NCERT Solutions For Class 9 Maths Chapter 9 Areas of Parallelograms And Triangles (Ex 9.4)

Q.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.

Ans

$\begin{array}{l}\text{Given}:\text{Parallelogram ABCD and rectangle ABEF have same}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}base AB and equal area.}\\ \text{To prove: Perimeter of ABCD > Perimeter of ABEF}\\ \text{Proof: Since, both Parallelogram ABCD and rectangle ABEF}\\ \text{have same base AB and equal area}.\text{So, CF}\parallel \mathrm{AB}\\ \mathrm{Now},\mathrm{AB}=\mathrm{DC}\left[\mathrm{Opposite}\text{sides of rectangle}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}and \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{AB}=\mathrm{EF}\left[\mathrm{Opposite}\text{sides of parallelogram}\right]\\ \Rightarrow \mathrm{CD}=\mathrm{EF}\\ \text{Then},\mathrm{AB}+\mathrm{CD}=\mathrm{AB}+\mathrm{EF}...\left(\mathrm{i}\right)\\ \text{Since},\text{the line segment drawn from an external point to a line}\\ \text{is the shortest line segment.}\\ \text{Then,}\mathrm{AD}>\mathrm{AF}\mathrm{and}\mathrm{BC}>\mathrm{BE}\\ \Rightarrow \mathrm{AD}+\mathrm{BC}>\mathrm{AF}+\mathrm{BE}...\left(\mathrm{ii}\right)\\ \text{From equation (i) and equation}\left(\mathrm{ii}\right),\text{we get}\\ \mathrm{AB}+\mathrm{CD}+\mathrm{AD}+\mathrm{BC}>\mathrm{AB}+\mathrm{EF}+\mathrm{AF}+\mathrm{BE}\\ \Rightarrow \text{Perimeter of ABCD > Perimeter of ABEF}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Hence proved.}\end{array}$

Q.2 In the figure given below, D and E are two points on BC such that BD = DE = EC. Show that ar(ABD)= ar(ADE)= ar(AEC). 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?

Ans

$\begin{array}{l}\mathrm{Given}:\mathrm{In}\mathrm{\Delta }\text{ABC,}\mathrm{BD}=\mathrm{DE}=\mathrm{EC}.\\ \text{To prove:}\mathrm{ar}\left(\mathrm{ABD}\right)=\mathrm{ar}\left(\mathrm{ADE}\right)=\mathrm{ar}\left(\mathrm{AEC}\right)\\ \text{Construction: Draw AL}\perp \text{BC}\\ \text{Proof: Since, Area of a triangle}=\frac{1}{2}\times \mathrm{perpendicular}\times \mathrm{base}\\ \therefore \mathrm{ar}\left(\mathrm{\Delta ABD}\right)=\frac{1}{2}\times \mathrm{BD}\times \mathrm{AM}\\ \mathrm{ar}\left(\mathrm{\Delta ADE}\right)=\frac{1}{2}\times \mathrm{DE}\times \mathrm{AM}\\ \mathrm{ar}\left(\mathrm{\Delta AEC}\right)=\frac{1}{2}\times \mathrm{EC}\times \mathrm{AM}\\ \text{Since},\mathrm{BD}=\mathrm{DE}=\mathrm{EC}\\ \mathrm{So},\frac{1}{2}\times \mathrm{BD}\times \mathrm{AM}=\frac{1}{2}\times \mathrm{DE}\times \mathrm{AM}=\frac{1}{2}\times \mathrm{EC}\times \mathrm{AM}\\ \Rightarrow \mathrm{ar}\left(\mathrm{ABD}\right)=\mathrm{ar}\left(\mathrm{ADE}\right)=\mathrm{ar}\left(\mathrm{AEC}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Hence proved.}\end{array}$

Q.3 In the figure below, ABCD, DCFE and ABFE are parallelograms. Show that ar (ADE) = ar (BCF). 

Ans

\begin{array}{l}\text{Given}:\text{ABCD},\text{DCFE and ABFE are parallelograms}.\\ \text{To prove}:\text{ar}\left(\text{ADE}\right)=\text{ar}\left(\text{BCF}\right)\\ \text{Proof}:\text{Since},\text{ABCD is a parallelogram}.\\ \text{So},\text{AD}=\text{BC}\left[\text{Opposite sides of parallelogram}.\right]\\ \text{Since},\text{DCFE is a parallelogram}.\\ \text{So},\text{DE}=\text{CF}\left[\text{Opposite sides of parallelogram}.\right]\\ \text{Since},\text{ABFE is a parallelogram}.\\ \text{So},\text{AE}=\text{BF}\left[\text{Opposite sides of parallelogram}.\right]\\ \text{In}\Delta \text{ADE and}\Delta \text{BCF}\\ \text{AD}=\text{BC}\left[Opposite\text{sides of parallelogram}\text{.}\right]\\ \text{DE}=\text{CF}\left[Opposite\text{sides of parallelogram}\text{.}\right]\\ \text{AE}=\text{BF}\left[Opposite\text{sides of parallelogram}\text{.}\right]\\ \Delta \text{ADE}\cong \Delta \text{BCF}\left[By\text{S}\text{.S}\text{.S}\text{.}\right]\\ ar\left(\Delta \text{ADE}\right)=ar\left(\Delta \text{BCF}\right)\left[Area\text{of congruent triangles is equal}\text{.}\right]\\ Hence\text{proved}\text{.}\end{array}

Q.4 In the figure below, 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).


Ans
Given : ABCD is a parallelogram and BC is produced upto Q such that AD = CQ.
To prove : ar(BPC) = ar (DPQ)
Construction : Join AC.

$\begin{array}{l}\text{Proof}:\text{Since,}\mathrm{\Delta APC}\text{and}\mathrm{\Delta BPC}\text{have same base and lie between}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}same parallels. So,}\\ \mathrm{ar}\left(\mathrm{\Delta APC}\right)=\mathrm{ar}\left(\mathrm{\Delta BPC}\right)...\left(\mathrm{i}\right)\\ \mathrm{In}\text{quadrilateral ACQD,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}AD}=\text{CQ and AD}=\mathrm{CQ}\\ \mathrm{So},\text{ACQD is a parallelogram.}\\ \mathrm{\Delta DCQ}\text{and}\mathrm{\Delta ACQ}\text{have same base QC and lie between}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}same parallels. So,}\\ \mathrm{ar}\left(\mathrm{\Delta DCQ}\right)=\mathrm{ar}\left(\mathrm{\Delta ACQ}\right)\\ \text{Substracting ar}\left(\mathrm{\Delta PCQ}\right)\text{from both sides, we get}\\ \mathrm{ar}\left(\mathrm{\Delta DCQ}\right)-\text{ar}\left(\mathrm{\Delta PCQ}\right)=\mathrm{ar}\left(\mathrm{\Delta ACQ}\right)-\text{ar}\left(\mathrm{\Delta PCQ}\right)\\ \Rightarrow \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}ar}\left(\mathrm{\Delta DPQ}\right)=\text{ar}\left(\mathrm{\Delta APC}\right)...\left(\mathrm{ii}\right)\\ \text{From equation}\left(\mathrm{i}\right)\text{ and equation}\left(\mathrm{ii}\right),\text{we get}\\ \mathrm{ar}\left(\mathrm{\Delta BPC}\right)=\text{ar}\left(\mathrm{\Delta DPQ}\right)\text{Hence proved.}\end{array}$

Q.5

$\begin{array}{l}\text{In Fig.9.33, ABC and BDE are two equilateral triangles such}\\ \text{that D is the mid-point of BC. If AE intersects BC at F, show}\\ \text{that}\\ \left(\text{i}\right)\text{\hspace{0.33em}ar\hspace{0.33em}}\left(\text{BDE}\right)\text{\hspace{0.33em}=\hspace{0.33em}}\frac{\text{1}}{\text{4}}\text{\hspace{0.33em}ar\hspace{0.33em}}\left(\text{ABC}\right)\\ \left(\text{ii}\right)\text{\hspace{0.33em}ar\hspace{0.33em}}\left(\text{BDE}\right)\text{\hspace{0.33em}=\hspace{0.33em}}\frac{\text{1}}{\text{2}}\text{\hspace{0.33em}ar\hspace{0.33em}}\left(\text{BAE}\right)\\ \left(\text{iii}\right)\text{\hspace{0.33em}ar}\left(\text{ABC}\right)\text{\hspace{0.33em}=\hspace{0.33em}2ar\hspace{0.33em}}\left(\text{BEC}\right)\\ \left(\text{iv}\right)\text{\hspace{0.33em}ar\hspace{0.33em}}\left(\text{BFE}\right)\text{\hspace{0.33em}=\hspace{0.33em}ar\hspace{0.33em}}\left(\text{AFD}\right)\\ \left(\text{v}\right)\text{\hspace{0.33em}ar\hspace{0.33em}}\left(\text{BFE}\right)\text{\hspace{0.33em}=\hspace{0.33em}2\hspace{0.33em}ar\hspace{0.33em}}\left(\text{FED}\right)\\ \left(\text{vi}\right)\text{\hspace{0.33em}ar\hspace{0.33em}}\left(\text{FED}\right)\text{\hspace{0.33em}=\hspace{0.33em}}\frac{\text{1}}{\text{8}}\text{\hspace{0.33em}ar\hspace{0.33em}}\left(\text{AFC}\right)\end{array}$

Ans

$\begin{array}{l}\mathrm{Given}:\mathrm{\Delta ABC}\text{and}\mathrm{\Delta BDE}\text{are two equilateral triangles. D is}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}mid-point of BC. AE intersects BC at F.}\\ \mathrm{To}\text{prove:}\left(\mathrm{i}\right)\mathrm{ar}\left(\mathrm{BDE}\right)=\frac{1}{4}\mathrm{ar}\left(\mathrm{ABC}\right)\\ \left(\mathrm{ii}\right)\mathrm{ar}\left(\mathrm{BDE}\right)=\frac{1}{2}\mathrm{ar}\left(\mathrm{BAE}\right)\\ \left(\mathrm{iii}\right)\mathrm{ar}\left(\mathrm{ABC}\right)=2\mathrm{ar}\left(\mathrm{BEC}\right)\\ \left(\mathrm{iv}\right)\mathrm{ar}\left(\mathrm{BFE}\right)=\mathrm{ar}\left(\mathrm{AFD}\right)\\ \left(\mathrm{v}\right)\mathrm{ar}\left(\mathrm{BFE}\right)=2\mathrm{ar}\left(\mathrm{FED}\right)\\ \left(\mathrm{vi}\right)\mathrm{ar}\left(\mathrm{FED}\right)=\frac{1}{8}\mathrm{ar}\left(\mathrm{AFC}\right)\\ \mathrm{Construction}:\mathrm{Join}\text{EC and AD. Let mid-points of AB and AC}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}as G and H respectively. Join G and H, G and D,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}H and D respectively.}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Proof:\hspace{0.17em}\hspace{0.17em}In}\mathrm{\Delta ABC},\text{G and H are the mid-points of AB and AC}\\ \text{respectively. So, by mid-point theorem}\\ \text{GH}\parallel \text{BC and GH}=\frac{1}{2}\mathrm{BC}\\ \Rightarrow \mathrm{GH}\parallel \text{B}\mathrm{D}\text{and GH}=\text{BD}\\ \Rightarrow \mathrm{GHDB}\text{is a parallelogram because one pair of}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}opposite sides is equal and parallel.}\\ \mathrm{Similarly},\text{}\mathrm{GHDC}\text{is a parallelogram because one pair of}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}opposite sides is equal and parallel.}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}In}\mathrm{\Delta CAB},\text{D and H are the mid-points of CB}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}and CA respectively. So, by mid-point theorem}\\ \text{DH}\parallel \mathrm{A}\text{B and DH}=\frac{1}{2}\mathrm{AB}\\ \Rightarrow \mathrm{D}\mathrm{H}\parallel \mathrm{A}\text{B and DH}=\text{AG}\\ \Rightarrow \mathrm{DHAG}\text{is a parallelogram because one pair of}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}opposite sides is equal and parallel.}\\ \text{Now, \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}GB}\parallel \mathrm{DE}\left[\mathrm{alternate}\text{angles are equal i.e., each 60°}\right]\\ \mathrm{and}\mathrm{GB}=\mathrm{DE}[\because \mathrm{DE}=\mathrm{BD}=\frac{1}{2}\mathrm{BC}=\frac{1}{2}\mathrm{AB}=\mathrm{GB}]\\ \mathrm{So},\mathrm{BEDG}\text{is a parallelogram because one pair}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}of opposite sides is equal and parallel.}\\ \mathrm{Since},\text{diagonal divides a parallelogram in two triangles of}\\ \mathrm{equal}\mathrm{area}.\\ \text{Then, ar}\left(\mathrm{DGH}\right)=\text{ar}\left(\mathrm{DGB}\right)\left[\mathrm{For}{\parallel }^{\mathrm{gm}}\mathrm{BDGH}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}ar}\left(\mathrm{DGH}\right)=\text{ar}\left(\mathrm{AGH}\right)\left[\mathrm{For}{\parallel }^{\mathrm{gm}}\mathrm{AHDG}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}ar}\left(\mathrm{DGH}\right)=\text{ar}\left(\mathrm{HDC}\right)\left[\mathrm{For}{\parallel }^{\mathrm{gm}}\mathrm{GHCD}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}ar}\left(\mathrm{DGB}\right)=\text{ar}\left(\mathrm{HDC}\right)\left[\mathrm{For}{\parallel }^{\mathrm{gm}}\mathrm{BEDG}\right]\\ \mathrm{and}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}ar}\left(\mathrm{DGB}\right)=\text{ar}\left(\mathrm{BDE}\right)\left[\mathrm{For}{\parallel }^{\mathrm{gm}}\mathrm{BEDG}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}ar}\left(\mathrm{ABC}\right)=\text{ar}\left(\mathrm{DGH}\right)+\text{ar}\left(\mathrm{DGB}\right)+\text{ar}\left(\mathrm{AGH}\right)+\text{ar}\left(\mathrm{HDC}\right)\\ =\text{ar}\left(\mathrm{DGB}\right)+\text{ar}\left(\mathrm{DGB}\right)+\text{ar}\left(\mathrm{DGB}\right)+\text{ar}\left(\mathrm{DGB}\right)\\ =4\text{\hspace{0.17em}ar}\left(\mathrm{DGB}\right)\\ =4\text{ar}\left(\mathrm{BDE}\right)\\ \therefore \text{ar}\left(\mathrm{BDE}\right)=\frac{1}{4}\text{ar}\left(\mathrm{ABC}\right)\\ \left(\mathrm{ii}\right)\mathrm{S}\mathrm{ince},\mathrm{\Delta BDE}\text{and parallelogram BEDG lie on the same base}\\ \text{and between same parallels, so}\\ \text{ar}\left(\mathrm{\Delta BDE}\right)=\frac{1}{2}\mathrm{ar}\left(\text{BEDG}\right)...\left(\mathrm{i}\right)\\ \because \mathrm{BE}\parallel \mathrm{DG}\text{and DG}\parallel \mathrm{HA}\\ \Rightarrow \mathrm{BE}\parallel \mathrm{AH}\text{and AB}\parallel \mathrm{HE}\\ \Rightarrow \mathrm{ABEH}\text{is a parallelogram.}\\ \text{Then, ar}\left(\mathrm{BEDG}\right)=\frac{1}{2}\mathrm{ar}\left(\mathrm{ABEH}\right)...\left(\mathrm{ii}\right)\\ \text{Since,}\mathrm{\Delta ABE}\text{and parallelogram ABEH lie on the same base}\\ \text{and between same parallels, so}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}ar}\left(\mathrm{\Delta ABE}\right)=\frac{1}{2}\mathrm{ar}\left(\text{ABEH}\right)...\left(\mathrm{iii}\right)\\ \mathrm{From}\text{equation}\left(\mathrm{i}\right),\left(\mathrm{ii}\right)\text{and}\left(\mathrm{iii}\right),\text{we have}\\ \text{ar}\left(\mathrm{\Delta BDE}\right)=\frac{1}{2}\text{\hspace{0.17em}ar}\left(\mathrm{\Delta ABE}\right)\end{array}$

$\begin{array}{l}\left(\mathrm{iii}\right)\text{ar}\left(\mathrm{\Delta ABE}\right)=\mathrm{ar}\left(\mathrm{\Delta BEC}\right)\left[\text{Common base BE and BE}\left|\right|\text{AC}\right]\\ \mathrm{ar}\left(\mathrm{\Delta ABF}\right)+\mathrm{ar}\left(\mathrm{\Delta BFE}\right)=\mathrm{ar}\left(\mathrm{\Delta BEC}\right)...\left(\mathrm{i}\right)\\ \mathrm{Since},\text{AB}\parallel \mathrm{ED}\\ \therefore \mathrm{ar}\left(\mathrm{\Delta BED}\right)=\mathrm{ar}\left(\mathrm{\Delta AED}\right)\left[\text{Common base BE and BE}\left|\right|\text{AC}\right]\\ \Rightarrow \mathrm{ar}\left(\mathrm{\Delta BFE}\right)+\mathrm{ar}\left(\mathrm{\Delta EFD}\right)=\mathrm{ar}\left(\mathrm{\Delta AFD}\right)+\mathrm{ar}\left(\mathrm{\Delta EFD}\right)\\ \Rightarrow \mathrm{ar}\left(\mathrm{\Delta BFE}\right)=\mathrm{ar}\left(\mathrm{\Delta AFD}\right)...\left(\mathrm{ii}\right)\\ \mathrm{From}\text{equation}\left(\mathrm{i}\right)\text{and}\left(\mathrm{ii}\right),\text{we have}\\ \mathrm{ar}\left(\mathrm{\Delta ABF}\right)+\mathrm{ar}\left(\mathrm{\Delta AFD}\right)=\mathrm{ar}\left(\mathrm{\Delta BEC}\right)\\ \Rightarrow \mathrm{ar}\left(\mathrm{\Delta ABD}\right)=\mathrm{ar}\left(\mathrm{\Delta BEC}\right)\\ \Rightarrow \frac{1}{2}\mathrm{ar}\left(\mathrm{\Delta ABC}\right)=\mathrm{ar}\left(\mathrm{\Delta BEC}\right)\left[\begin{array}{l}\mathrm{AD}\text{is median, so}\\ \mathrm{ar}\left(\mathrm{\Delta ABD}\right)=\frac{1}{2}\mathrm{ar}\left(\mathrm{\Delta ABC}\right)\end{array}\right]\\ \Rightarrow \mathrm{ar}\left(\mathrm{ABC}\right)=2\mathrm{ar}\left(\mathrm{BEC}\right)\\ \left(\mathrm{iv}\right)\mathrm{Since},\text{AB}\parallel \mathrm{ED}\\ \therefore \mathrm{ar}\left(\mathrm{\Delta BED}\right)=\mathrm{ar}\left(\mathrm{\Delta AED}\right)\left[\text{Common base BE and BE}\left|\right|\text{AC}\right]\\ \Rightarrow \mathrm{ar}\left(\mathrm{\Delta BFE}\right)+\mathrm{ar}\left(\mathrm{\Delta EFD}\right)=\mathrm{ar}\left(\mathrm{\Delta AFD}\right)+\mathrm{ar}\left(\mathrm{\Delta EFD}\right)\\ \Rightarrow \mathrm{ar}\left(\mathrm{\Delta BFE}\right)=\mathrm{ar}\left(\mathrm{\Delta AFD}\right)\\ \left(\mathrm{v}\right)\mathrm{Let}\text{height of}\mathrm{\Delta BDE}\text{be h and height of}\mathrm{\Delta ABC}\text{be H, then}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}ar}\left(\mathrm{\Delta BDE}\right)=\frac{1}{2}\times \mathrm{BD}\times \mathrm{h}\\ \mathrm{ar}\left(\mathrm{\Delta ABC}\right)=\frac{1}{2}\times \mathrm{BC}\times \mathrm{H}\\ \mathrm{Since},\mathrm{ar}\left(\mathrm{\Delta BDE}\right)=\frac{1}{4}\mathrm{ar}\left(\mathrm{\Delta ABC}\right)\\ \frac{1}{2}\times \mathrm{BD}\times \mathrm{h}=\frac{1}{4}(\frac{1}{2}\times \mathrm{BC}\times \mathrm{H})\\ \Rightarrow \frac{1}{2}\times \mathrm{BD}\times \mathrm{h}=\frac{1}{4}(\frac{1}{2}\times 2\mathrm{BD}\times \mathrm{H})[\because \mathrm{BC}=2\mathrm{BD}]\\ \Rightarrow \mathrm{h}=\frac{1}{2}\mathrm{H}\\ \mathrm{Now},\because \mathrm{ar}\left(\mathrm{BEF}\right)=\mathrm{ar}\left(\mathrm{AFD}\right)\left[\mathrm{From}\left(\mathrm{iv}\right)\right]\\ =\frac{1}{2}\times \mathrm{FD}\times \mathrm{H}\\ =\frac{1}{2}\times \mathrm{FD}\times 2\mathrm{h}\\ =2(\frac{1}{2}\times \mathrm{FD}\times \mathrm{h})\\ \therefore \mathrm{ar}\left(\mathrm{BEF}\right)=2\mathrm{ar}\left(\mathrm{\Delta FED}\right)\\ \left(\mathrm{vi}\right)\text{Since, ar}\left(\mathrm{AFC}\right)=\mathrm{ar}\left(\mathrm{AFD}\right)+\mathrm{ar}\left(\mathrm{ADC}\right)\\ =\mathrm{ar}\left(\mathrm{BFE}\right)+\frac{1}{2}\mathrm{ar}\left(\mathrm{ABC}\right)\left[\begin{array}{l}\mathrm{from}\left(\mathrm{iv}\right)\text{and D is}\\ \text{mid-point of BC.}\end{array}\right]\\ =\mathrm{ar}\left(\mathrm{BFE}\right)+\frac{1}{2}\times 4\mathrm{ar}\left(\mathrm{BDE}\right)\left[\mathrm{from}\left(\mathrm{i}\right)\right]\\ =\mathrm{ar}\left(\mathrm{BFE}\right)+2\mathrm{ar}\left(\mathrm{BDE}\right)\\ =\mathrm{ar}\left(\mathrm{BFE}\right)+2\{\mathrm{ar}\left(\mathrm{BEF}\right)+\mathrm{ar}\left(\mathrm{FED}\right)\}\\ =2\mathrm{ar}\left(\mathrm{FED}\right)+2[2\mathrm{ar}\left(\mathrm{FED}\right)+\mathrm{ar}\left(\mathrm{FED}\right)]\\ \left[\mathrm{from}\text{}\left(\mathrm{v}\right)\right]\\ =2\mathrm{ar}\left(\mathrm{FED}\right)+6\mathrm{ar}\left(\mathrm{FED}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}ar}\left(\mathrm{AFC}\right)=8\mathrm{ar}\left(\mathrm{FED}\right)\\ \Rightarrow \mathrm{ar}\left(\mathrm{FED}\right)=\frac{1}{8}\text{ar}\left(\mathrm{AFC}\right)\end{array}$

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

Ans

$\begin{array}{l}\text{Given}:\mathrm{ABCD}\text{is a quadrilateral in which diagonals AC and BD}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}intersect at P.}\\ \text{To prove:\hspace{0.17em}}\mathrm{ar}\left(\mathrm{APB}\right)\times \mathrm{ar}\left(\mathrm{CPD}\right)=\mathrm{ar}\left(\mathrm{APD}\right)\times \mathrm{ar}\left(\mathrm{BPC}\right)\\ \text{Construction}:\text{Draw AM}\perp \text{BD and CN}\perp \text{BD.}\\ \text{Proof:L.H.S.}=\mathrm{ar}\left(\mathrm{APB}\right)\times \mathrm{ar}\left(\mathrm{CPD}\right)\\ =(\frac{1}{2}\times \mathrm{PB}\times \mathrm{AM})\times (\frac{1}{2}\times \mathrm{PD}\times \mathrm{CN})\\ =(\frac{1}{2}\times \mathrm{PD}\times \mathrm{AM})\times (\frac{1}{2}\times \mathrm{PB}\times \mathrm{CN})\\ =\mathrm{ar}\left(\mathrm{APD}\right)\times \mathrm{ar}\left(\mathrm{BPC}\right)\\ =\mathrm{R}.\mathrm{H}.\mathrm{S}.\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Hence proved.}\end{array}$

Q.7

$\begin{array}{l}\mathrm{P}\mathrm{and}\mathrm{Q}\mathrm{are}\mathrm{respectively}\mathrm{the}\mathrm{mid}–\mathrm{points}\mathrm{of}\mathrm{sides}\mathrm{AB}\mathrm{and}\mathrm{BC}\\ \mathrm{of}\mathrm{a}\mathrm{triangle}\mathrm{ABC}\mathrm{and}\mathrm{R}\mathrm{is}\mathrm{the}\mathrm{mid}–\mathrm{point}\mathrm{of}\mathrm{AP},\mathrm{show}\mathrm{that}\\ \left(\mathrm{i}\right)\mathrm{ar}\left(\mathrm{PRQ}\right)=\frac{1}{2}\mathrm{ar}\left(\mathrm{ARC}\right)\\ \left(\mathrm{ii}\right)\mathrm{ar}\left(\mathrm{RQC}\right)=\frac{3}{8}\mathrm{ar}\left(\mathrm{ABC}\right)\\ \left(\mathrm{iii}\right)\mathrm{ar}\left(\mathrm{PBQ}\right)=\mathrm{ar}\left(\mathrm{ARC}\right)\end{array}$

Ans

$\begin{array}{l}\mathrm{Given}:\mathrm{In}\mathrm{\Delta ABC},\text{P and Q are the mid-points of AB and BC}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}respectively. R is the mid-point of AP.}\\ \mathrm{To}\text{prove:\hspace{0.17em}}\left(\mathrm{i}\right)\mathrm{ar}\left(\mathrm{PRQ}\right)=\frac{1}{2}\mathrm{ar}\left(\mathrm{ARC}\right)\\ \left(\mathrm{ii}\right)\mathrm{ar}\left(\mathrm{RQC}\right)=\frac{3}{8}\mathrm{ar}\left(\mathrm{ABC}\right)\\ \left(\mathrm{iii}\right)\mathrm{ar}\left(\mathrm{PBQ}\right)=\mathrm{ar}\left(\mathrm{ARC}\right)\\ \mathrm{Construction}:\text{Take S as a mid-point of AC and join PS.}\mathrm{Draw}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}CT}\parallel \mathrm{AP}\text{and produce PQ upto T as PQ}=\text{QT.Join PC.}\\ \text{Proof: In}\mathrm{\Delta }\text{ABC, P and Q are the mid-points of AB and BC points}\\ \text{of AB and BC respectively}.\text{Hence},\text{by using mid-point}\\ \text{theorem},\text{we}\mathrm{get}\\ \mathrm{PQ}\parallel \mathrm{AC}\text{and PQ}=\frac{1}{2}\mathrm{AC}\\ \Rightarrow \mathrm{PQ}\parallel \mathrm{AC}\text{and PQ}=\mathrm{AS}\\ \therefore \mathrm{PQSA}\text{is a parallelogram.\hspace{0.17em}PQSA is a parallelogram}.\text{We know}\\ \text{that diagonals of a parallelogram bisect it into equal areas of}\\ \text{triangles}.\\ \therefore \mathrm{ar}\left(\mathrm{\Delta PAS}\right)=\mathrm{ar}\left(\mathrm{\Delta PAQ}\right)=\mathrm{ar}\left(\mathrm{\Delta SQP}\right)=\mathrm{ar}\left(\mathrm{\Delta SQA}\right)\\ \mathrm{Now},\mathrm{PQ}\parallel \mathrm{SC}\text{and PQ}=\mathrm{SC}\\ \therefore \mathrm{PQCS}\text{is a parallelogram.\hspace{0.17em}PQCS is a parallelogram}.\text{We know}\\ \text{that diagonals of a parallelogram bisect it into equal areas of}\\ \text{triangles}.\\ \therefore \mathrm{ar}\left(\mathrm{\Delta PQS}\right)=\mathrm{ar}\left(\mathrm{\Delta QSC}\right)\\ \mathrm{Similarly},\text{QSCT and PSQB are also parallelogram.}\\ \therefore \mathrm{ar}\left(\mathrm{\Delta QSC}\right)=\mathrm{ar}\left(\mathrm{\Delta CTQ}\right)\\ \mathrm{and}\mathrm{ar}\left(\mathrm{\Delta PSQ}\right)=\mathrm{ar}\left(\mathrm{\Delta QBP}\right)\\ \mathrm{Thus},\mathrm{ar}\left(\mathrm{\Delta PAS}\right)=\mathrm{ar}\left(\mathrm{\Delta SQP}\right)=\mathrm{ar}\left(\mathrm{\Delta SQA}\right)=\mathrm{ar}\left(\mathrm{\Delta PAQ}\right)=\mathrm{ar}\left(\mathrm{\Delta QSC}\right)\\ =\mathrm{ar}\left(\mathrm{\Delta CTQ}\right)=\mathrm{ar}\left(\mathrm{\Delta CTQ}\right)...\left(\mathrm{i}\right)\\ \because \mathrm{ar}\left(\mathrm{\Delta ABC}\right)=\mathrm{ar}\left(\mathrm{\Delta PBQ}\right)+\mathrm{ar}\left(\mathrm{\Delta PAS}\right)+\mathrm{ar}\left(\mathrm{\Delta PQS}\right)+\mathrm{ar}\left(\mathrm{\Delta QSC}\right)\\ =\mathrm{ar}\left(\mathrm{\Delta PBQ}\right)+\mathrm{ar}\left(\mathrm{\Delta PBQ}\right)+\mathrm{ar}\left(\mathrm{\Delta PBQ}\right)+\mathrm{ar}\left(\mathrm{\Delta PBQ}\right)\\ =4\mathrm{ar}\left(\mathrm{\Delta PBQ}\right)\\ \Rightarrow \mathrm{ar}\left(\mathrm{\Delta PBQ}\right)=\frac{1}{4}\mathrm{ar}\left(\mathrm{\Delta ABC}\right)...\left(\mathrm{ii}\right)\\ \left(\mathrm{i}\right)\mathrm{In}\mathrm{\Delta PAQ},\text{ QR is median. So,}\\ \text{ar}\left(\mathrm{\Delta PQR}\right)=\frac{1}{2}\mathrm{ar}\left(\mathrm{\Delta PAQ}\right)\\ =\frac{1}{2}\mathrm{ar}\left(\mathrm{\Delta PBQ}\right)[\because \mathrm{QP}\text{is a median in}\mathrm{\Delta QBA}]\\ =\frac{1}{2}\times \frac{1}{4}\mathrm{ar}\left(\mathrm{\Delta ABC}\right)\left[\mathrm{From}\text{equation}\left(\mathrm{ii}\right)\right]\\ =\frac{1}{8}\mathrm{ar}\left(\mathrm{\Delta ABC}\right)...\left(\mathrm{iii}\right)\\ \left(\mathrm{ii}\right)\mathrm{In}\text{}\mathrm{\Delta ABC},\text{P and Q are the mid-points of AB and BC res}\\ \text{So, by mid-point theorem}\\ \text{PQ}=\frac{1}{2}\mathrm{AC}\text{and PQ}\parallel \text{AC}\\ \Rightarrow \text{2PQ}=\mathrm{AC}\text{and PQ}\parallel \text{AC}\\ \Rightarrow \text{PT}=\mathrm{AC}\text{and PT}\parallel \text{AC\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Q}\text{is mid-point of PT.}\right]\\ \Rightarrow \mathrm{PACT}\text{is a parallelogram.}\\ \mathrm{ar}\left(\mathrm{PACT}\right)=\mathrm{ar}\left(\mathrm{PACQ}\right)+\mathrm{ar}\left(\mathrm{\Delta QTC}\right)\\ =\mathrm{ar}\left(\mathrm{PACQ}\right)+\mathrm{ar}\left(\mathrm{\Delta PBQ}\right)\left[\mathrm{From}\text{equation}\left(\mathrm{i}\right)\right]\\ \mathrm{ar}\left(\mathrm{PACT}\right)=\mathrm{ar}\left(\mathrm{\Delta ABC}\right)...\left(\mathrm{iv}\right)\end{array}$

$\begin{array}{l}\mathrm{ar}\left(\mathrm{\Delta ARC}\right)=\frac{1}{2}\mathrm{ar}\left(\mathrm{\Delta PAC}\right)[\because \mathrm{CR}\text{is a median.}]\\ =\frac{1}{2}\times \frac{1}{2}\mathrm{ar}\left(\mathrm{\Delta ABC}\right)[\because \mathrm{CP}\text{is a median.}]\\ =\frac{1}{4}\mathrm{ar}\left(\mathrm{\Delta ABC}\right)\\ =2\times \frac{1}{8}\mathrm{ar}\left(\mathrm{\Delta ABC}\right)\\ \mathrm{ar}\left(\mathrm{\Delta ARC}\right)=2\text{\hspace{0.17em}ar}\left(\mathrm{\Delta PQR}\right)...\left(\mathrm{v}\right)\left[\mathrm{From}\text{equation}\left(\mathrm{iii}\right)\right]\\ \mathrm{Now},\text{}\\ \mathrm{ar}\left(\mathrm{\Delta RQC}\right)=\mathrm{ar}\left(\mathrm{PACQ}\right)-\mathrm{ar}\left(\mathrm{\Delta ARC}\right)-\text{ar}\left(\mathrm{\Delta PQR}\right)\\ =\mathrm{ar}\left(\mathrm{PACQ}\right)-2\text{\hspace{0.17em}ar}\left(\mathrm{\Delta PQR}\right)-\text{ar}\left(\mathrm{\Delta PQR}\right)\\ \left[\mathrm{From}\text{equation}\left(\mathrm{v}\right)\right]\\ =\mathrm{ar}\left(\mathrm{PACT}\right)-\mathrm{ar}\left(\mathrm{\Delta QTC}\right)-3\text{\hspace{0.17em}ar}\left(\mathrm{\Delta PQR}\right)\\ =\mathrm{ar}\left(\mathrm{\Delta ABC}\right)-\mathrm{ar}\left(\mathrm{\Delta QTC}\right)-3\text{\hspace{0.17em}ar}\left(\mathrm{\Delta PQR}\right)\\ \left[\mathrm{From}\text{equation}\left(\mathrm{iv}\right)\right]\\ =\mathrm{ar}\left(\mathrm{\Delta ABC}\right)-\mathrm{ar}\left(\mathrm{\Delta PBQ}\right)-3\text{\hspace{0.17em}ar}\left(\mathrm{\Delta PQR}\right)\\ \left[\mathrm{From}\text{equation}\left(\mathrm{i}\right)\right]\\ =\mathrm{ar}\left(\mathrm{\Delta ABC}\right)-\frac{1}{4}\mathrm{ar}\left(\mathrm{\Delta ABC}\right)-\frac{3}{8}\mathrm{ar}\left(\mathrm{\Delta ABC}\right)\\ \left[\mathrm{From}\text{equation}\left(\mathrm{ii}\right)\text{and equation}\left(\mathrm{iii}\right)\right]\\ \mathrm{ar}\left(\mathrm{\Delta RQC}\right)=\frac{3}{8}\mathrm{ar}\left(\mathrm{\Delta ABC}\right)\\ \left(\mathrm{iii}\right)\mathrm{R}.\mathrm{H}.\mathrm{S}.=\mathrm{ar}\left(\mathrm{\Delta ARC}\right)\\ =\mathrm{ar}\left(\mathrm{PACQ}\right)-\mathrm{ar}\left(\mathrm{\Delta PRQ}\right)-\mathrm{ar}\left(\mathrm{\Delta QRC}\right)\end{array}$

$\begin{array}{l}=\{\mathrm{ar}\left(\mathrm{PACT}\right)-\mathrm{ar}\left(\mathrm{\Delta QTC}\right)\}-\mathrm{ar}\left(\mathrm{\Delta PRQ}\right)-\mathrm{ar}\left(\mathrm{\Delta QRC}\right)\\ =\{\mathrm{ar}\left(\mathrm{\Delta ABC}\right)-\mathrm{ar}\left(\mathrm{\Delta PBQ}\right)\}-\mathrm{ar}\left(\mathrm{\Delta PRQ}\right)-\mathrm{ar}\left(\mathrm{\Delta QRC}\right)\\ \left[\mathrm{From}\text{equation\hspace{0.33em}}\left(\mathrm{i}\right)\mathrm{and}\left(\mathrm{iv}\right)\right]\\ =\{\mathrm{ar}\left(\mathrm{\Delta ABC}\right)-\frac{1}{4}\mathrm{ar}\left(\mathrm{\Delta ABC}\right)\}-\mathrm{ar}\left(\mathrm{\Delta PRQ}\right)-\mathrm{ar}\left(\mathrm{\Delta QRC}\right)\\ =\frac{3}{4}\mathrm{ar}\left(\mathrm{\Delta ABC}\right)-\frac{1}{8}\mathrm{ar}\left(\mathrm{\Delta ABC}\right)-\frac{3}{8}\mathrm{ar}\left(\mathrm{\Delta ABC}\right)\\ \left[\mathrm{From}\text{equation}\left(\mathrm{ii}\right)\text{and equation\hspace{0.33em}}\left(\mathrm{iii}\right)\right]\\ =\frac{1}{4}\mathrm{ar}\left(\mathrm{\Delta ABC}\right)\\ =\mathrm{ar}\left(\mathrm{\Delta PBQ}\right)\left[\mathrm{From}\text{equation\hspace{0.33em}}\left(\mathrm{ii}\right)\right]\\ =\mathrm{L}.\mathrm{H}.\mathrm{S}.\\ \mathrm{Thus},\\ \mathrm{ar}\left(\mathrm{\Delta PBQ}\right)=\mathrm{ar}\left(\mathrm{\Delta ARC}\right)\end{array}$

Q.8

$\begin{array}{l}\mathrm{In}\mathrm{the}\mathrm{figure}\mathrm{below},\mathrm{ABC}\mathrm{is}\mathrm{a}\mathrm{right}\mathrm{triangle}\mathrm{right}\mathrm{angled}\mathrm{at}\mathrm{A}.\mathrm{BCED},\\ \mathrm{ACFG}\mathrm{and}\mathrm{ABMN}\mathrm{are}\mathrm{squares}\mathrm{on}\mathrm{the}\mathrm{sides}\mathrm{BC},\mathrm{CA}\mathrm{and}\mathrm{AB}\\ \mathrm{respectively}.\mathrm{Line}\mathrm{segment}\mathrm{AX}\perp \mathrm{DE}\mathrm{meets}\mathrm{BC}\mathrm{at}\mathrm{Y}.\\ \mathrm{Show}\mathrm{that}:\\ \left(\mathrm{i}\right)\mathrm{\Delta MBC}\cong \mathrm{\Delta ABD}\\ \left(\mathrm{ii}\right)\mathrm{ar}\left(\mathrm{BYXD}\right)=2\mathrm{ar}\left(\mathrm{MBC}\right)\\ \left(\mathrm{iii}\right)\mathrm{ar}\left(\mathrm{BYXD}\right)=\mathrm{ar}\left(\mathrm{ABMN}\right)\\ \left(\mathrm{iv}\right)\mathrm{\Delta FCB}\cong \mathrm{\Delta ACE}\\ \left(\mathrm{v}\right)\mathrm{ar}\left(\mathrm{CYXE}\right)=2\mathrm{ar}\left(\mathrm{FCB}\right)\\ \left(\mathrm{vi}\right)\mathrm{ar}\left(\mathrm{CYXE}\right)=\mathrm{ar}\left(\mathrm{ACFG}\right)\\ \left(\mathrm{vii}\right)\mathrm{ar}\left(\mathrm{BCED}\right)=\mathrm{ar}\left(\mathrm{ABMN}\right)+\mathrm{ar}\left(\mathrm{ACFG}\right)\end{array}$

Ans

$\begin{array}{l}\left(\mathrm{i}\right)\mathrm{In}\mathrm{\Delta MBC}\text{and}\mathrm{\Delta ABD}\\ \mathrm{MB}=\mathrm{AB}\left[\mathrm{Sides}\text{of square ABMN}\right]\\ \angle \mathrm{MBC}=\angle \mathrm{ABD}[\angle \mathrm{MBC}=90\mathrm{°}+\angle \mathrm{ABC}=\angle \mathrm{ABD}]\\ \mathrm{BC}=\mathrm{BD}\left[\text{Sides of square BCED}\right]\\ \therefore \mathrm{\Delta MBC}\cong \mathrm{\Delta ABD}\left[\mathrm{By}\text{S.A.S.}\right]\\ \text{(ii)ΔMBC and parallelogram BYXD lie on the same base BD and}\\ \text{lie between same parallels BD and AX. So,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}ar}\left(\mathrm{\Delta MBC}\right)=\frac{1}{2}\mathrm{ar}\left(\mathrm{BYXD}\right)\\ \Rightarrow 2\text{\hspace{0.17em}ar}\left(\mathrm{\Delta MBC}\right)=\mathrm{ar}\left(\mathrm{BYXD}\right)\\ \mathrm{or}\mathrm{ar}\left(\mathrm{BYXD}\right)=2\text{\hspace{0.17em}ar}\left(\mathrm{\Delta MBC}\right)\\ \text{(iii)ΔMBC and parallelogram ABMN lie on the same base MB and}\\ \text{lie between same parallels MB and NC. So,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}ar}\left(\mathrm{\Delta MBC}\right)=\frac{1}{2}\mathrm{ar}\left(\mathrm{ABMN}\right)\\ \mathrm{But}\text{\hspace{0.17em}\hspace{0.17em}ar}\left(\mathrm{\Delta MBC}\right)=\frac{1}{2}\mathrm{ar}\left(\mathrm{BYXD}\right)\left[\mathrm{From}\left(\mathrm{ii}\right)\right]\\ \Rightarrow \frac{1}{2}\mathrm{ar}\left(\mathrm{ABMN}\right)=\frac{1}{2}\mathrm{ar}\left(\mathrm{BYXD}\right)\\ \Rightarrow \mathrm{ar}\left(\mathrm{ABMN}\right)=\mathrm{ar}\left(\mathrm{BYXD}\right)\\ \text{(iv)In}\mathrm{\Delta FCB}\text{and}\mathrm{\Delta ACE}\\ \mathrm{F}\mathrm{C}=\mathrm{AC}\left[\mathrm{Sides}\text{of square ACFG}\right]\\ \angle \mathrm{FCB}=\angle \mathrm{ACE}[\angle \mathrm{FCB}=90\mathrm{°}+\angle \mathrm{ACB}=\angle \mathrm{ACE}]\\ \mathrm{BC}=\mathrm{CE}\left[\mathrm{Sides}\text{of square BCED}\right]\\ \therefore \mathrm{\Delta FCB}\cong \mathrm{\Delta ACE}\left[\mathrm{By}\text{S.A.S.}\right]\\ \left(\mathrm{v}\right)\mathrm{S}\mathrm{ince},\mathrm{\Delta FCB}\cong \mathrm{\Delta ACE}\\ \text{So, ar}\left(\mathrm{\Delta FCB}\right)=\mathrm{ar}\left(\mathrm{\Delta ACE}\right)...\left(\mathrm{i}\right)\\ \mathrm{\Delta ACE}\text{and parallelogram CYXE lie on the same base CE and}\\ \text{lie between same parallels CE and AX. So,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}ar}\left(\mathrm{\Delta ACE}\right)=\frac{1}{2}\mathrm{ar}\left(\text{CYXE}\right)\\ \mathrm{or}\mathrm{ar}\left(\text{CYXE}\right)=2\text{\hspace{0.17em}ar}\left(\mathrm{\Delta ACE}\right)\\ \mathrm{or}\mathrm{ar}\left(\text{CYXE}\right)=2\text{\hspace{0.17em}ar}\left(\mathrm{\Delta FCB}\right)\left[\mathrm{From}\text{equation}\left(\mathrm{i}\right)\right]\\ \left(\mathrm{vi}\right)\mathrm{\Delta }\mathrm{FCB}\text{and parallelogram ACFG lie on the same base CF and}\\ \text{lie between same parallels CF and BG. So,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}ar}\left(\mathrm{\Delta FCB}\right)=\frac{1}{2}\mathrm{ar}\left(\mathrm{ACFG}\right)\\ \mathrm{or}\mathrm{ar}\left(\mathrm{ACFG}\right)=2\text{\hspace{0.17em}ar}\left(\mathrm{\Delta FCB}\right)\\ \because \mathrm{ar}\left(\text{CYXE}\right)=2\text{\hspace{0.17em}ar}\left(\mathrm{\Delta FCB}\right)\\ \mathrm{So},\text{\hspace{0.17em}}\mathrm{ar}\left(\text{CYXE}\right)=\mathrm{ar}\left(\mathrm{ACFG}\right)\\ \left(\mathrm{vii}\right)\mathrm{ar}\left(\mathrm{BCED}\right)=\mathrm{ar}\left(\mathrm{BYXD}\right)+\mathrm{ar}\left(\mathrm{CYXE}\right)\\ =\text{\hspace{0.17em}ar}\left(\mathrm{ABMN}\right)+\text{\hspace{0.17em}ar}\left(\mathrm{ACFG}\right)\left[\mathrm{From}\left(\mathrm{iii}\right)\text{and}\left(\mathrm{vi}\right)\right]\\ \mathrm{He}\text{nce proved.}\end{array}$