Question 14 Marks
In $\triangle\text{ABC},$ $D$ and $E$ are the mid-points of $AB$ and $AC$ respectively. Find the ratio of the areas of $\triangle\text{ADE}$ and $\triangle\text{ABC.}$
Answer
We have, $D$ and $E$ as the mid-points of $AB$ and $AC$
So, according to the mid-point theorem
$DE || BC$ and DE $=\frac{1}{2}\text{BC}\ \ ....(\text{i})$
In $\triangle\text{ADE and }\triangle\text{ABC}$
$\angle\text{A}=\angle\text{A}$ [Common]
$\angle\text{ADE}=\angle\text{B}$ [Corresponding angles]
Then, $\triangle\text{ADE}\sim\triangle\text{ABC}$ [By AA similarity]
By area of similar triangle theorem
$\frac{\text{ar}\ (\triangle\text{ADE})}{\text{ar }(\triangle\text{ABC})}=\frac{\text{DE}^2}{\text{BC}^2}$
$=\frac{\Big(\frac{1}{2}\text{BC}\Big)^2}{\text{BC}^2}\ \ \ [\text{From (i)}]$
$=\frac{\frac{1}{4}\text{BC}^2}{\text{BC}^2}$
$=\frac{1}{4}$ View full question & answer→Question 24 Marks
$ABCD$ is a rectangle. Points $M$ and $N$ are on $BD$ such that $\text{AM}\perp\text{BD}$ and $\text{CN}\perp\text{BD}.$ Prove that $BM^2+ BN^2= DM^2+ DN^2$.
AnswerGiven: A rectangle $ABCD$ where $\text{AM}\perp\text{BD}$ and $\text{CN}\perp\text{BD}.$
To prove: $BM^2+ BN^2= DM^2+ DN^2$
Proof:
Apply Pythagoras Theorem in $\triangle\text{AMB}$ and $\triangle\text{CND},$
$\mathrm{AB}^2=\mathrm{AM}^2+\mathrm{MB}^2 $
$ \mathrm{CD}^2=\mathrm{CN}^2+\mathrm{ND}^2$
Since $A B=C D, A M^2+M B^2=C N^2+N D^2$
$\Rightarrow \mathrm{AM}^2-\mathrm{CN}^2={ND}^2-\mathrm{MB}^2 \ldots \text { (i) }$
Again apply Pythagoras Theorem in $\triangle \mathrm{AMD}$ and $\triangle \mathrm{CNB}$,
$ A D^2=A M^2+M D^2 $
$ C B^2=C N^2+N B^2$
Since $A D=B C, A M^2+M D^2=C N^2+N B^2$
$\Rightarrow \mathrm{AM}^2-\mathrm{CN}^2=\mathrm{NB}^2-\mathrm{MD}^2....(ii)$
Equating (i) and (ii),
$N D^2-M B^2=N B^2-M D^2$
l.e., $\mathrm{BM}^2+\mathrm{BN}^2=\mathrm{DM}^2+\mathrm{DN}^2$
This proves the given relation. View full question & answer→Question 34 Marks
In the given figure, $AB || CD$, if $OA = 3x - 19, OB = x - 4, OC = x - 3$ and $OD = 4$, find $x.$
AnswerIn figure, $AB || CD$, if $OA = 3x - 19, OB = x - 4, OC = x - 3$ and $OD = 4$, find $x.$
we have,
$AB || CD$ and $OA = 3x - 19, OB = x - 4, OC = x - 3$ and $OD = 4$
Now, $\frac{\text{AO}}{\text{OC}}=\frac{\text{BO}}{\text{OD}}$
$\Rightarrow\frac{(3\text{x}-19)}{(\text{x}-3)}=\frac{(\text{x}-4)}{4}$
$ \Rightarrow 4(3 x-19)=(x-4)(x-3) $
$ \Rightarrow 12 x-76=x^2-4 x-3 x+12 $
$ \Rightarrow 12 x-76=x^2-7 x+12 $
$ \Rightarrow x^2-7 x-12 x=-76-12 $
$ \Rightarrow x^2-19 x=-88 $
$ \Rightarrow x^2-19 x+88=0 $
$ \Rightarrow x^2-11 x-8 x+88=0 $
$ \Rightarrow x(x-11)-8(x-11)=0 $
$ \Rightarrow(x-11)(x-8)=0 $
$ \Rightarrow x-11=0 \text { or } x-8=0 $
$ \Rightarrow x=11 \text { or } x=8$
$\text { Thus, } x=11,8$ View full question & answer→Question 44 Marks
A point $D$ is on the side $BC$ of an equilateral triangle $ABC$ such that $\text{DC}=\frac{1}{4}\text{BC}.$ Prove that $AD^2= 13\ CD^2$
AnswerGiven: In the equilateral $\triangle\text{ABC,}$ $D$ is a point on $BC$ such that $\text{DC}=\frac{1}{4}\text{BC}$
To prove: $AD^2= 13CD^2$
Construction: Draw $\text{AE}\perp\text{BC}$
Proof: $\because\text{AE}\perp\text{BC}$
$\therefore$ E is the mid-point of BC and D is a point such that $\text{DC}=\frac{1}{4}\text{BC}$
Now in right $\triangle\text{ADE,}$
$A D^2=A E^2+E D^2$ (Pythagoras Theorem)
and in right $\triangle \mathrm{AEC}$,
$A C^2=A E^2+E C^2 $
$\Rightarrow A E^2=A C^2-E C^2$
$ \therefore A D^2=A C^2-E C^2+E D^2$
$=\text{BC}^2-\Big(\frac{1}{2}\text{BC}\Big)^2+\text{ED}^2$ {$\because$ E is mid point of $BC\}$
$=\text{BC}^2-\frac{1}{4}\text{BC}^2+\text{ED}^2$
$=\frac{3}{4}\text{BC}^2+\text{ED}^2$
$=\frac{3}{4}(4\text{DC})^2+\text{DC}^2\ \ \ \ \ (\because\text{ED}=\text{DC})$
$=\frac{3}{4}\times16\text{DC}^2+\text{DC}^2$
$= 12DC^2+ DC^2= 13DC^2$
Hence $AD^2= 13DC^2$ or $13CD^2$
Hence proved. View full question & answer→Question 54 Marks
A guy wire attached to a vertical pole of height $18\ m$ is $24\ m$ long has a stake attached to the other end. How far from the base of pole should the stake be driven so that the wire will be taut?
AnswerWe will draw the figure from the given information as below,
Let AB be the vertical pole of length $18\ m$ and let the stake be at the point $C$ so the wire will be taut.
Therefore, we have $AB = 18m, AC = 24m$ and we have to find $BC.$
Now we will use Pythagoras theorem,
$A C^2=A B^2+B C^2$
Let us substitute the values we get,
$24^2=18^2+B C^2$
Subtracting 324 from both sides of the equation we get,
$ B C^2=576-324 $
$\mathrm{BC}^2=252$
We can rewrite the 252 as $36 \times 7$, therefore, our equation becomes,
$B C^2=36 \times 7$
Now we will take the square root,
$\text{BC}=6\times\sqrt{7}$
Therefore, the stake should be $6\sqrt{7}\text{ m}$ far from the base of the pole so that the wire will be taut. View full question & answer→Question 64 Marks
The corresponding altitudes of two similar triangles are $6\ cm$ and $9\ cm$ respectively. Find the ratio of their areas.
Answer
We have,
$\triangle\text{ABC}\sim\triangle\text{PQR}$
$AD = 6cm$
And, $PS = 9cm$
By area of similar triangle theorem
$\frac{\text{Area}(\triangle\text{ABC})}{\text{Area}(\triangle\text{PQR})}=\frac{\text{AB}^2}{\text{PQ}^2}\ ...(\text{i})$
In $\triangle\text{ABD}$ and $\triangle\text{PQS}$
$\angle\text{B}=\angle\text{Q}\ \ \ \ \ [\triangle\text{ABC}\sim\triangle\text{PQR}]$
$\angle\text{ADB}=\angle\text{PSQ}$ [Each 90°]
Then, $\triangle\text{ABD}\sim\triangle\text{PQS}$ [by AA similarity]
$\therefore\frac{\text{AB}}{\text{PQ}}=\frac{\text{AD}}{\text{PS}}$ [Corresponding parts of similar $\triangle$ are proportional]
$\Rightarrow\frac{\text{AB}}{\text{PQ}}=\frac{6}{9}$
$\Rightarrow\frac{\text{AB}}{\text{PQ}}=\frac{2}{3}\ \ \ ....(\text{ii})$
Compare equations (i) and (ii)
$\frac{\text{Area}(\triangle\text{ABC})}{\text{Area}(\triangle\text{PQR})}=\Big(\frac{2}{3}\Big)^2=\frac{4}{9}$ View full question & answer→Question 74 Marks
$AD$ is an altitude of an equilateral triangle $ABC$. On $AD$ as base, another equilateral triangle $ADE$ is constructed.
Prove that Area $(\triangle\text{ADE)}$ : Area $(\triangle\text{ABC)} = 3 : 4.$
Answer
We have,
$\triangle\text{ABC}$ is an equilateral triangle
Then,$ AB = BC = AC$
Let, $AB = BC = AC = 2x$
Since, $\text{AD}\perp\text{BC}$ then $BD = DC = x$
$A B^2=(2 x)^2-(x)^2 $
$ \Rightarrow A D^2=4 x^2-x^2=3 x^2$
$\Rightarrow\text{AD}=\sqrt{3}\text{x cm}$
Since, $\triangle\text{ABC}$ and $\triangle\text{ADE}$ both are equilateral triangles then they are equiangular
$\therefore\triangle\text{ABC}\sim\triangle\text{ADE}$ [By AA similarity]
By area of similar triangle theorem
$\frac{\text{ar}(\triangle\text{ADE)}}{\text{ar}(\triangle\text{ABC})}=\frac{\text{AD}^2}{\text{AB}^2}$
$=\frac{(\sqrt{3}\text{x})^2}{(2\text{x})^2}$
$=\frac{3\text{x}^2}{4\text{x}^2}$
$=\frac{3}{4}$ View full question & answer→Question 84 Marks
Nazima is fly Ashing in a stream. The tip of her Ashing rod is $1.8\ m$ above the surface of the water and the fly at the end of the string rests on the water $3.6\ m$ away and $2.4\ m$ from a point directly under the tip of the rod. Assuming that her string (from the tip of her rod to the fly) is taut, how much string does she have out? If she pulls the string at the rate of $5\ cm$ per second, what will the horizontal distance of the fly from her after $12$ seconds.
AnswerHeight of the rod from stream level $=1.8 \mathrm{~m}$ and of string from the point under the tip of rod $=2.4 \mathrm{~m}$
Let the length of string $=x$
$ x^2=(1.8)^2+(2.4)^2=3.24+5.76=9.00=(3.0)^2 $
$x=3.0$
Length of string $=3 \mathrm{~m}$
Rate of pulling the string $=5 \mathrm{~cm}$ per second
Distance covered in $12$ seconds $=5 \times 12=60 \mathrm{~cm}$.
At this stage, length of string $=3.0-0.6=2.4 \mathrm{~m}$
Height $=1.8 \mathrm{~m}$
Let base $=y$ then
$(2.4)^2=y^2+(1.8)^2 $
$\Rightarrow 5.76=y^2+3.24 $
$\Rightarrow y^2=5.76-3.24=2.52 $
$y=1.59$
and distance from her $=1.59+1.2=2.79 \mathrm{~m}$.
View full question & answer→Question 94 Marks
The diagonals of quadrilateral $ABCD$ intersect at $O$. Prove that $\frac{\text{ar}(\triangle\text{ACB})}{\text{ar}(\triangle\text{ACD})}=\frac{\text{BO}}{\text{DO}}.$
AnswerWe are given the following quadrilateral with $O$ as the intersection point of diagonals
To Prove: $\frac{\text{ar}(\triangle\text{ACB})}{\text{ar}(\triangle\text{ACD})}=\frac{\text{BO}}{\text{DO}}$
Given: $ACB$ and $ACD$ are two triangles on the same base $AC$
Consider h as the distance between two parallel sides
Now we see that the height of these two triangles $ACB$ and $ACD$ are same and are equal to h
So $\frac{\text{ar}(\triangle\text{ACB})}{\text{ar}(\triangle\text{ACD})}=\frac{\frac{1}{2}\times\text{AB}\times\text{h}}{\frac{1}{2}\times\text{CD}\times\text{h}}$
$=\frac{\text{AB}}{\text{CD}}\ \ \ ....(1)$
Now consider the triangles $AOB$ and $COD$ in which
$\angle\text{AOB}=\angle\text{COD}$
$\angle\text{ABO}=\angle\text{ODC}$ (alternative angle)
$\angle\text{BAO}=\angle\text{DCA}$ (alternative angle)
Therefore, $\triangle\text{ODC}\sim\triangle\text{OBA}$
$\Rightarrow\frac{\text{AO}}{\text{OC}}=\frac{\text{BO}}{\text{DO}}=\frac{\text{AB}}{\text{CD}}$
$\Rightarrow\frac{\text{BO}}{\text{DO}}=\frac{\text{AB}}{\text{CD}}\ \ \ ....(2)$
From equation (1) and (2) we get
$\frac{\text{ar}(\triangle\text{ACB})}{\text{ar}(\triangle\text{ACD})}=\frac{\text{BO}}{\text{DO}}$
Hence prove that $\frac{\text{ar}(\triangle\text{ACB})}{\text{ar}(\triangle\text{ACD})}=\frac{\text{BO}}{\text{DO}}$ View full question & answer→Question 104 Marks
In an isosceles $\triangle\text{ABC,}$ the base $AB$ is produced both the ways to $P$ and $Q$ such that $AP × BQ = AC^2$. Prove that $\triangle\text{APC}\sim \triangle\text{BCQ}.$
Answer
Given: In $\triangle\text{ABC},$ CA = CB and $AP × BQ = AC^2$
To prove: $\triangle\text{APC}\sim\triangle\text{BCQ}$
Proof:
$\text{AP}\times\text{BQ}=\text{AC}^2\ \ \ \ \ [\text{Given}]$
$\Rightarrow\text{AP}\times\text{BQ}=\text{AC}\times\text{AC}$
$\Rightarrow\text{AP}\times\text{BQ}=\text{AC}\times\text{BC}\ \ \ \ [\text{AC = BC given}]$
$\Rightarrow\frac{\text{AP}}{\text{BC}}=\frac{\text{AC}}{\text{BQ}}\ \ ....(\text{i})$
Since, $CA = CB $[Given]
Then, $\angle\text{CAB}=\angle\text{CBA}\ .....(\text{ii})$ [Opposite angles to equal sides]
Now, $\angle\text{CAB}+\angle\text{CAP}=180^\circ\ .....(\text{iii})$ [Linear pair of angles]
And, $\angle\text{CBA}+\angle\text{CBQ}=180^\circ\ .....(\text{iv})$ [Linear pair of angles]
Compare equation (ii) (iii) & (iv)
$\angle\text{CAP}=\angle\text{CBQ}\ \ \ ....(\text{v})$
In $\triangle\text{APC}$ and $\triangle\text{BCQ}$
$\angle\text{CAP}=\angle\text{CBQ}$ [From (V)]
$\frac{\text{AP}}{\text{BC}}=\frac{\text{AC}}{\text{BQ}}$ [From (i)]
Then, $\triangle\text{APC}\sim\triangle\text{BCQ}$ [By SAS similarity] View full question & answer→Question 114 Marks
In $\triangle\text{ABC},\ \angle\text{ABC}=135^\circ.$ Prove that $AC^2= AB^2+ BC^2+ 4 \text{ar}(\triangle\text{ABC}).$
AnswerWe have the following figure.
Here $\triangle\text{ADB}$ is a right triangle right angled at D. therefore by Pythagoras theorem we have
$A B^2=A D^2+D B^2$
Again $\triangle \mathrm{ADC}$ is a right triangle right angled at D .
Therefore, by Pythagoras theorem, we have
$ A C^2=A D^2+D C^2 $
$ A C^2=A D^2+(D B+B C)^2 $
$ A C^2=A D^2+D B^2+B C^2+2 \times B C \times B D$
Since angle $A B D$ is $45^{\circ}$ and therefore angle $B A D$ is also $45^{\circ}$.
Hence $A B=D B$
So,
$A C^2=A D^2+D B^2+B C^2+2 B C \times A D$
$= \text{AD}^2 + \text{DB}^2 + \text{BC}^2 + 2\times2\times\frac{1}{2}\text{BC}\times\text{AD}$
$= \text{AD}^2 + \text{DB}^2 + \text{BC}^2 + 4\text{ar}(\triangle\text{ABC})$
Since $A B^2=A D^2+D B^2$
So, $\text{AC}^2 = \text{AB}^2 + \text{BC}^2 + 4\text{ar}(\triangle\text{ABC})$
Hence we have proved that $\text{AC}^2 = \text{AB}^2 + \text{BC}^2 + 4\text{ar}(\triangle\text{ABC})$ View full question & answer→Question 124 Marks
$A B C D$ is a parallelogram and $A P Q$ is a straight line meeting $B C$ at $P$ and $D C$ produced at $Q$. Prove that the rectangle obtained by $B P$ and $D Q$ is equal to the rectangle contained by $A B$ and $B C$.
Answer
Given: $ABCD$ is a parallelogram
To prove: $BP \times DQ = AB \times BC$
Proof: In $\triangle\text{ABP}$ and $\triangle\text{QDA}$
$\angle\text{B}=\angle\text{D}$ [Opposite angles of parallelogram]
$\angle\text{BAP}=\angle\text{AQD}$ [Alternate interior angles]
Then, $\triangle\text{ABP}\sim\triangle\text{QDA}$ [By AA similarity]
$\therefore\frac{\text{AB}}{\text{QD}}=\frac{\text{BP}}{\text{DA}}$ [Corresponding parts of similar $\triangle$ are proportional]
But, $DA = BC$ [Opposite sides of parallelogram]
Then, $\frac{\text{AB}}{\text{QD}}=\frac{\text{BP}}{\text{BC}}$
$\Rightarrow\text{AB}\times\text{BC}=\text{QD}\times\text{BP}$ View full question & answer→Question 134 Marks
$D, E$ and $F$ are the points on sides $ BC,CA$ and $AB$ respectively of $\triangle\text{ABC}$ such that $AD$ bisects $\angle\text{A},$ $BE$ bisects $\angle\text{B}$ and $CF$ bisects $\angle\text{C}.$ If $AB = 5cm, BC = 8cm$ and $CA = 4cm$, determine $AF, CE$ and $BD.$
AnswerIn $\triangle\text{ABC},$
AD, BE and CE bisects $\angle\text{A},\angle\text{B}\ \text{and}\ \angle\text{C}$ respectively.
AB = 5cm, BC = 8cm and CA = 4cm
$\because$ AD bisects $\angle\text{A}$ so,
$\frac{\text{AB}}{\text{BD}}=\frac{\text{AC}}{\text{CD}}$
$\Rightarrow\frac{\text{AB}}{\text{BD}}=\frac{\text{AC}}{\text{BC}-\text{BD}}$
$\Rightarrow\frac{5}{\text{BD}}=\frac{4}{8-\text{BD}}$
$\Rightarrow40-5\text{BD}=4\text{BD}$
$\Rightarrow9\text{BD}=40$
$\Rightarrow\text{BD}=\frac{40}{9}=4.44\text{cm}$
and, BE bisects $\angle\text{B}$ so,
$\Rightarrow\frac{\text{BC}}{\text{CE}}=\frac{\text{AB}}{\text{AE}}$
$\Rightarrow\frac{\text{BC}}{\text{CE}}=\frac{\text{AB}}{\text{AC}-\text{CE}}$
$\Rightarrow\frac{8}{\text{CE}}=\frac{5}{4-\text{CE}}$
$\Rightarrow32-8\text{CE}=5\text{CE}$
$\Rightarrow13\text{CE}=32$
$\Rightarrow\text{CE}=\frac{32}{13}$
$\Rightarrow\text{CE}=2.46\text{cm}$
and, CF bisects $\angle\text{C}$ so,
$\Rightarrow\frac{\text{BC}}{\text{BF}}=\frac{\text{AC}}{\text{AF}}$
$\Rightarrow\frac{\text{BC}}{\text{AB}-\text{AF}}=\frac{\text{AC}}{\text{AF}}$
$\Rightarrow\frac{8}{5-\text{AF}}=\frac{4}{\text{AF}}$
$\Rightarrow8\text{AF}=20-4\text{AF}$
$\Rightarrow12\text{AF}=20$
$\Rightarrow\text{AF}=\frac{20}{12}$
$\Rightarrow\text{AF}=\frac{5}{3}$
$\Rightarrow\text{AF}=1.66$
$\Rightarrow\text{AF}=1.7\text{cm}$
View full question & answer→Question 144 Marks
In $\triangle\text{ABC}, AD$ is median. Prove that $AB^2+ AC^2= 2AD^2+ 2DC^2$.
AnswerGiven: In $\triangle\text{ABC}, AD$ is the median of $BC$
To prove: $AB^2+ AC^2= 2AD^2+ 2DC^2$
Construction: Draw $\text{AE}\perp\text{BC}$
Proof: In $\triangle\text{ABE},$
$AB^2= AE^2+ BE^2....(i)$(Pythagoras Theorem)
Similarly in right $\triangle\text{ACE},$
$AC^2= AE^2+ EC^2.....(ii)$
and in right $\triangle\text{AED},$
$AD^2= AE^2+ ED^2....(iii)$
Adding $(i)$ and $(ii)$
$ A B^2+A C^2=A E^2+B E^2+A E^2+E C^2 $
$ =2 A E^2+(B D-E D)^2+(D C+E D)^2 $
$ =2 A E^2+B D^2+E D^2-2 B D \times E D+D C^2+E D^2+2 B C \times E D $
$ =2 A E^2+B D^2+2 E D^2+B D^2+2 B D \times E D-2 B D \times E D(\because D \text { is mid points }) $
$ =2 A E^2+2 B D^2+2 E D^2 $
$ =2\left(A E^2+E D^2\right)+2 D C^2(\because D C=B D) $
$ =2 A D^2+2 D C^2\{\text { From (iii) }\}$
Hence proved. View full question & answer→Question 154 Marks
In the given figure, if $AB || CD$, find the value of $x.$
AnswerIn the figure, $AB || CD$
The diagonals of a trapezium divides each other proportionally
$\therefore\frac{\text{AO}}{\text{OC}}=\frac{\text{BO}}{\text{OD}}\Rightarrow\frac{4}{4\text{x}-2}=\frac{\text{x}+1}{2\text{x}+4}$
$⇒ 4x(2x + 4) = (4x - 2)(x + 1)$ (by cross multiplication)
$ \Rightarrow 8 x+16=4 x^2+4 x-2 x-2$
$\Rightarrow 8 x+16=4 x^2+2 x-2 $
$ \Rightarrow 4 x^2+2 x-2-8 x-16=0 $
$ \Rightarrow 4 x^2-6 x-18=0 $
$ \Rightarrow 2 x^2-3 x-9=0$
$⇒ 2x^2 - 6x + 3x - 9 = 0$
$\begin{cases}\because-9\times2=-18\\\therefore-18=-6\times3\\-3=-6+3\end{cases}$
$⇒ 2x(x - 3) + 3(x - 3) = 0$
$⇒ (x - 3)(2x + 3) = 0$
Either $x - 3 = 0$, then $x = 3$
or 2x + 3 = 0, then 2x = -3 $\Rightarrow\text{x}=\frac{-3}{2}$
But it is negative
$\therefore\text{x}=3$
View full question & answer→Question 164 Marks
In the given figure,$ l || m$
- Name three pairs of similar triangle with proper correspondence; write similarities.
- prove that $\frac{\text{AB}}{\text{PQ}}=\frac{\text{AC}}{\text{PR}}=\frac{\text{BC}}{\text{RQ}}$
AnswerIn the figure,$ l || m$
$\therefore\angle\text{A}=\angle\text{P}$ and $\angle\text{C}=\angle\text{R}$ (Alternate angles)
$\angle\text{AKB}=\angle\text{PKQ}$
$\angle\text{CKB}=\angle\text{RKQ}$ (Vertically opposite angles)
$\therefore\triangle\text{ABK}\sim\triangle\text{PQR}\ \ \ ... (\text{i})$ (AA criterion)
and $\triangle\text{CBK}\sim\triangle\text{RQK}\ \ \ ... (\text{ii})$ (AA criterion)
and $\triangle\text{AKC}\sim\triangle\text{PKR}\ \ \ ... (\text{iii})$ (AAA criterion)
(i) From (ii)
$\because\triangle\text{AKB}\sim\triangle\text{PKQ}$
$\therefore\frac{\text{AB}}{\text{PQ}}=\frac{\text{BK}}{\text{QK}}=\frac{\text{AK}}{\text{PK}}\ \ ....(\text{iv})$
and from (ii)
$\because\frac{\text{BC}}{\text{RQ}}=\frac{\text{BK}}{\text{QK}}=\frac{\text{CK}}{\text{RK}}\ \ ....(\text{v})$
and from (iii)
$\triangle\text{AKC}=\angle\text{PKR}$
$\therefore\frac{\text{AK}}{\text{PK}}=\frac{\text{CK}}{\text{RK}}=\frac{\text{AC}}{\text{PR}}\ \ ....(\text{vi})$
From (iv), (v) and (vi)
$\frac{\text{AB}}{\text{PQ}}=\frac{\text{AC}}{\text{PR}}=\frac{\text{BC}}{\text{RQ}}$
Hence proved. View full question & answer→Question 174 Marks
In $\triangle\text{ABC},$ AD and BE are altitudes. Prove that: $\frac{\text{ar}(\triangle\text{DEC})}{\text{ar}(\triangle\text{ABC})}=\frac{\text{DC}^2}{\text{AC}^2}$
Answer
Given: $\triangle\text{ABC}$ in which AD and BE are altitudes on sides BC and AC respectively.
Since $\angle\text{ADB}=\angle\text{AEB}=90^\circ,$ there must be a circle passing through point D and E having AB as diameter.
We also know that, angle in a semi-circle is a right angle.
Now, join DE.
So ABDE is a cyclic quadrilateral with AB being the diameter of the circle.
$\angle\text{A}+\angle\text{BDE}=180^\circ$ [Opposite angles in a cyclic quadrilateral are supplementary]
$\Rightarrow\angle\text{A}+(\angle\text{BDA}+\angle\text{ADE})=180^\circ$
$\Rightarrow\angle\text{BDA}+\angle\text{ADE}=180^\circ-\angle\text{A}\ ....(1)$
Again
$\angle\text{BDA}+\angle\text{ADC}=180^\circ$ [Linear pair]
$\Rightarrow\angle\text{BDA}+\angle\text{ADE}+\angle\text{EDC}=180^\circ$
$\Rightarrow\angle\text{BDA}+\angle\text{ADE}=180^\circ-\angle\text{EDC}\ \ ...(2)$
Equating (1) and (2), we get
$180^\circ-\angle\text{A}=180^\circ-\angle\text{EDC}$
$\Rightarrow\angle\text{A}=\angle\text{EDC}$
Similarly, $\angle\text{B}=\angle\text{CED}$
Now, in $\triangle\text{ABC}$ and $\triangle\text{DEC}$, we have
$\angle\text{A}=\angle\text{EDC}$
$\angle\text{B}=\angle\text{CED}$
$\angle\text{C}=\angle\text{C}$
$\therefore\triangle\text{ABC}\sim\triangle\text{DEC}$
$\Rightarrow\frac{\text{Area of }\triangle\text{DEC}}{\text{Area of }\triangle\text{ABC}}=\Big(\frac{\text{DC}}{\text{AC}}\Big)^2$ View full question & answer→Question 184 Marks
$ABCD$ is a square. $F$ is the mid-point of $AB. BE$ is one third of $BC$. If the area of $\triangle\text{FBE}=108\text{cm}^2,$ find the length of $AC.$
Answer
We have,
$\text{BE}=\frac{1}{3}\text{BC},\ \ \text{AF}=\text{FB}=\frac{1}{2}\text{AB}$
$\text{AB} = \text{BC} = \text{AC} = \text{AD}$ (sides of square)
In $\triangle\text{FBE},$
Area $(\triangle\text{FBE})=\frac{1}{2}\times\text{EB}\times\text{FB}$
$=\frac{1}{2}\times\frac{1}{3}\text{AC}\times\frac{1}{2}\text{AB}=\frac{1}{12}\text{AC}\times\text{AB}$
$108=\frac{1}{12}\text{AC}\times\text{AB}$
$\text{AC}\times\text{AB}=108\times12\text{cm}^2\ \ ...(1)$
Area (square ABCD) = (side)$^2$
$=\text{AB}^2$
$=\text{AB}\times\text{AC}\ \ ...(2)$
From $(1)$ and $(2)$
$108\times12 = \text{AB}^2$
$\text{AB}=36\text{cm}$
Now. in $\triangle\text{ABC},$
$\text{AC}^2=\text{AB}^2+\text{BC}^2$
$\text{AC}^2=(36)^2+(36)^2$
$\text{AC}^2=2\times(36)^2$
$\text{AC}=\sqrt{2(36)^2}$
$\text{AC}=36\sqrt{2}\text{cm}$
$\text{AC}=36\times1.414$
$\text{AC}=50.904$
Thus, $\text{AC} = 50.90\text{cm.}$ View full question & answer→Question 194 Marks
$ABCD$ is a quadrilateral in which $AD = BC$. If $P, Q, R, S$ be the mid-points of $AB, AC, CD$ and $BD$ respectively, show that $PQRS$ is a rhombus.
AnswerGiven: In quadrilateral $ABCD, AD = BC$
$P, Q, R$ and $S$ are the mid points of $AB, AC, CD$ and $AD$ respectively
$PQ, QR, RS, SP$ are joined
To prove: $PQRS$ is a rhombus
Proof: In $\triangle\text{ABC},$
$P$ and $Q$ are mid points of AB and $AC$
$\therefore\text{PQ}||\text{BC and }\frac{1}{2}\text{BC}\ ....(\text{i})$
Similarly in $\triangle\text{ABD},$
$P$ and $S$ are mid points of $AB$ and $BD$
$\therefore\text{PS}||\text{AD and }\frac{1}{2}\text{AD}\ ....(\text{ii})$
From (i) and (ii)
$\text{PQ} = \text{PS}\ \ \ \ \ \ \ \ (\text{AD} = \text{BC})$
Similarly $\text{QR}=\text{SR}=\Big(\frac{1}{2}\text{BC or AD}\Big)$
$\therefore PQ = QR = RS = PS$
$\therefore PQRS$ is a rhombus. View full question & answer→Question 204 Marks
A vertical stick $10\ cm$ long casts a shadow $8\ cm$ long. At the same time a tower casts a shadow $30\ m$ long. Determine the height of the tower.
Answer
Length of stick $= 10cm$
Length of shadow of stick $= 8cm$
Length of shadow of tower $= h cm$
In $\triangle\text{ABC}$ and $\triangle\text{PQR}$
$\angle\text{B}=\angle\text{Q}=90^\circ$
And, $\angle\text{C}=\angle\text{R}$ [Angular elevation of sum]
Then, $\triangle\text{ABC}\sim\triangle\text{PQR}$ [By AA similarity]
$\therefore\frac{\text{AB}}{\text{PQ}}=\frac{\text{BC}}{\text{QR}}$
$\Rightarrow\frac{10\text{cm}}{8\text{cm}}=\frac{\text{h cm}}{3000}$
$\Rightarrow\text{h}=\frac{10}{8}\times3000=3750\text{cm}=37.5\text{m}$ View full question & answer→Question 214 Marks
In the figures given below, an altitude is drawn to the hypotenuse by a right-angled triangle. The length of different line-segment are marked in figure. Determine $x, y, z$ in case.
Answer$\triangle\text{ABC}$ is right angled triangle right angled at $B$
$A B^2+B C^2=A C^2 $
$ x^2+z^2=(4+5)^2$
$ x^2+z^2=9^2$
$ x^2+z^2=81 \ldots \ldots . .(i)$
$\triangle \mathrm{BDA}$ is right triangle right angled at $D$
$ \mathrm{BD} 2+\mathrm{AD} 2=\mathrm{AB} 2 $
$ \mathrm{y}^2+4^2=\mathrm{x}^2 $
$ \mathrm{y}^2+16=\mathrm{x}^2$
$16=\mathrm{x}^2-\mathrm{y}^2 \ldots \ldots . . .(\mathrm{ii})$
$\triangle \mathrm{BDC}$ is right triangle right angled at $D$
$\mathrm{BD}^2+\mathrm{DC}^2=\mathrm{BC}^2 $
$ \mathrm{y}^2+25=\mathrm{z}^2$
$ 25=\mathrm{z}^2-\mathrm{y}^2 \ldots . . . \text { (iii) }$
By canceling equation (i) and (ii) by elimination method, we get
y canceling and by elimination method we get,
$\ \text{z}^2-\text{y}^2=25\\\underline{\ \text{z}^2+\text{y}^2=65}\\2\text{z}^2\ \ \ \ \ \ \ \ =90$
$\text{z}^2=\frac{90}{2}$
$\text{z}^2=45$
$\text{z}=\sqrt{45}$
$\text{z}=\sqrt{3\times3\times5}$
$\text{z}=3\sqrt{5}$
Now, substituting $z^2=45$ in equation (iv) we get
$ y^2+z^2=65$
$ y^2+45=65 $
$ y^2=65-45$
$ y^2=20$
$ y=\sqrt{20}$
$ y=\sqrt{2 \times 2 \times 5}$
$y=2 \sqrt{5}$
Now, substituting $\mathrm{y}^2=20$ in equation (ii) we get,
$ x^2-y^2=16 $
$ x^2-20=16 $
$ x^2=16+20 $
$x^2=36$
$\text{x}=\sqrt{36}$
$\text{x}=\sqrt{6\times6}$
$\text{x}=6$
Hence the values of $\text{x, y, z}$ is $6,2\sqrt{5},3\sqrt{5}.$ View full question & answer→Question 224 Marks
In the given figure, if $AB || CD$. find the value of $x.$
AnswerIn the below fig., If $AB || CD$, find the value of$ x.$
$\Rightarrow\frac{3\text{x}-1}{5\text{x}-3}=\frac{2\text{x}+1}{6\text{x}-5}$
$ \Rightarrow(3 x-1)(6 x-5)=(2 x+1)(5 x-3)$
$ \Rightarrow 3 x(6 x-5)-1(6 x-5)=2 x(5 x-3)+1(5 x-3)$
$ \Rightarrow 18 x^2-15 x-6 x+5=10 x^2-6 x+5 x-3$
$ \Rightarrow 8 x^2-20 x+8=0$
$ \Rightarrow 4\left(2 x^2-5 x+2\right)=0$
$ \Rightarrow 2 x^2-4 x-1 x+2=0$
$ \Rightarrow 2 x(x-2)-1(x-2)=0$
$ \Rightarrow(2 x-1)(x-2)=0$
$ \Rightarrow 2 x-1=0 \text { or } x-2=0$
$\Rightarrow\text{x}=\frac{1}{2}\ \text{or x}=2$
$\text{x}=\frac{1}{2}$ is not possible, because, $OC = 5x - 3$
$=5\Big(\frac{1}{2}\Big)-3$
$=\frac{5-6}{2}=-\frac{1}{2}$ View full question & answer→Question 234 Marks
In $\triangle\text{ABC},\ \angle\text{A}$ is obtuse, $\text{PB}\perp\text{AC}$ and $\text{QC}\perp\text{AB}$ Prove that:
$BC^2= (AC × CP + AB × BQ)$
Answer
In $\triangle\text{BPC},$ by pythagoras theorem
$ B C^2=B P^2+P C^2 $
$ \Rightarrow B C^2=A B^2-A P^2+(A P+A C)^2[B y \text { pythagoras theorem }] $
$ \Rightarrow B C^2=A B^2+A C^2+2 A P \times A C \ldots . \text { (ii) }$
In $\triangle \mathrm{BQC}$, by pythagoras theorem,
$B C^2=C Q^2+B Q^2 $
$ \Rightarrow \mathrm{BC}^2=\mathrm{AC}^2-\mathrm{AQ}{ }^2+(\mathrm{AB}+\mathrm{AQ})^2[\mathrm{By} \text { pythagoras theorem }] $
$ \Rightarrow B C^2=A C^2-A Q^2+A B^2+A Q^2+2 A B \times A Q $
$ \Rightarrow B C^2=A C^2+A B^2+2 A B \times A Q$
Add equations (ii) \$ (iii)
$ 2 B C^2=2 A C^2+2 A B^2+2 A P \times A C+2 A B \times A Q $
$ \Rightarrow 2 B C^2=2 A C^2+2 A B^2+2 A P \times A C+2 A B \times A Q $
$ \Rightarrow 2 B C^2=2 A C[A C+A P]+A B[A B+A Q] $
$ \Rightarrow 2 B C^2=2 A C \times P C+2 A B \times B Q $
$ \Rightarrow B C^2=A C \times P C+A B \times B Q[\text { Divide by } 2]$ View full question & answer→Question 244 Marks
In the figures given below, an altitude is drawn to the hypotenuse by a right-angled triangle. The length of different line-segment are marked in figure. Determine$ x, y, z$ in case.
Answer$\triangle\text{PQR}$ is a right triangle, right angled at $Q$
$6 + z^2 = (4 + x)^2$
$36 + z^2 = 16 + x^2 + 8x$
$z^2 - x^2 - 8x = 16 - 36$
$z^2 - x^2 - 8x = -20 ........(i)$
$\triangle\text{QSP}$ is a right triangle right angled at S
$QS^2 + PS^2 = PQ^2$
$y^2 + 4^2 = 6^2$
$y^2 + 16 = 36$
$y^2 = 36 - 16$
$y^2 = 20$
$\text{y}=\sqrt{20}$
$\text{y}=\sqrt{2\times2\times5}$
$\text{y}=2\sqrt{5}$
$\triangle\text{QSR}$ is a right triangle right angled at S
$QS^2+ RS^2 = QR^2$
$y^2 + x^2 = z^2 ......(ii)$
Now substituting $y^2 + x^2 = z^2$ in equation $(i)$ we get
$y^2 + x^2 - x^2 - 8x = -20$
$y^2 + x^2 - x^2 - 8x = -20$
$y^2 - 8x = -20 .......(iii)$
Now substituting y2 = 20 in equation $(iii)$ we get
$y^2 - 8x = -20$
$20 - 8x = -20$
$-8x = -20 - 20$
$-8x = -40$
$\text{x}=\frac{40}{8}$
$\text{x}=5$
Now substituting x = 5 and y2 = 20 in equation $(ii)$ we get
$y^2 + x^2 = z^2$
$20 + 52 = z^2$
$20 + 25 = z^2$
$45 = z2$
$\sqrt{3\times3\times5}=\text{z}^2$
$3\sqrt{5}=\text{z}$
Hence the value of x, y and z are $5,2\sqrt{5},3\sqrt{5}.$ View full question & answer→Question 254 Marks
In $\triangle\text{ABC}, P$ and $Q$ are points on sides $AB$ and $AC$ respectively such that $PQ || BC$. If $AP = 4cm, PB = 6cm$ and $PQ = 3cm$, determine $BC.$
AnswerIn $\triangle\text{ABC}, P$ and $Q$ are points on A$B$ and $AC$ respectively such that $PQ || BC AP = 4cm, PB = 6cm, PQ = 3cm$
Let $BC = x cm$
$\because PQ || BC$
$\therefore\triangle\text{APQ}\sim\triangle\text{ABC}$
$\therefore\frac{\text{AP}}{\text{AB}}=\frac{\text{AQ}}{\text{AC}}=\frac{\text{PQ}}{\text{BC}}$
$\Rightarrow\frac{\text{AP}}{\text{AP}+\text{PB}}=\frac{\text{PQ}}{\text{BC}}$
$\Rightarrow\frac{4}{4+6}=\frac{3}{\text{x}}\Rightarrow\frac{4}{10}=\frac{3}{\text{x}}$
$\Rightarrow\text{x}=\frac{10\times3}{4}=\frac{15}{2}$
$\therefore\text{BC}=\frac{15}{2}\text{cm}=7.5\text{cm}$ View full question & answer→Question 264 Marks
The areas of two similar triangles are $121 \mathrm{~cm}^2$ and $64 \mathrm{~cm}^2$ respectively. If the median of the first triangle is $12.1cm$, find the corresponding median of the other.
Answer
We have,
$\triangle\text{ABC}\sim\triangle\text{PQR}$
Area $(\triangle\text{ABC})= 121cm^2,$
Area $(\triangle\text{PQR}) = 64cm^2$
$AD = 12.1cm$
And $AD$ and $PS$ are the medians
By area of similar triangle theorem
$\frac{\text{Area}(\triangle\text{ABC})}{\text{Area}(\triangle\text{PQR})}=\frac{\text{AB}^2}{\text{PQ}^2}$
$\Rightarrow\frac{121}{64}=\frac{\text{AB}^2}{\text{PQ}^2}$
$\Rightarrow\frac{11}{8}=\frac{\text{AB}}{\text{PQ}}\ \ ...(\text{i})$
Since, $\triangle\text{ABC}\sim\triangle\text{PQR}$
Then, $\frac{\text{AB}}{\text{PQ}}=\frac{\text{BC}}{\text{QR}}$ [Corresponding parts of similar $\triangle$ are proportional]
$\Rightarrow\frac{\text{AB}}{\text{PQ}}=\frac{\text{2BD}}{\text{2QS}}$ [AD and PS are medians]
$\Rightarrow\frac{\text{AB}}{\text{PQ}}=\frac{\text{BD}}{\text{QS}}\ \ ...(\text{ii})$
In $\triangle\text{ABD}$ and $\triangle\text{PQS}$
$\angle\text{B}=\angle\text{Q}\ \ \ \ [\triangle\text{ABC}\sim\triangle\text{PQS}]$
$\frac{\text{AB}}{\text{PQ}}=\frac{\text{BD}}{\text{QS}}$ [From (ii)]
Then, $\triangle\text{ABD}\sim\triangle\text{PQS}$ [By SAS similarity]
$\therefore\frac{\text{AB}}{\text{PQ}}=\frac{\text{AD}}{\text{PS}}\ \ \ ...\text{(iii)}$ [Corresponding parts of similar $\triangle$ are proportional]
Compare $(i)$ and $(iii)$
$\frac{11}{8}=\frac{\text{AD}}{\text{PS}}$
$\Rightarrow\frac{11}{8}=\frac{12.1}{\text{PS}}$
$\Rightarrow\text{PS}=\frac{8\times12.1}{\text{PS}}=8.8\text{cm}$ View full question & answer→Question 274 Marks
In the given figure, $\angle\text{B}<90^\circ$ and segment $\text{AD}\perp\text{BC}$ show that
- $b^2=h^2+a^2+x^2-2 a x$
- $b^2=a^2+c^2-2 a x$
Answer
- Since AD perpendicular to BC we obtained two right angled triangles, triangle ADB and triangle ADC.
We will use Pythagoras theorem in the right angled triangle ADC
$A C^2=A D^2+D C^2 \ldots(1)$
Let us substitute $A D=h, A C=b$ and $D C=(a-x)$ in equation (1) we get,
$ b^2=h^2+(a-x)^2 $
$ b^2=h^2+a^2-2 a x+x^2 $
$ b^2=h^2+a^2+x^2-2 a x....(2)$
ii. Let us use Pythagoras theorem in the right angled triangle $A D B$ as shown below
$A B^2=A D^2+B D^2 \ldots \ldots(3)$
Let us substitute $A B=c, A D=h$ and $B D=x$ in equation (3) we get,
$c^2=h^2+x^2$
Let us rewrite the equation (2) as below,
$b^2=h^2+x^2+a^2-2 a x \ldots \ldots(4)$
Now we will substitute $h^2+x^2=c^2$ in equation (4) we get,
$b^2=c^2+a^2-2 a x$
Therefore, $\mathrm{b}^2=\mathrm{c}^2+\mathrm{a}^2-2 \mathrm{ax}$. View full question & answer→Question 284 Marks
$D$ is the mid-point of side $B C$ of a $\triangle A B C$. $A D$ is bisected at the point $E$ and $B E$ produced cuts $A C$ at the point $X$. Prove that $B E=E X=3: 1$.
AnswerGiven: $A B C$ is a triangle in which $D$ is the mid point of $B C, E$ is the mid point of $A D, B E$ produced meets $A C$ at $X$.
To Prove: $BE : EX =3: 1$.
Construction: We draw a line $DY$ parallel to $BX.$
Prrof:
In $\triangle\text{BCX}$ and $\triangle\text{DCY,}$
$\angle\text{CBX}=\angle\text{CDY}$ (Corresponding angles)
$\angle\text{CXB}=\angle\text{CYD}$ (Corresponding angles)
$\triangle\text{BCX}\sim\triangle\text{DCY}$ (AA similarity)
We know that corresponding sides of similar triangles are proportional.
Thus, $\frac{\text{BC}}{\text{DC}}=\frac{\text{BX}}{\text{DY}}=\frac{\text{CX}}{\text{CY}}$
$\Rightarrow\frac{\text{BX}}{\text{DY}}=\frac{\text{BC}}{\text{DC}}$
$\Rightarrow\frac{\text{BX}}{\text{DY}}=\frac{2\text{DC}}{\text{DC}}$ (As D is the mid point of BC)
$\Rightarrow\frac{\text{BX}}{\text{DY}}=\frac{2}{1}\ ....(1)$
In $\triangle\text{AEX}$ and $\triangle\text{ADY,}$
$\angle\text{AEX}=\angle\text{ADY}$ (Corresponding angles)
$\angle\text{AXE}=\angle\text{AYD}$ (Corresponding angles)
$\triangle\text{AEX}\sim\triangle\text{ADY}$ (AA similarity)
We know that corresponding sides of similar triangles are proportional.
Thus, $\frac{\text{AE}}{\text{AD}}=\frac{\text{EX}}{\text{DY}}=\frac{\text{AX}}{\text{AY}}$
$\Rightarrow\frac{\text{EX}}{\text{DY}}=\frac{\text{AE}}{\text{AD}}$
$\Rightarrow\frac{\text{EX}}{\text{DY}}=\frac{\text{AE}}{2\text{AE}}$ (As D is the mid point of BC)
$\Rightarrow\frac{\text{EX}}{\text{DY}}=\frac{1}{2}\ ...(2)$
Dividing (1) by (2), we get
$\frac{\text{BX}}{\text{EX}}=4$
$\Rightarrow\text{BX}=4\text{EX}$
$\Rightarrow\text{BE}+\text{EX}=4\text{EX}$
$\Rightarrow\text{BE}=3\text{EX}$
$\Rightarrow\text{BE}:\text{EX}=3:1$ View full question & answer→Question 294 Marks
In the given figure, given that $\triangle\text{ABC}\sim\triangle\text{PQR}$ and quad ABCD ∼ quad $PQRS$. Determine the value of $x, y, z$ in each case.
-
-
-
Answer
- We have, $\triangle\text{ABC}\sim\triangle\text{PQR}$
So the ratio of the sides of the triangles will be proportional to each other.
$\frac{\text{AB}}{\text{PQ}}=\frac{\text{BC}}{\text{QR}}=\frac{\text{AC}}{\text{PR}}$
Therefore put the values of the known terms in the above equation to get,
$\frac{12}{9}=\frac{7}{\text{x}}=\frac{10}{\text{y}}$
On solving these simultaneous equations, we get
$\text{x}=\frac{21}{4}$
$\text{y}=\frac{30}{4}$
- We have, $\Box\text{ABCD}\sim\Box\text{PQRS}$
So the ratio of the sides of the quadrilaterals will be proportional to each other.
$\frac{\text{AB}}{\text{PQ}}=\frac{\text{BC}}{\text{QR}}=\frac{\text{CD}}{\text{RS}}=\frac{\text{DA}}{\text{SP}}$
Therefore put the values of the known terms in the above equation to get,
$\frac{20}{7}=\frac{16}{\text{x}}=\frac{50}{\text{y}}=\frac{50}{3\text{z}}$
On solving these simultaneous equations, we get
$\text{x}=\frac{28}{5}$
$\text{y}=\frac{35}{2}$
$\text{z}=\frac{35}{6}$ View full question & answer→Question 304 Marks
In the given figure, $D$ is the mid-point of side $BC$ and $\text{AE}\perp\text{BC}.$ If $BC = a, AC = b, AB = c, ED = x, AD = p$ and $AE = h$, prove that:
- $\text{b}^2=\text{p}^2+\text{ax}+\frac{\text{a}^2}{4}$
- $\text{c}^2=\text{p}^2-\text{ax}+\frac{\text{a}^2}{4}$
- $\text{b}^2+\text{c}^2=2\text{p}^2+\frac{\text{a}^2}{4}$
AnswerGiven: In $\triangle\text{ABC},$
AD is median of BC and $\text{AE}\perp\text{BC}$
$AB = c, BC = a, CA = b$
$ED = x, AD = p, AE = h.$
To prove:
- $\text{b}^2=\text{p}^2+\text{ax}+\frac{\text{a}^2}{4}$
- $\text{c}^2=\text{p}^2-\text{ax}+\frac{\text{a}^2}{4}$
- $\text{b}^2+\text{c}^2=2\text{p}^2+\frac{\text{a}^2}{4}$
Proof: In right $\triangle\text{AEC},$
- $A C^2=A E^2+E C^2$(pythagoras theorem)
$A C^2=A E^2+(E D+D C)^2$
$\text{b}^2=\text{h}^2+\Big(\text{x}+\frac{1}{2}\text{a}\Big)^2$ (D is mid-points of BC)
$=\text{h}^2+\text{x}^2+\text{ax}+\frac{\text{a}^2}{4}\ \ ....(\text{i})$
But in right $\triangle\text{AED},$
$ A D^2=A E^2+E D^2 $
$ p^2=h^2+x^2 \ldots \ldots \text { (ii) }$
$\therefore\text{b}^2=\text{p}^2+\text{ax}+\frac{\text{a}^2}{4}$
- Similarly in right $\triangle\text{AEB},$
$ A B^2=A E^2+B E^2$
$ \Rightarrow A B^2=A E^2+(B D-E D)^2$
$\Rightarrow\text{c}^2=\text{h}^2+\Big(\frac{\text{1}}{2}\text{BC}-\text{x}\Big)^2$
$\Rightarrow\text{c}^2=\text{h}^2+\Big(\frac{\text{a}}{2}-\text{x}\Big)^2$
$\Rightarrow\text{c}^2=\text{h}^2+\text{x}^2+\frac{\text{a}^2}{4}-\text{ax}$
$\Rightarrow\text{c}^2=\text{p}^2-\text{ax}+\frac{\text{a}^2}{4}\ \ \ [\because\text{From (ii)}]$
- Adding results of $(i)$ and $(ii)$
$\text{b}^2+\text{c}^2=\text{p}^2+\text{ax}+\frac{\text{a}^2}{4}+\text{p}^2-\text{ax}+\frac{\text{a}^2}{4}$
$=2\text{p}^2+2\times\frac{\text{a}^2}{4}=2\text{p}^2+\frac{\text{a}^2}{2}$
Hence proved. View full question & answer→