp BC$ . It is given that $\Delta$ ABC and $\Delta$ BDE is an equilateral triangle.
So, AB =BC = CA = $a$ and D i sthe midpoint of BC
therefore, $BD = \frac{a}{2}= DE = BE$
(i) Area of $\Delta$ ABC = $\frac{\sqrt{3}}{4}a^2$ and
Area of $\Delta$ BDE = $\frac{\sqrt{3}}{4}(\frac{a}{2})^2= \frac{\sqrt{3}}{16}a^2$

Hence, $\small ar(BDE)=\frac{1}{4}ar(ABC)$

Q5 (ii) In Fig. $\small 9.33$ , ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, show that
$\small ar(BDE)=\frac{1}{2}ar(BAE)$

[ Hint: Join EC and AD. Show that $\small BE\parallel AC$ and $\small DE\parallel AB$ , etc.]

Answer:

Since $\Delta$ ABC and $\Delta$ BDE are equilateral triangles.
Therefore, $\angle$ ACB = $\angle$ DBE = $60^0$
$\Rightarrow$ BE || AC

$\Delta$ BAE and $\Delta$ BEC are on the same base BE and between same parallels BE and AC.
Therefore, ar ( $\Delta$ BAE) = ar( $\Delta$ BEC)
$\Rightarrow$ ar( $\Delta$ BAE) = 2 ar( $\Delta$ BED) [since D is the meian of $\Delta$ BEC ]
$\Rightarrow$ $\small ar(BDE)=\frac{1}{2}ar(BAE)$
Hence proved.

Q5 (iii) In Fig. $\small 9.33$ , ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, show that
$\small ar(ABC)=2ar(BEC)$

[ Hint : Join EC and AD. Show that $\small BE\parallel AC$ and $\small DE\parallel AB$ , etc.]

Answer:

We already proved that,
ar( $\Delta$ ABC) = 4.ar( $\Delta$ BDE) (in part 1)
and, ar( $\Delta$ BEC) = 2. ar( $\Delta$ BDE) (in part ii )

$\Rightarrow$ ar( $\Delta$ ABC) = 2. ar( $\Delta$ BEC)

Hence proved.

Q5 (iv) In Fig.9.33, ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, show that
$\small ar(BFE)=ar(AFD)$
[ Hint: Join EC and AD. Show that $\small BE\parallel AC$ and $\small DE\parallel AB$ , etc.]

Answer:

Since $\Delta$ ABC and $\Delta$ BDE are equilateral triangles.
Therefore, $\angle$ ACB = $\angle$ DBE = $60^0$
$\Rightarrow$ BE || AC

$\Delta$ BDE and $\Delta$ AED are on the same base ED and between same parallels AB and DE.
Therefore, ar( $\Delta$ BED) = ar( $\Delta$ AED)

On subtracting $\Delta$ EFD from both sides we get

$\small ar(BFE)=ar(AFD)$

Hence proved.

Q5 (v) In Fig. $\small 9.33$ , ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, show that
$\small ar(BFE)=2ar(FED)$

[ Hint : Join EC and AD. Show that $\small BE\parallel AC$ and $\small DE\parallel AB$ , etc.]

Answer:

In right angled triangle $\Delta$ ABD, we get

$\\AB^2 = AD^2+BD^2\\ AD^2 = AB^2-BD^2$
$=a^2 - \frac{a^2}{4}=\frac{3a^2}{4}$
$AD=\frac{\sqrt{3}a}{2}$

So, in $\Delta$ PED,
$PE^2 = DE^2-DP^2$
$\\=(\frac{a}{2})^2-(\frac{a}{4})^2\\ =\frac{3a^2}{16}$
So, $PE=\frac{\sqrt{3}a}{4}$

Therefore, the Area of $\Delta AFD =1/2 (FD)\frac{\sqrt{3}a}{2}$ ..........(i)

And, Area of triangle $\Delta EFD =1/2 (FD)\frac{\sqrt{3}a}{4}$ ...........(ii)

From eq (i) and eq (ii), we get
ar( $\Delta$ AFD) = 2. ar( $\Delta$ EFD)
Since ar( $\Delta$ AFD) = ar( $\Delta$ BEF)

$\Rightarrow$ $\small ar(BFE)=2ar(FED)$

Q5 (vi) In Fig. $9.33$ , ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, show that
$ar (FED) = \frac{1}{8} ar (AFC)$

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

Answer:

$ar(\Delta AFC) =$
$\\= ar(\Delta AFD) + ar(\Delta ADC)\\ = ar(\Delta BFE) + 1/2. ar(\Delta ABC)\\ = ar(\Delta BFE) + 1/2\times [4\times ar(\Delta BDE)]\\ = ar(\Delta BFE) + 2\times ar(\Delta BDE)$
$= 2\times ar (\Delta FED) + 2\times [ar(\Delta BFE) + ar(\Delta FED)]$ .....(from part (v) ar( $\Delta$ BFE) = 2. ar( $\Delta$ FED) ]
$\\=2ar(\Delta FED) + 2[3\times ar(\Delta FED)]\\ =8\times ar(\Delta FED)$

$ar (FED) = \frac{1}{8} ar (AFC)$
Hence proved.

Q6 Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that $ar (APB) \times ar (CPD) = ar (APD) \times ar (BPC).$

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

Answer:

Given that,
A quadrilateral ABCD such that it's diagonal AC and BD intersect at P. Draw $AM \perp BD$ and $CN \perp BD$

Now, ar( $\Delta$ APB) = $\frac{1}{2}\times BP\times AM$ and,
$ar(\Delta CDP)=\frac{1}{2}\times DP\times CN$
Therefore, ar( $\Delta$ APB) $\times$ ar( $\Delta$ CDP)=
$\\=(\frac{1}{2}\times BP\times AM).(\frac{1}{2}\times DP \times CN)\\ =\frac{1}{4}\times BP \times DP\times AM\times CN$ ....................(i)

Similarly, ar( $\Delta$ APD) $\times$ ar ( $\Delta$ BPC) =
$=\frac{1}{4}\times BP \times DP\times AM\times CN$ ................(ii)

From eq (i) and eq (ii), we get

$ar (APB) \times ar (CPD) = ar (APD) \times ar (BPC).$
Hence proved.

Q7 (i) 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

$\small ar(PRQ)=\frac{1}{2}ar(ARC)$

Answer:

We have $\Delta$ ABC such that P, Q and R are the midpoints of the side AB, BC and AP respectively. Join PQ, QR, AQ, PC and RC as shown in the figure.
Now, in $\Delta$ APC,
Since R is the midpoint. So, RC is the median of the $\Delta$ APC
Therefore, ar( $\Delta$ ARC) = 1/2 . ar ( $\Delta$ APC)............(i)

Also, in $\Delta$ ABC, P is the midpoint. Thus CP is the median.
Therefore, ar( $\Delta$ APC) = 1/2. ar ( $\Delta$ ABC)............(ii)

Also, AQ is the median of $\Delta$ ABC
Therefore, 1/2. ar ( $\Delta$ ABC) = ar (ABQ)............(iii)

In $\Delta$ APQ, RQ is the median.
Therefore, ar ( $\Delta$ PRQ) = 1/2 .ar ( $\Delta$ APQ).............(iv)

In $\Delta$ ABQ, PQ is the median
Therefore, ar( $\Delta$ APQ) = 1/2. ar( $\Delta$ ABQ).........(v)

From eq (i),
$\frac{1}{2}ar(\Delta ARC)= \frac{1}{4}ar(\Delta APC)$ ...........(vi)
Now, put the value of ar( $\Delta$ APC) from eq (ii), we get
Taking RHS;
$= \frac{1}{8}[ar(\Delta ABC)]$
$= \frac{1}{4}[ar(\Delta ABQ)]$ (from equation (iii))

$= \frac{1}{2}[ar(\Delta APQ)]$ (from equation (v))

$=ar(\Delta PRQ)$ (from equation (iv))

$\Rightarrow$ $\small ar(PRQ)=\frac{1}{2}ar(ARC)$

Hence proved.

Q7 (ii) 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
$\small ar(RQC)=\frac{3}{8}ar(ABC)$

Answer:

In $\Delta$ RBC, RQ is the median
Therefore ar( $\Delta$ RQC) = ar( $\Delta$ RBQ)
= ar (PRQ) + ar (BPQ)
= 1/8 (ar $\Delta$ ABC) + ar( $\Delta$ BPQ) [from eq (vi) & eq (A) in part (i)]
= 1/8 (ar $\Delta$ ABC) + 1/2 (ar $\Delta$ PBC) [ since PQ is the median of $\Delta$ BPC]
= 1/8 (ar $\Delta$ ABC) + (1/2).(1/2)(ar $\Delta$ ABC) [CP is the medain of $\Delta$ ABC]
= 3/8 (ar $\Delta$ ABC)

Hence proved.

Q7 (iii) 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

$ar (PBQ) = ar (ARC)$

Answer:

QP is the median of $\Delta$ ABQ
Therefore, ar( $\Delta$ PBQ) = 1/2. (ar $\Delta$ ABQ)

= (1/2). (1/2) (ar $\Delta$ ABC) [since AQ is the median of $\Delta$ ABC
= 1/4 (ar $\Delta$ ABC)
= ar ( $\Delta$ ARC) [from eq (A) of part (i)]

Hence proved.

Q8 (i) In Fig. $\small 9.34$ , 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 $\small AX\perp DE$ meets BC at Y. Show that: $\small \Delta MBC\cong \Delta ABD$

Answer:

We have, a $\Delta$ ABC such that BCED, ACFG and ABMN are squares on its side BC, CA and AB respec. Line segment $\small AX\perp DE$ meets BC at Y

(i) $\angle CBD = \angle MBA$ [each 90]
Adding $\angle ABC$ on both sides, we get
$\angle ABC + \angle CBD = \angle MBA + \angle ABC$
$\Rightarrow \angle ABD = \angle MBC$
In $\Delta$ ABD and $\Delta$ MBC, we have
AB = MB
BD = BC
$\angle ABD = \angle MBC$
Therefore, By SAS congruency
$\Delta$ ABD $\cong$ $\Delta$ MBC

Hence proved.

Q8 (ii) In Fig. $\small 9.34$ , 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 $\small AX\perp DE$ meets BC at Y. Show that: $\small ar(BYXD)=2ar(MBC)$

Answer:

SInce ||gm BYXD and $\Delta$ ABD are on the same base BD and between same parallels BD and AX
Therefore, ar( $\Delta$ ABD) = 1/2. ar(||gm BYXD)..........(i)

But, $\Delta$ ABD $\cong$ $\Delta$ MBC (proved in 1st part)
Since congruent triangles have equal areas.
Therefore, By using equation (i) we get
$\small ar(BYXD)=2ar(MBC)$
Hence proved.

Q8 (iii) In Fig. $\small 9.34$ , 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 $\small AX\perp DE$ meets BC at Y. Show that: $ar(BYXD)=ar(ABMN)$

Answer:

Since, ar (||gm BYXD) = 2 .ar ( $\Delta$ MBC) ..........(i) [already proved in 2nd part]
and, ar (sq. ABMN) = 2. ar ( $\Delta$ MBC)............(ii)
[Since ABMN and AMBC are on the same base MB and between same parallels MB and NC]
From eq(i) and eq (ii), we get

$\small ar(BYXD)=ar(ABMN)$

Q8 (iv) In Fig. $\small 9.34$ , 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 $\small AX\perp DE$ meets BC at Y. Show that: $\small \Delta FCB\cong \Delta ACE$

Answer:

$\angle FCA = \angle BCE$ [both 90]
By adding $\angle$ ABC on both sides we get
$\angle$ ABC + $\angle$ FCA = $\angle$ ABC + $\angle$ BCE
$\angle$ FCB = $\angle$ ACE

In $\Delta$ FCB and $\Delta$ ACE
FC = AC [sides of square]
BC = AC [sides of square]
$\angle$ FCB = $\angle$ ACE
$\Rightarrow$$\Delta$ FCB $\cong$ $\Delta$ ACE

Hence proved.

Q8 (v) In Fig. $\small 9.34$ , 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 $\small AX\perp DE$ meets BC at Y. Show that: $\small ar(CYXE)=2ar(FCB)$

Answer:

Since ||gm CYXE and $\Delta$ ACE lie on the same base CE and between the same parallels CE and AX.
Therefore, 2. ar( $\Delta$ ACE) = ar (||gm CYXE)
B ut, $\Delta$ FCB $\cong$ $\Delta$ ACE (in iv part)
Since the congruent triangle has equal areas. So,
$\small ar(CYXE)=2ar(FCB)$
Hence proved.

Q8 (vi) In Fig. $\small 9.34$ , 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 $\small AX\perp DE$ meets BC at Y. Show that: $\small ar(CYXE)=ar(ACFG)$

Answer:

Since ar( ||gm CYXE) = 2. ar( $\Delta$ ACE) {in part (v)}...................(i)
Also, $\Delta$ FCB and quadrilateral ACFG lie on the same base FC and between the same parallels FC and BG.
Therefore, ar (quad. ACFG) = 2.ar( $\Delta$ FCB ) ................(ii)

From eq (i) and eq (ii), we get

$\small ar(CYXE)=ar(ACFG)$
Hence proved.

Q8 (vii) In Fig. $\small 9.34$ , 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 $\small AX\perp DE$ meets BC at Y. Show that: $\small ar(BCED)=ar(ABMN)+ar(ACFG)$

Answer:

We have,
ar(quad. BCDE) = ar(quad, CYXE) + ar(quad. BYXD)

= ar(quad, CYXE) + ar (quad. ABMN) [already proved in part (iii)]
Thus, $\small ar(BCED)=ar(ABMN)+ar(ACFG)$

Hence proved.