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 $\mathrm{\Delta }$ BEC and $\mathrm{\Delta }$ 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. $9.30$ , D and E are two points on BC such that $BD=DE=EC$ . Show that $ar(ABD)=ar(ADE)=ar(AEC)$ .

Answer:

In $\mathrm{\Delta }$ 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. $9.31$ , ABCD, DCFE and ABFE are parallelograms. Show that $ar(ADE)=ar(BCF)$ .

Answer:

Given,

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

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

Hence proved.

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

Answer:

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

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

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

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

$ar(BPC)=ar(DPQ)$

Hence proved.

Q5 (i) 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

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

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

Answer:

Let join the CE and AD and draw $EP\perp BC$ . It is given that $\mathrm{\Delta }$ ABC and $\mathrm{\Delta }$ 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 $\mathrm{\Delta }$ ABC = $\frac{\sqrt{3}}{4}{a}^2$ and
Area of $\mathrm{\Delta }$ BDE = $\frac{\sqrt{3}}{4}(\frac{a}{2}{)}^2=\frac{\sqrt{3}}{16}{a}^2$

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

Q5 (ii) 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
$ar(BDE)=\frac{1}{2}ar(BAE)$

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

Answer:

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

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

Q5 (iii) 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
$ar(ABC)=2ar(BEC)$

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

Answer:

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

$\Rightarrow$ ar( $\mathrm{\Delta }$ ABC) = 2. ar( $\mathrm{\Delta }$ 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
$ar(BFE)=ar(AFD)$
[ Hint: Join EC and AD. Show that $BE\parallel AC$ and $DE\parallel AB$ , etc.]

Answer:

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

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

On subtracting $\mathrm{\Delta }$ EFD from both sides we get

$ar(BFE)=ar(AFD)$

Hence proved.

Q5 (v) 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
$ar(BFE)=2ar(FED)$

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

Answer:

In right angled triangle $\mathrm{\Delta }$ ABD, we get

$$\begin{gathered} A{B}^2=A{D}^2+B{D}^2 \\ A{D}^2=A{B}^2-B{D}^2 \end{gathered}$$
$=a^2-\frac{a^2}{4}=\frac{3a^2}{4}$
$AD=\frac{\sqrt{3}a}{2}$

So, in $\mathrm{\Delta }$ PED,
$P{E}^2=D{E}^2-D{P}^2$
$$\begin{gathered} =(\frac{a}{2}{)}^2-(\frac{a}{4}{)}^2 \\ =\frac{3a^2}{16} \end{gathered}$$
So, $PE=\frac{\sqrt{3}a}{4}$

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

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

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

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

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:

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

$ar(FED)=\frac{1}{8}ar(AFC)$
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( $\mathrm{\Delta }$ APB) = $\frac{1}{2}\times BP\times AM$ and,
$ar(\mathrm{\Delta }CDP)=\frac{1}{2}\times DP\times CN$
Therefore, ar( $\mathrm{\Delta }$ APB) $\times$ ar( $\mathrm{\Delta }$ CDP)=
$$\begin{gathered} =(\frac{1}{2}\times BP\times AM).(\frac{1}{2}\times DP\times CN) \\ =\frac{1}{4}\times BP\times DP\times AM\times CN \end{gathered}$$ ....................(i)

Similarly, ar( $\mathrm{\Delta }$ APD) $\times$ ar ( $\mathrm{\Delta }$ 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

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

Answer:

We have $\mathrm{\Delta }$ 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 $\mathrm{\Delta }$ APC,
Since R is the midpoint. So, RC is the median of the $\mathrm{\Delta }$ APC
Therefore, ar( $\mathrm{\Delta }$ ARC) = 1/2 . ar ( $\mathrm{\Delta }$ APC)............(i)

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

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

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

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

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

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

$=ar(\mathrm{\Delta }PRQ)$ (from equation (iv))

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

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

Answer:

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

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

Hence proved.

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

Answer:

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

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

Hence proved.

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

Answer:

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

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

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

Answer:

Since, ar (||gm BYXD) = 2 .ar ( $\mathrm{\Delta }$ MBC) ..........(i) [already proved in 2nd part]
and, ar (sq. ABMN) = 2. ar ( $\mathrm{\Delta }$ 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

$ar(BYXD)=ar(ABMN)$

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

Answer:

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

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

Hence proved.

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

Answer:

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

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

Answer:

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

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

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

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

Answer:

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

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

Hence proved.