NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles (Ex 9.4) Exercise 9.4

Exercise 9.4

1. Parallelogram ${\text{ABCD}}$ and rectangle ${\text{ABEF}}$ are on the same base ${\text{AB}}$ and have equal areas. Show that the perimeter of the parallelogram is greater than that of the rectangle.

Ans:  Given that the parallelogram and the rectangle have the same base and equal area, So, these will also lie between the same parallels.

Consider the parallelogram ${\text{ABCD}}$ and rectangle ABEF as follows.

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So, parallelogram ${\text{ABCD}}$ and rectangle ${\text{ABEF}}$ are between the same parallels ${\text{AB}}$ and ${\text{CF}}$.

Since, opposite sides of a parallelogram or a rectangle are of equal lengths.

Therefore,

${\text{AB}} = {\text{EF}}$ (For rectangle)

${\text{AB}} = {\text{CD}}$ (For parallelogram)

$\therefore CD = EF$

$\therefore {\text{AB}} + {\text{CD}} = {\text{AB}} + {\text{EF}}$

Of all the line segments that can be drawn to a given line from a point not lying on it, the perpendicular line segment is the shortest.

$\therefore AF < AD$

Similarly, ${\text{BE}} < {\text{BC}}$

$\therefore {\text{AF}} + {\text{BE}} < {\text{AD}} + {\text{BC}}$

${\text{AB}} + {\text{EF}} + {\text{AF}} + {\text{BE}} < {\text{AD}} + {\text{BC}} + {\text{AB}} + {\text{CD}}$

Perimeter of rectangle ${\text{ABEF}} < $ Perimeter of parallelogram ${\text{ABCD}}$.

Hence, proved

2. In the following figure, ${\text{D}}$ and ${\text{E}}$ are two points on ${\text{BC}}$ such that, ${\text{BD}} = {\text{DE}} = {\text{EC}}$. Show that $\operatorname{ar} ({\text{ABD}}) = \operatorname{ar} ({\text{ADE}}) = \operatorname{ar} ({\text{AEC}})$

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Can you now answer the question that you have left in the 'Introduction' of this chapter, whether the field of Budhia has been actually divided into three parts of equal area?

(Remark: Note that by taking, ${\text{BD}} = {\text{DE}} = {\text{EC}}$, the triangle ${\text{ABC}}$ is divided into three triangles ${\text{ABD}},{\text{ADE}}$ and AEC of equal areas. In the same way, by dividing ${\text{BC}}$ into ${\text{n}}$ equal parts and joining the points of division so obtained to the opposite vertex of BC, you can divide $\Delta ABC$ into $n$ triangles of equal areas.)

Ans: Let us draw a line segment ${\text{AL}} \bot {\text{BC}}$.

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We know that, ${\text{ Area of a triangle }} = \dfrac{1}{2} \times {\text{ Base }} \times {\text{ Altitude }}$

$\Rightarrow {\text{Area }}(\Delta {\text{ADE}}) = \dfrac{1}{2} \times {\text{DE}} \times {\text{AL }}$

$\Rightarrow {\text{Area }}(\Delta {\text{ABD}}) = \dfrac{1}{2} \times {\text{BD}} \times {\text{AL }}$

$\Rightarrow {\text{Area }}(\Delta {\text{AEC}}) = \dfrac{1}{2} \times {\text{EC}} \times {\text{AL}}$

It is given that ${\text{DE}} = {\text{BD}} = {\text{EC}}$

$\Rightarrow \dfrac{1}{2} \times {\text{DE}} \times {\text{AL}} = \dfrac{1}{2} \times {\text{BD}} \times {\text{AL}} = \dfrac{1}{2} \times {\text{EC}} \times {\text{AL}}$

$\Rightarrow \operatorname{Area} (\Delta {\text{ADE}}) = \operatorname{Area} (\Delta {\text{ABD}}) = {\text{ Area }}(\Delta {\text{AEC}})$

It can be observed that Budhia has divided her field into 3 equal parts.

3. In the following figure, ${\text{ABCD}},{\text{DCFE}}$ and ${\text{ABFE}}$ are parallelograms. Show that  the ar$(\Delta {\text{ADE}}) = \operatorname{ar} (\Delta {\text{BCF}})$.

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Ans: It is given that ${\text{ABCD}}$ is a parallelogram. We know that opposite sides of a parallelogram are equal.

$\therefore {\text{AD}} = {\text{BC}}$

Similarly, for parallelograms DCEF and ABFE

 $EA = FB$

$DE = CF$

ln $\Delta {\text{ADE}}$ and $\Delta {\text{BCF}}$

${\text{AD}} = {\text{BC}}$

$EA = FB$

$DE = CF$

$\Delta {\text{ADE}} \cong \Delta {\text{BCF }}(By\:SSS)$

So, ar$(\Delta {\text{ADE}}) = \operatorname{ar} (\Delta {\text{BCF}})$ (By cpct)

Hence, proved

4. In the following figure, ${\text{ABCD}}$ is parallelogram and ${\text{BC}}$ is produced to a point ${\text{Q}}$ such that ${\text{AD}} = $ ${\text{CQ}}$. If ${\text{AQ}}$ intersect ${\text{DC}}$ at ${\text{P}}$, show that ar $(\Delta {\text{BPC}}) = \operatorname{ar} (\Delta {\text{DPQ}})$ 

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(Hint: Join ${\text{AC}}$)

Ans: It is given that ${\text{ABCD}}$ is a parallelogram.

So, and  (Opposite sides of a parallelogram are parallel to each other) 

Join the point $A$ to point $C$.

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Consider $\Delta {\text{APC}}$ and $\Delta {\text{BPC}}$

$\Delta {\text{APC}}$and $\Delta {\text{BPC}}$ are lying on the same base ${\text{PC}}$ and between the same parallels ${\text{PC}}$ and AB. Therefore,

Area $(\Delta {\text{APC}}) = \operatorname{Area} (\Delta {\text{BPC}})$………(1)

In quadrilateral ${\text{ACDQ}}$, it is given that

${\text{AD}} = {\text{CQ}}$

Since ABCD is a parallelogram,

 (Opposite sides of a parallelogram are parallel)

${\text{CQ}}$ is a line segment which is obtained when line segment ${\text{BC}}$ is produced.

We have,

$AC = DQ$ and 

Hence, ${\text{ACQD}}$ is a parallelogram.

Consider ${\text{BDCQ}}$ and ${\text{BACQ}}$

These are on the same base ${\text{CQ}}$ and between the same parallels ${\text{CQ}}$ and ${\text{AD}}.$

$\therefore $ Area $(\Delta DPQ) = $ Area $(\Delta APC)$………(2)

From equations (1) and (2), we obtain

Area $(\Delta BPC) = $ Area $(\Delta DPQ)$

Hence, proved

5. In the following figure, ${\text{ABC}}$ and ${\text{BDE}}$ are two equilateral triangles such that ${\text{D}}$ is the mid-point of ${\text{BC}}$. If ${\text{AE}}$ intersects ${\text{BC}}$ at ${\text{F}}$, show that,

(i) $\quad \operatorname{ar} ({\text{BDE}}) = \dfrac{1}{4}\operatorname{ar} ({\text{ABC}})$

(ii) $\operatorname{ar} ({\text{BDE}}) = \dfrac{1}{2}\operatorname{ar} ({\text{BAE}})$

(iii) $\operatorname{ar} ({\text{ABC}}) = 2\operatorname{ar} ({\text{BEC}})$

(iv) $\quad \operatorname{ar} ({\text{BFE}}) = \operatorname{ar} ({\text{AFD}})$

(v) $\quad {\text{ar(BFE) = 2}}{\text{ar}}\left( {{\text{FED}}} \right)$ 

(vi) $\operatorname{ar} ({\text{FED}}) = \dfrac{1}{8}\operatorname{ar} ({\text{AFC}})$

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(Hint: Join EC and AD. Show that  and DE || AB etc.)

Ans: (i) Let ${\text{G}}$ and ${\text{H}}$ be the mid-points of side ${\text{AB}}$ and ${\text{AC}}$ respectively. Line segment ${\text{GH}}$ is joining the mid-points and is parallel to third side. Therefore, BD will be half of the length of BC (Using midpoint theorem).

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\[\therefore GH = \dfrac{1}{2}BCandGH||BC\]

$\therefore {\text{GH}} = {\text{BD}} = {\text{DC}}$ and  (D is the mid-point of ${\text{BC}}$ )

Similarly, ${\text{GD}} = {\text{HC}} = {\text{HA}}$ and ${\text{HD}} = {\text{AG}} = {\text{BG}}$

Therefore, $\Delta {\text{ABC}}$ is divided into 4 equal equilateral triangles viz $\Delta {\text{BGD}},\Delta {\text{AGH}},\Delta {\text{DHC}}$ and $\Delta {\text{GHD}}$.

Thus, $\Delta {\text{BGD}} = \dfrac{1}{4}\Delta {\text{ABC}}$

Now consider, $\Delta {\text{BDG}}$ and $\Delta {\text{BDE}}$

${\text{BD}} = {\text{BD}}$ (Common base)

As both triangles are equilateral triangles, So

${\text{BG}} = {\text{BE}};{\text{DG}} = {\text{DE}}$

Therefore, \[\Delta {\text{BDG}} \cong \Delta {\text{BDE}}\] (By SSS congruency)

Thus, area \[(\Delta BDG) = \] area $(\Delta BDE)$

$\operatorname{ar} (\Delta {\text{BDE}}) = \dfrac{1}{4}\operatorname{ar} (\Delta {\text{ABC}})$

Hence proved

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(ii)  $ar(\Delta BDE) = (\Delta AED)$

$ar(\Delta BDE) - ar(\Delta FED) = ar(\Delta AED) - ar(\Delta FED)$

$ar(\Delta BEF) = ar(\Delta AFD).....(1)$

$ar(\Delta ABD) = ar(\Delta ABF) + ar(\Delta AFD)$

$ar(\Delta ABD) = ar(\Delta ABF) + ar(\Delta BEF)$

$ar(\Delta ABD) = ar(\Delta ABE).......(2)$

AD is median of \[\Delta ABC\]

$ar(\Delta ABD) = \dfrac{1}{2}ar(\Delta ABC) = \dfrac{4}{2}ar(\Delta BDE)$

$ar(\Delta ABD) = 2ar(\Delta BDE)...........(3)$

From (2) and (3), we obtain

$2ar(\Delta BDE) = ar(\Delta ABE)$

$ar(\Delta BDE) = \dfrac{1}{2}ar(\Delta ABE)$

(iii) $ar(\Delta ABE) = ar(\Delta BEC)$

  $ar(\Delta ABF) + ar(\Delta BEF) = ar(\Delta BEC)$

  $ar(\Delta ABF) + ar(\Delta AFD) = ar(\Delta BEC)$

  $ar(\Delta ABD) = ar(\Delta BEC)$

  $\dfrac{1}{2}ar(\Delta ABC) = ar(\Delta BEC)$

  $ar(\Delta ABC) = 2ar(\Delta BEC)$

(iv) It is seen that area($\Delta {\text{BDE}}$)  and area( $\Delta {\text{AED}}$) lie on the same base DE and between the parallels ${\text{DE}}$ and ${\text{AB}}$.

\[\therefore ar(\Delta BDE) = ar(\Delta AED)\]

\[\therefore ar(\Delta BDE) - ar(\Delta FED) = ar(\Delta AED) - ar(\Delta FED)\]

\[\therefore ar(\Delta BFE) = ar(\Delta AFD)\]

(v) Let $h$ be the height of vertex $E$, corresponding to the side BD  in $\Delta BDE$. Let ${\text{H}}$ be the height of vertex ${\text{A}}$, corresponding to the side ${\text{BC}}$ in $\Delta {\text{ABC}}.$

In (i), it was shown that ar (BDE) $ = \dfrac{1}{4}\operatorname{ar} (ABC)$

In (iv), it was shown that ar $(\Delta {\text{BFE}}) = \operatorname{ar} (\Delta {\text{AFD}})$.

Hence, Proved

(vi) $ar(\Delta AFC) = ar(\Delta AFD) + ar(\Delta ADC) = 2ar(\Delta FED) + \dfrac{1}{2}ar(\Delta ABC)$

 $= 2ar(\Delta FED) + \dfrac{1}{2}\left[ {4 \times ar(\Delta BDE)} \right]$

$= 2ar(\Delta FED) + 2ar(\Delta BDE)$

$= 2ar(\Delta FED) + 2ar(\Delta AED)$

$= 2ar(\Delta FED) + 2\left[ {ar(\Delta AFD) + \,ar(\Delta FED)} \right]$

$= 4ar(\Delta FED) + 4ar(\Delta FED)$

$\Rightarrow ar\left( {\Delta AFC} \right) = 8ar(\Delta FED)$

$\Rightarrow ar(\Delta FED) = \dfrac{1}{8}ar(\Delta AFC)$ 

6. Diagonals ${\text{AC}}$ and ${\text{BD}}$ of a quadrilateral ${\text{ABCD}}$ intersect each other at ${\text{P}}$. Show that \[ar(\Delta AED) \times ar(\Delta BEC) = ar(\Delta ABE) \times ar(\Delta CDE)\].

 (Hint: From A and C, draw perpendiculars to BD)

Ans: Given: A quadrilateral ${\text{ABCD}}$ in which diagonals ${\text{AC}}$ and BD intersect each other at point ${\text{E}}$

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${\text{To Prove: }}\operatorname{ar} (\Delta {\text{AED}}) \times \operatorname{ar} (\Delta BEC) = \operatorname{ar} (\Delta ABE) \times \operatorname{ar} (\Delta CDE)$

Construction: From ${\text{A}}$ draw $AM \bot BD$ and from ${\text{C}}$ draw $CN \bot BD.$

${\text{ Proof}}{\text{: }}\operatorname{ar} (\Delta {\text{ABE}}) = \dfrac{1}{2} \times {\text{BE}} \times {\text{AM}}............{\text{(i)}}$

 $\operatorname{ar} (\Delta {\text{AED}}) = \dfrac{1}{2} \times {\text{DE}} \times {\text{AM}}................{\text{(ii)}}$ 

Dividing eq. (I) by eq. (ii)

$\dfrac{{\operatorname{ar} (\Delta {\text{AED}})}}{{\operatorname{ar} (\Delta {\text{ABE}})}} = \dfrac{{\dfrac{1}{2} \times {\text{DE}} \times {\text{AM}}}}{{\dfrac{1}{2} \times {\text{BE}} \times {\text{AM}}}}$

$\Rightarrow \dfrac{{\operatorname{ar} (\Delta {\text{AED}})}}{{\operatorname{ar} (\Delta {\text{ABE}})}} = \dfrac{{{\text{DE}}}}{{{\text{BE}}}}................(iii)$ 

Similarly $\dfrac{{\operatorname{ar} (\Delta {\text{CDE}})}}{{\operatorname{ar} (\Delta {\text{BEC}})}} = \dfrac{{{\text{DE}}}}{{{\text{BE}}}}................(iv)$

From eq. (iii) and (iv), we get

$\dfrac{{\operatorname{ar} (\Delta {\text{AED}})}}{{\operatorname{ar} (\Delta {\text{ABE}})}} = \dfrac{{\operatorname{ar} (\Delta {\text{CDE}})}}{{\operatorname{ar} (\Delta {\text{BEC}})}}$

$\Rightarrow \operatorname{ar} (\Delta {\text{AED}}) \times \operatorname{ar} (\Delta {\text{BEC}}) = \operatorname{ar} (\Delta {\text{ABE}}) \times \operatorname{ar} (\Delta {\text{CDE}})$

Hence, proved.

7. ${\text{P}}$ and ${\text{Q}}$ are respectively the mid-points of sides ${\text{AB}}$ and ${\text{BC}}$ of a triangle ${\text{ABC}}$ and ${\text{R}}$ is the mid-point of AP, show that

(i) ${\text{ar(PRQ) = }}\dfrac{{\text{1}}}{{\text{2}}}{\text{ar(ARC)}}$

(ii) $ar(RQC) = \dfrac{3}{8}\operatorname{ar} (ABC)$

(iij) $ar\{ PBQ) = ar(ARC)$

Ans: (i) ${\text{PC}}$ is the median of $\Delta {\text{ABC}}$.

$\therefore \operatorname{ar} (\Delta {\text{BPC}}) = \operatorname{ar} (\Delta {\text{APC}})$.......... (i)

${\text{RC}}$ is the median of $\Delta {\text{APC}}$.

$\therefore \operatorname{ar} \left( {\Delta ARC} \right) = \dfrac{1}{2}\operatorname{ar} (\Delta {\text{APC}})$…………..(ii)

(Median divides the triangle into two triangles of equal area) 

PQ is the median of $\Delta BPC$.

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\[\therefore \operatorname{ar} \left( {\Delta PQC} \right) = \dfrac{1}{2}\operatorname{ar} (\Delta B{\text{PC}})\]……………(iii)

From eq. (i) and (iii), we get,

$\operatorname{ar} (\Delta {\text{PQC}}) = \dfrac{1}{2}\operatorname{ar} (\Delta {\text{APC}}) \ldots  \ldots  \ldots {\text{ (iv) }}$

From eq. (ii) and (iv), we get

$\operatorname{ar} (\Delta {\text{PQC}}) = \operatorname{ar} \left( {\Delta ARC} \right) \ldots  \ldots  \ldots .({\text{v}})$

We are given that ${\text{P}}$ and ${\text{Q}}$ are the mid-points of ${\text{AB}}$ and ${\text{BC}}$ respectively.

$\therefore {\text{PQ}}{\text{||}}{\text{AC and PA}} = \dfrac{1}{2}{\text{AC}}$

$ \Rightarrow \operatorname{ar} (\Delta {\text{APQ}}) = \operatorname{ar} (\Delta {\text{PQC}})$ …………(vi)

 (Triangles between same parallel are equal in area)

From eq. (v) and (vi) we get

$\operatorname{ar} \left( {\Delta APQ} \right) = ar(\Delta ARC)$ ………..(vii)

${\text{R}}$ is the mid-point of ${\text{AP}}$. Therefore ${\text{RQ}}$ is the median of $\Delta {\text{APQ}}$.

$\therefore \operatorname{ar} (\Delta {\text{PRQ}}) = \dfrac{1}{2}\operatorname{ar} \left( {\Delta APQ} \right)$

From (vii) and (viii) we get

$\operatorname{ar} (\Delta P{\text{RQ}}) = \dfrac{1}{2}{\text{ar}}(\Delta {\text{ARC}})$

(ii) PQ is the median of $\Delta ABC$

$\therefore \operatorname{ar} (\Delta {\text{PQC}}) = \dfrac{1}{2}\operatorname{ar} (\Delta {\text{BPC}}) = \dfrac{1}{2} \times \dfrac{1}{2}\operatorname{ar} (\Delta {\text{ABC}}) = \dfrac{1}{4}\operatorname{ar} (\Delta {\text{ABC}}) \ldots  \ldots  \ldots ({\text{ix}})$

${\text{ar}}(A{\text{PRC}}) = \dfrac{1}{2}\operatorname{ar} (\Delta {\text{APC}})[{\text{Using}}({\text{iv}})]$

$\Rightarrow {\text{ ar }}(\Delta {\text{PRC}}) = \dfrac{1}{2} \times \dfrac{1}{2}\operatorname{ar} (\Delta {\text{ABC}}) = \dfrac{1}{4}\operatorname{ar} (\Delta {\text{ABC}}) \ldots  \ldots .({\text{x}})$ 

Adding (ix) and (x), we get

$\operatorname{ar} (\Delta {\text{PQC}}) + \operatorname{ar} (\Delta {\text{PRC}}) = \left( {\dfrac{1}{4} + \dfrac{1}{4}} \right)\operatorname{ar} (\Delta {\text{ABC}})$

$\Rightarrow \operatorname{ar} ({\text{PQCR}}) = \dfrac{1}{2}\operatorname{ar} (\Delta {\text{ABC}}) \ldots  \ldots .({\text{xi}})$ 

Subtracting ar$(\Delta {\text{PRQ}})$ from the both sides,

$\operatorname{ar} ($ quad. ${\text{PQCR}}) - \operatorname{ar} (\Delta {\text{PRQ}}) = \dfrac{1}{2}\operatorname{ar} \left( {\Delta ABC} \right) - \operatorname{ar} (\Delta {\text{PRQ}})$

$ \Rightarrow \operatorname{ar} \left( {\Delta RQC} \right) = \dfrac{1}{2}\operatorname{ar} \left( {\Delta ABC} \right) - \dfrac{1}{2}\operatorname{ar} (\Delta {\text{ARC}})[U\operatorname{sing} \operatorname{result} ({\text{i}})]$

$ \Rightarrow \operatorname{ar} (\Delta {\text{ARC}}) = \dfrac{1}{2}\operatorname{ar} (\Delta {\text{ABC}}) - \dfrac{1}{2} \times \dfrac{1}{2}\operatorname{ar} (\Delta {\text{APC}})$

$ \Rightarrow \operatorname{ar} (\Delta {\text{RQC}}) = \dfrac{1}{2}\operatorname{ar} (\Delta {\text{ABC}}) - \dfrac{1}{4}\operatorname{ar} (\Delta {\text{APC}})$

\[ \Rightarrow \operatorname{ar} \left( {\Delta RQC} \right) = \dfrac{1}{2}\operatorname{ar} \left( {\Delta ABC} \right) - \dfrac{1}{4} \times \dfrac{1}{2}\operatorname{ar} \left( {\Delta ABC} \right)\] \[\left [ PC\:is\:the\:median\:of\:\triangle ABC \right ]\]

$ \Rightarrow \operatorname{ar} \left( {\Delta RQC} \right) = \dfrac{1}{2}\operatorname{ar} \left( {\Delta ABC} \right) - \dfrac{1}{8}\operatorname{ar} (\Delta {\text{ABC}})$

$ \Rightarrow \operatorname{ar} \left( {\Delta RQC} \right) = \left( {\dfrac{1}{2} - \dfrac{1}{8}} \right) \times \operatorname{ar} (\Delta {\text{ABC}})$

$ \Rightarrow \operatorname{ar} \left( {\Delta RQC} \right) = \dfrac{3}{8}\operatorname{ar} \left( {\Delta ABC} \right)$

(iii) ar $(\Delta {\text{PRQ}}) = \dfrac{1}{2}$ ar $(\Delta {\text{ARC}})$  [Using result (i) $]$ 

$ \Rightarrow 2\operatorname{ar} (\Delta {\text{PRQ}}) = \operatorname{ar} \left( {\Delta ABC} \right).$…….(xii)

 ar $(\Delta {\text{PRQ}}) = \dfrac{1}{2}\operatorname{ar} (\Delta {\text{APQ}})[RQ$ is the median of $\Delta {\text{APQ}}] \ldots  \ldots ..$ (xiii)

But ar$(\Delta {\text{APQ}}) = \operatorname{ar} (\Delta {\text{PQC}})$ [Using reason of eq. (vi) ]

From eq. (xiii) and (xiv), we get

$\operatorname{ar} (\Delta {\text{PRQ}}) = \dfrac{1}{2}\operatorname{ar} (\Delta {\text{PQC}}) \ldots  \ldots  \ldots ({\text{xv}})$

But ar $(\Delta BPQ) = $ ar $(\Delta PQC)[PQ$ is the median of $\Delta BPC] \ldots  \ldots $(xvi) 

From eq. (xv) and (xvi), we get,

$\operatorname{ar} (\Delta {\text{PRQ}}) = \dfrac{1}{2}\operatorname{ar} \left( {\Delta BPQ} \right) \ldots  \ldots  \ldots {\text{ (xvii) }}$

Now, from eq. (xii) and (xvii), we get

$2 \times \dfrac{1}{2}ar(\Delta BPQ) = \operatorname{ar} \left( {\Delta ARC} \right)$

$\Rightarrow \operatorname{ar} (\Delta {\text{BPQ}}) = \operatorname{ar} \left( {\Delta ARC} \right)$ 

8. In the following figure, ${\text{ABC}}$ is a right triangle right angled at ${\text{A}}.{\text{BCED}},{\text{ACFG}}$ and ${\text{ABMN}}$ are squares on the sides ${\text{BC}},{\text{CA}}$ and ${\text{AB}}$ respectively. Line segment ${\text{AX}} \bot {\text{DE}}$ meets ${\text{BC}}$ at Y.

(Image will be uploaded soon)

Show that:

(i) $\Delta {\text{MBC}} \cong \Delta {\text{ABD}}$

(ii) $\operatorname{ar} ({\text{BYXD}}) = 2{\text{ar}}({\text{MBC}})$

(iii) $\operatorname{ar} ({\text{BYXD}}) = 2\operatorname{ar} ({\text{ABMN}})$

(iv) $\Delta {\text{FCB}} \cong \Delta {\text{ACE}}$

(v) $\operatorname{ar} ({\text{CYXE}}) = 2\operatorname{ar} ({\text{FCB}})$

(vi) $\operatorname{ar} ({\text{CYXE}}) = \operatorname{ar} ({\text{ACF}}G)$

(vii) $\operatorname{ar} ({\text{BCED}}) = \operatorname{ar} ({\text{ABMN}}) + \operatorname{ar} ({\text{ACFG}})$

Note: Result (vii) is the famous Theorem of Pythagoras. You shall learn a simpler proof of this theorem in class ${\text{X}}$.

Ans: (i) Each angle of a square is ${90^\circ }$.

Hence, $\angle {\text{ABM}}{\text{ = }}\angle {\text{DBC}}{\text{ = }}{90^\circ }$

$\therefore \angle {\text{ABM}} + \angle {\text{ABC}}{\text{ = }}\angle {\text{DBC}} + \angle {\text{ABC}}$

$\therefore \angle {\text{MBC}}{\text{ = }}\angle {\text{ABD}}$

In \[\Delta MBC\] and \[\Delta ABD\]

\[\angle MBC = \angle ABD\] (Proved above)

\[MB = AB\] (sides of square ABMN)

\[BC = BD\](sides of square BCED)

\[\therefore \Delta MBC \cong \Delta ABD\]( SAS congruence rule)

(ii) We have \[\Delta MBC \cong \Delta ABD\]

\[\therefore ar(\Delta MBC) = ar(\Delta ABD)......(1)\]

It is given that \[AX \bot DE\] and \[BD \bot DE\](Adjacent sides of square BCED)

\[\therefore BD||AX\] (Two lines perpendicular to same line are parallel to each other)

\[\Delta ABD\] and parallelogram \[BXYD\] are on same base BD and between same parallels BD and AX.

\[ar(\Delta YXD) = 2ar(\Delta MBC)........(2)\] (using equation 1)

(iii) \[\Delta MBC\]and parallelogram \[AMBN\] are on same base MB and between same parallels MB and NC.

\[2ar(\Delta MBC) = ar(AMBN)\]

\[ar(\Delta YXD) = ar(AMBN)\]………(3) (using equation 2)

(iv) Each angle of a square is ${90^\circ }$.

Hence, $\angle FCA{\text{ = }}\angle {\text{BCE}}{\text{ = }}{90^\circ }$

$\therefore \angle FCA + \angle {\text{ACB}}{\text{ = }}\angle {\text{BCE}} + \angle {\text{ACB}}$

$\therefore \angle F{\text{CB}}{\text{ = }}\angle {\text{ACE}}$

In \[\Delta FCB\] and \[\Delta ACE\]

\[\angle FCB = \angle ACE\] (Proved above)

\[FC = AC\] (sides of square ABMN)

\[CB = CE\](sides of square BCED)

\[\therefore \Delta FCB \cong \Delta ACE\]( SAS congruence rule)

(v) It is given that \[AX \bot DE\] and \[CE \bot DE\](Adjacent sides of square BCED)

\[\therefore CE||AX\] (Two lines perpendicular to same line are parallel to each other)

\[BACE\] and parallelogram \[CYXE\] are on same base CE and between same parallels CE and AX.

\[ar(\Delta YXE) = 2ar(\Delta ACE)........(4)\]

Also, we had proved \[\Delta FCB \cong \Delta ACE\]……(5)

Comparing eq. (4) and (5), we get

\[ar(CYXE) = 2ar(\Delta FCB)........(6)\]

(vi)  Consider BFCB and parallelogram ACFG

BFCB and parallelogram ACFG are lying on the same base CF and between parallels ${\text{CF}}$ and ${\text{BG}}$.

$\therefore \operatorname{ar} ({\text{ACFG}}) = \operatorname{ar} ({\text{CYXE}})[$ Using equation $(6)] \ldots (7)$

(vii) From the figure, it is evident that

\[ar(\Delta CED) = ar(\Delta YXD) + ar(CYXE)\]

$\therefore \operatorname{ar} (\Delta {\text{CED}}) = \operatorname{ar} ({\text{ABMN}}) + \operatorname{ar} ({\text{ACFG}})$ (using equations (3) and (7) ).

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