Question 14 Marks
In the given figure, if $\angle\text{ADE}=\angle\text{B},$ show that $\triangle\text{ADE}\sim\triangle\text{ABC}.$ If $AD = 3.8\ cm, AE = 3.6\ cm, BE = 2.1\ cm$ and $BC = 4.2\ cm$, find $DE.$
Answer
Given: $\angle\text{ADE}=\angle\text{B},$ $AD = 3.8\ cm, AE = 3.6\ cm, BE = 2.1\ cm, BC = 4.2\ cm$
Proof:
In $\triangle\text{ADE}$ and $\triangle\text{ABC},$
$\angle\text{A}=\angle\text{A}$ (common)
$\angle\text{ADE}=\angle\text{B}$ (given)
Therefore, $\triangle\text{ADE}\sim\triangle\text{ABC}$ ($AA$ Criterion)
$\Rightarrow\frac{\text{AD}}{\text{AB}}=\frac{\text{DE}}{\text{BC}}$
$\Rightarrow\frac{3.8}{(3.6+2.1)}=\frac{\text{x}}{4.2 }(\text{DE}=\text{x})$
$\Rightarrow\frac{3.8}{5.7}=\frac{\text{x}}{4.2}$
$\text{x}=\frac{3.8\times4.2}{5.7}=2.8\text{cm}$
Hence, $DE = 2.8\ cm$ View full question & answer→Question 24 Marks
State the $AAA$-similarity criterion.
AnswerIf is any two triangles, the corresponding angles are equal, then their corresponding sides are proportional and hence the triangles are similar.
View full question & answer→Question 34 Marks
$\triangle\text{ABC}$ is right-angled at $A$ and $\text{AD}\perp\text{BC}.$ If $BC = 13\ cm$ and $AC =5\ cm$, find the ratio of the areas of $\triangle\text{ABC}$ and $$$\triangle\text{ADC}.$
AnswerIn $\triangle\text{ABC}$ and $\triangle\text{ADC},$ we have:
$\angle\text{BAC}=\angle\text{ADC}=90^\circ$
$\angle\text{ACB}=\angle\text{ACD}$ (common)
By AA similarity, we can conclude that $\triangle\text{BAC}\sim\triangle\text{ADC}.$
Hence, the ratio of the areas of these triangles is equal to the ratio of squares of their corresponding sides.
$\therefore\frac{\text{ar}(\triangle\text{BAC})}{\text{ar}(\triangle\text{ADC})}=\frac{\text{BC}^2}{\text{AC}^2}$
$\Rightarrow\frac{\text{ar}(\triangle\text{BAC})}{\text{ar}(\triangle\text{ADC})}=\frac{\text{13}^2}{\text{5}^2}$
$=\frac{169}{25}$
Hence, the ratio of both the triangles is $169 : 25$
View full question & answer→Question 44 Marks
A ladder is placed in such a way that its foot is at a distance of $15\ m$ from a wall and its top reaches a window $20\ m$ above the ground. Find the length of the ladder.
AnswerLet $AB$ be the well where window is at $B, CB$ be the ladder and $AC$ be the distance between the foot of the ladder and wall.
Then,
$\text{AB}=20\text{m},\text{AC}=15\text {m},$ and $\angle\text{CAB}=90^\circ$
By Pythagoras theorem, we have
$\text{CB}^2=\text{AB}^2+\text{AC}^2$
$=\Big[(20)^2+(15)^2\Big]\text{m}^2$
$=(400+225)\text{m}^2$
$=625\text{m}^2$
$\text{CB}=\sqrt{625}\text{m}=25\text{m}$
Hence, the length of ladder is $25 \ m.$ View full question & answer→Question 54 Marks
For the following statments state whether true $(T)$ or false$(F):$
Two circles with different radii are similar.
AnswerTrue.
Solution:
Similar figures have the same shape but need not have the same size.
Since all circles irrespective of the radii will have the same shape, all be similar.
View full question & answer→Question 64 Marks
$\triangle\text{ABC}\sim\triangle\text{DEF}$ and their areas are respectively $100cm^2$ and $49cm^2$. If the altitude of $\triangle\text{ABC}$ is 5cm, the corresponding altitude of $\triangle\text{DEF}.$
Answer
It is given that $\triangle\text{ABC}\sim\triangle\text{DEF}.$
Therefore, the ratio of the areas of these triangles will be equal to the ratio of squares of their corresponding sides.
Also, the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding altitudes.
Let the altitude of $\triangle\text{ABC}$ be $AP$, drawn from $A$ to $BC$ to meet $BC$ at $P$ and the altitude of $\triangle\text{DEF}$ be $DQ$, drawn from $D$ to meet $EF$ at $Q$
Then,
$\frac{\text{ar}(\triangle\text{ACB})}{\text{ar}(\triangle\text{DEF})}=\frac{\text{AP}^2}{\text{DQ}^2}$
$\Rightarrow\frac{100}{49}=\frac{\text{5}^2}{\text{DQ}^2}$
$\Rightarrow\frac{100}{49}=\frac{\text{25}}{\text{DQ}^2}$
$\Rightarrow\text{DQ}^2=\frac{49\times25}{100}$
$\Rightarrow\text{DQ}=\sqrt{\frac{49\times25}{100}}$
$\Rightarrow\text{DQ}=3.5\text{cm}$
Hence, the altitude of $\triangle\text{DEF}$ is $3.5\ cm$ View full question & answer→Question 74 Marks
State the converse of Thales' theorem.
AnswerIf a line divides any two sides of a triangle in the same ratio then the line must be parallel to the third side.
View full question & answer→Question 84 Marks
In the given figure, $\triangle\text{ABC}$ is an obtuse triangle, obtuse-angled at B. If $\text{AD}\perp\text{CB}$ (produced) prove that $AC^2= AB^2+ BC^2+2BC.BD.$
Answer
Applying Pythagoras theorem in right-angled triangle $ADC$, we get:
$A C^2=A D^2+D C^2 $
$ \Rightarrow A C^2-D C^2=A D^2 $
$ \Rightarrow A D^2=A C^2-D C^2...(1)$
Applying Oythagoras theorem in right-triangle $ADB$, we get:
$ A B^2=A D^2+D B^2 $
$ \Rightarrow A B^2-D B^2=A D^2 $
$ \Rightarrow A D^2=A B^2-D B^2...(2)$
From equation $(1)$ and $(2)$,, we have:
$A C^2-D C^2=A B^2-D B^2 $
$ \Rightarrow A C^2=A B^2+D C^2-D B^2 $
$ \Rightarrow A C^2=A B^2+(D B+B C)^2-D B^2(\therefore D B+B C=D C) $
$ \Rightarrow A C^2=A B^2+D B^2+B C^2+2 D B \cdot B C-D B^2 $
$ \Rightarrow A C^2=A B^2+B C^2+2 B C \cdot B D$
This completes the proof. View full question & answer→Question 94 Marks
In the given figure, $\angle\text{ACB}=90^\circ$ and $\text{CD}\perp\text{AB}.$
Prove that $\frac{\text{BC}^2}{\text{AC}^2}=\frac{\text{BD}}{\text{AD}}.$
AnswerGiven: $\angle\text{ACB}=90^\circ$ and $\text{CD}\perp\text{AB}$
To Prove: $\frac{\text{BC}^2}{\text{AC}^2}=\frac{\text{BD}}{\text{AD}}$
Proof:
In $\triangle\text{ACB}$ and $\triangle\text{CDB}$
$\angle\text{ACB}=\angle\text{CDB}=90^\circ$ (Given)
$\angle\text{ABC}=\angle\text{CBD}$ (common)
By AA similarity-criterion $\triangle\text{ACB}\sim\triangle\text{CDB}$
When two triangles are similar, the ratios of the lengths of their corresponding sides are proportional.
$\therefore\frac{\text{BC}}{\text{BD}}=\frac{\text{AB}}{\text{BC}}$
$\Rightarrow\text{BC}^2=\text{BD}.\text{AB}\ .....(1)$
In $\triangle\text{ACB}$ and $\triangle\text{ADC}$
$\angle\text{ACB}=\angle\text{ADC}=90^\circ$ (Given)
$\angle\text{CAB}=\angle\text{DAC}$ (common)
By AA similarity-criterion $\triangle\text{ACB}\sim\triangle\text{ADC}$
When two triangles are similar, then the ratios of the lengths of their corresponding sides are proportional.
$\therefore\frac{\text{AC}}{\text{AD}}=\frac{\text{AB}}{\text{AC}}$
$\Rightarrow\text{AC}^2=\text{AD}.\text{AB}\ .....(2)$
Divinding $(1)$ by $(2)$,, we get
$\frac{\text{BC}^2}{\text{AC}^2}=\frac{\text{BD}}{\text{AD}}$
View full question & answer→Question 104 Marks
In an equilateral triangle with side a, prove that area $=\frac{\sqrt{3}}{4}\text{a}^2.$
Answer
Let ABC be the equilateral triangle with each side equal to $a$.
Let $A D$ be the altitude from $A$, meeting $B C$ at $D$.
Therefore, $D$ is the midpoint of $B C$.
Let $A D$ be $h$.
Applying Pythagoras theorem in right-angled triangle $ABD$, we have:
$A B^2=A D^2+B D^2$
$\Rightarrow\text{a}^2=\text{h}^2+\big(\frac{\text{a}}{2}\big)^2$
$\Rightarrow\text{h}^2=\text{a}^2-\frac{\text{a}}{4}=\frac{3}{4}\text{a}^2$
$\Rightarrow\text{h}=\frac{\sqrt{3}}{2}\text{a}$
Therefore,
Area of triangle $\text{ABC}=\frac{1}{2}\times\text{base}\times\text{height}=\frac{1}{2}\times\text{a}\times\frac{\sqrt{3}}{2}\text{a}$
$=\frac{\sqrt{3}}{4}\text{a}^2$
This completes the proof. View full question & answer→Question 114 Marks
State the $AA-$similarity criterion.
AnswerIf two angles of one triangle are respectively equal to two angles of another triangle then the two triangles are similar.
View full question & answer→Question 124 Marks
$\triangle\text{ABC}\sim\triangle\text{DEF}$ such that $\text{ar}(\triangle\text{ABC})=64\text{cm}^2$ and $\text{ar}(\triangle\text{DEF})=169\text{cm}^2.$ If $BC = 4\ cm$, find $EF.$
Answer$\triangle\text{ABC}\sim\triangle\text{DEF}$
$\Rightarrow\frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{DEF})}=\frac{\text{BC}^2}{\text{EF}^2}$
$\Rightarrow\frac{64}{169}=\frac{4^2}{\text{EF}^2}$
$\Rightarrow\text{EF}^2=\frac{16\times169}{64}$
$\Rightarrow\text{EF}=\sqrt{\frac{16\times169}{64}}$
$\Rightarrow\text{EF}=\frac{4\times13}{8}$
$\Rightarrow\text{EF}=6.5\text{cm}$
View full question & answer→Question 134 Marks
In the adjoining figure, $A B C D$ is a trapezium in which $C D \| A B$ and its digonals intersect at $O$. If $A O=(2 x+1) cm , O C=(5 x-7) cm , D O$ $=(7 x-5) cm$ and $O B=(7 x+1) cm$, find the value of $x$.
AnswerIn trapezium $ABCD, AB || CD$ and the diagonals $AC$ and $BD$ intersect $O.$
Therefore,
$\frac{\text{AO}}{\text{OC}}=\frac{\text{BO}}{\text{OD}}$
$\Rightarrow\frac{5\text{x}-7}{2\text{x}+1}=\frac{7\text{x}-5}{7\text{x}+1}$
$\Rightarrow(5\text{x}-7)(7\text{x}+1)=(7\text{x}-5)(2\text{x}+1)$
$\Rightarrow35\text{x}^2+5\text{x}-49\text{x}-7=14\text{x}^2-10\text{x}+7\text{x}-5$
$\Rightarrow21\text{x}^2-41\text{x}-2=0$
$\Rightarrow21\text{x}^2-42\text{x}+\text{x}-2=0$
$\Rightarrow21\text{x}(\text{x}-2)+1(\text{x}-2)=0$
$\Rightarrow(\text{x}-2)+(21\text{x}+1)=0$
$\Rightarrow\text{x}=2,-\frac{1}{21}$
$\therefore\text{x}\not=-\frac{1}{21}$
$\therefore\text{x}=2$
View full question & answer→Question 144 Marks
$D$ and $E$ are points on the sides $A B$ and $A C$ respectively of a $\triangle A B C$. In the following cases, determine whether $D E \| B C$ or not.
$AB=11.7 cm, AC=11.2 cm, BD=6.5 cm \text { and } AE=4.2 cm .$
AnswerWe have:
$AB = 11.7cm, DB = 6.5cm$
Therefore,
$AD = 11.7 - 6.5 = 5.2cm$
Similarly,
$AC = 11.2cm, AE = 4.2cm$
Therefore,
$EC = 11.2 - 4.2 = 7cm$
Now,
$\frac{\text{AD}}{\text{DB}}=\frac{5.2}{6.5}=\frac{4}{5}$
$\frac{\text{AE}}{\text{EC}}=\frac{4.2}{7}$
Thus, $\frac{\text{AD}}{\text{DB}}\not=\frac{\text{AE}}{\text{EC}}$
Applying the converse of Thalse' theorem, we conclude thet $DE$ is not parallel to $BC.$
View full question & answer→Question 154 Marks
$P$ and $Q$ are points on the sides $A B$ and $A C$ respectively of a $\triangle A B C$. If $A P=2 \ cm, P B=4 \ cm, A Q=3 \ cm$ and $Q C=6 \ cm$, show that $B C=3 P Q$.
Answer
Given: $P$ is a point on $A B$.
Then, $A B=A P+P B=(2+4) cm =6 cm$
Also $Q$ is a point on $A C$.
Then, $A C=A Q+Q C=(3+6) cm =9 cm$
$\therefore\frac{\text{AP}}{\text{AB}}=\frac{2}{6}=\frac{1}{3}$
and $\frac{\text{AQ}}{\text{AC}}=\frac{3}{9}=\frac{1}{3}$
$\therefore\frac{\text{AP}}{\text{AB}}=\frac{\text{AQ}}{\text{AC}}$
Thus, in $\triangle\text{APQ}$ and $\triangle\text{ABC}$
$\angle\text{A}=\angle\text{A}$ (common)
And $\frac{\text{AP}}{\text{AB}}=\frac{\text{AQ}}{\text{AC}}$
$\therefore\triangle\text{APQ}\sim\triangle\text{ABC}$ (by $SAS$ similarity)
$\Rightarrow\frac{\text{AP}}{\text{AB}}=\frac{\text{PQ}}{\text{BC}}=\frac{\text{AQ}}{\text{AC}}$
$\therefore\frac{\text{PQ}}{\text{BC}}=\frac{\text{AQ}}{\text{AC}}\Rightarrow\frac{\text{PQ}}{\text{BC}}=\frac{3}{9}=\frac{1}{3}$
$\therefore\text{BC}=3\text{PQ}$
Hence proved. View full question & answer→Question 164 Marks
In the given figure, $D$ is the midpoint of side $B C$ and $A E \perp B C$. If $B C=a, A C=b, A B=c, E D=x, A D=p$ and $A E=h$, prove that.
$\text{c}^2=\text{p}^2-\text{ax}+\frac{\text{a}^2}{4}$ AnswerGiven: $D$ is the midpoint of side $\text{BC},\text{AE}\perp\text{BC},\text{BC}=\text{a},\text{AC}=\text{b},\text{AB}=\text{c},\text{ED}=\text{x},\text{AD}=\text{p}$ and $\text{AE}=\text{h}$
In $\triangle\text{AEC},\angle\text{AEC}=90^\circ$
$\text{AD}^2=\text{AE}^2+\text{ED}^2$ (by pythagoras theorem)
$\Rightarrow\text{p}^2=\text{h}^2+\text{x}^2$
In $\triangle\text{ABE},\angle\text{ABE}=90^\circ$
$\text{AB}^2=\text{AE}^2+\text{BE}^2$ (by pythagoras theorem)
$\Rightarrow\text{c}^2=\text{h}^2+\Big(\frac{\text{a}}{2}-\text{x}\Big)^2\dots(2)$
$=\big(\text{h}^2+\text{x}^2\big)-\text{ax}+\frac{\text{a}^2}{4}$
$=\text{p}^2-\text{ax}+\frac{\text{a}^2}{4}$
Hence, $\text{c}^2=\text{p}^2-\text{ax}+\frac{\text{a}^2}{4}$
View full question & answer→Question 174 Marks
In an equilateral triangle with side a, prove that area $=\frac{\sqrt{3}}{4}\text{a}^2.$
Answer
Let $\triangle\text{ABC}$ be an equilateral triangle with side a.
Then, $AB = AC = BC = a.$
Draw $\text{AD}\perp\text{BC}.$
In $\triangle\text{ADB}$ and $\triangle\text{A},$ we have
$\text{AB}=\text{AC}(\text{given}),\angle\text{B}=\angle\text{C}=60^\circ$
and $\angle\text{ADB}=\angle\text{ADC}=90^\circ$
$\therefore\triangle\text{ADB}\cong\triangle\text{ADC}$
$\therefore\text{BD}=\text{DC}=\frac{\text{a}}{2}$
From right $\triangle\text{ADB},$ we have
$\text{AB}^2=\text{AD}^2+\text{BD}^2$ ....(By Pythagoras theorem)
$\Rightarrow\text{AD}=\sqrt{\text{AB}^2-\text{AB}^2}$
$\Rightarrow\text{AD}=\sqrt{\text{a}^2-\Big(\frac{\text{a}}{2}\Big)^2}$
$\Rightarrow\text{AD}=\sqrt{\text{a}^2-\frac{\text{a}}{4}^2}$
$\Rightarrow\text{AD}=\sqrt{\frac{3\text{a}}{4}^2}$
$\Rightarrow\text{AD}=\frac{\sqrt{3}\text{a}}{2}$
So, the altitude is $\frac{\sqrt{3\text{a}}}{2},$
Area of $\triangle\text{ABC}=\frac{1}{2}\times\text{BC}\times\text{AD}$
$=\frac{1}{2}\times\text{a}\times\frac{\sqrt{3\text{a}}}{2}$
$=\frac{\sqrt{3}\text{a}^2}{4}$
Hence proved. View full question & answer→Question 184 Marks
Find the lenght of the altitude of an equilateral triangle of side $2a \ cm.$
Answer
Let $\triangle\text{ABC}$ be an equilateral triangle.
We know that,
in an equilateral triangle the altitube is same as the median.
So, $BD = DC = a \ cm$
By Pythagoras theorem,
$ A C^2=A D^2+D C^2 $
$ \Rightarrow A D^2=A C^2-D C^2 $
$ \Rightarrow A D^2=(2 a)^2-a^2 $
$ \Rightarrow A D^2=4 a^2-a^2 $
$ \Rightarrow A D^2=3 a^2$
$\Rightarrow\text{AD}=\sqrt{3}\text{a cm}$
So, lendth of the altitude is $\sqrt{3}\text{a cm}.$ View full question & answer→Question 194 Marks
The lengths of the diagonals of a rhobbus are $40\ cm$ and $42\ cm$. find the length of each side of the rhombus.
Answer
In an rhombus, the diagonals are perpendicular bisectors of each other, and sides are equal to eachother.
So, $\text{AO}=\frac{1}{2}\text{AC}=21\text{cm}$
$\text{OD}=\frac{1}{2}\text{BD}=20\text{cm}$
In right-angled $\triangle\text{AOD},$
$\text{AD}^2=\text{AO}^2+\text{OD}^2$
$\Rightarrow\text{AD}^2=21^2+20^2$
$\Rightarrow\text{AD}^2=441+400$
$\Rightarrow\text{AD}^2=841$
$\Rightarrow\text{AD}=29\text{cm}$
So, the length of the each side of the rhombus is $29\ cm.$ View full question & answer→Question 204 Marks
A ladder 10m long reaches the window of a house $8\ m$ above the ground. Find the distance of the foot of the ladder from the base of the wall.
Answer
Let BW be the ladder and $OB$ be the house.
$\triangle\text{BOW}$ forms a right-angled triangle.
By Pythagoras theorem,
$ \mathrm{BW}^2=\mathrm{OW}^2+\mathrm{OB}^2 $
$ \mathrm{OW}^2=\mathrm{BW}^2-\mathrm{OB}^2 $
$ \mathrm{OW}^2=10^2-8^2 $
$ \mathrm{OW}=100-64 $
$ \mathrm{OW}=6 \mathrm{~m}$
So the distance of the foot of the ladder from the house is $6\ m.$ View full question & answer→Question 214 Marks
$A B C D$ is a parallelogram and $E$ is a point on $B C$. If the diagonal $B D$ intersect $A E$ at $F$, prove that $A F \times F B=E F \times F D$.
Answer
Given: $A B C D$ is a parallelogram and $E$ is point on $B C$. Diagonals $D B$ intersects $A E$ at $F$.
To Prove: $AF \times FB = EF \times FD$
Proof: In $\triangle\text{AFD}$ and $\triangle\text{EFD}$
$\angle\text{AFD}=\angle\text{EFB}$ $($vertically opposite $\angle\text{s})$
$\angle\text{DAF}=\angle\text{BEF}$ $($ Alternate $\angle\text{s})$
$\therefore\triangle\text{AFD}\approx\triangle\text{EFD}$ [By AAA similarity]
$\therefore\frac{\text{AF}}{\text{EF}}=\frac{\text{FD}}{\text{FB}}$
$\text{AF}\times\text{FB}=\text{EF}\times\text{FD}$
Hence proved. View full question & answer→Question 224 Marks
In the given figure, $\angle\text{ABC}=90^\circ$ and $\text{BD}\perp\text{AC}.$ If $AB = 5.7\ cm$, $BD = 3.8\ cm$, and $CD = 5.4\ cm$ find $BC.$
Answer
Given that $AB = 5.7\ cm, BD = 3.8\ cm$, and $CD = 5.4\ cm$
In $\triangle\text{CBA}$ and $\triangle\text{CDB}$
$\angle\text{CBA}=\angle\text{CDB}=90^\circ$
$\angle\text{C}=\angle\text{C}$ (common)
Therefore, $\triangle\text{CBA}\sim\triangle\text{CDB}$ (by $AA$ similarities)
$\Rightarrow\frac{\text{BC}}{\text{CD}}=\frac{\text{BA}}{\text{BD}}$
$\Rightarrow\frac{\text{BC}}{\text{5.4}}=\frac{\text{5.7}}{\text{3.8}}$
$\Rightarrow\text{BC}=\frac{\text{5.7}\times5.4}{\text{3.8}}$
$\therefore\text{BC}=8.1\text{cm}$
Hence, $BC= 8.1\ cm$ View full question & answer→Question 234 Marks
In the given pairs of triangles, find which pair of triangles are similar. State the similarity criterior and write the similarity relation in symbolic from.
Answer$\triangle\text{CAB}\sim\triangle\text{QRP}$ ($SAS$ Similarity)
as $\frac{\text{CA}}{\text{QR}}=\frac{\text{CB}}{\text{QP}}$ and $\angle\text{C}=\angle\text{Q}$
View full question & answer→Question 244 Marks
The sides of certain triangles are given below. Determine them are right triangles:
$7\ cm, 24\ cm, 25\ cm.$
AnswerFor a given triangle to be a right angled, the sum of the squares of the two sides must be equal to the square of the largest side.
Let $a = 7\ cm, b = 24\ cm$ and $c = 25\ cm$, Then
$\big(\text{a}^2+\text{b}^2\big)=\big[7^2+(24)^2\big]\text{cm}^2$
$=(49+576)\text{cm}^2$
$=625\text{cm}^2$
$\text{c}^2=(25\text{cm})^2=625\text{cm}^2$
$\therefore\big(\text{a}^2+\text{b}^2\big)-\text{c}^2$
Hence, the given triangle is a right triangle.
View full question & answer→Question 254 Marks
The areas of two similar triangles are $169cm^2$ and $121cm^2$ respectively. If the longest side of the larger triangle is $26\ cm$, find the longest side of the smaller triangle.
AnswerIt is given that the triangle are similar.
Therefore, the ratio of the areas of these triangle will be equal to the ratio of squares of their corresponding sides.
Let the longest side of smaller triangle be $x \ cm.$
$\frac{\text{ar(Larger triangle)}}{\text{ar(Smalles triangle)}}=\frac{(\text{Longest side of larger triangle})^2}{(\text{Longest side of smaller triangle})^2}$
$\Rightarrow\frac{169}{121}=\frac{26^2}{\text{x}^2}$
$\Rightarrow\text{x}=\sqrt{\frac{26\times26\times121}{169}}$
$=22$
Hence, the longest side of the smaller triangle is $22\ cm.$
View full question & answer→Question 264 Marks
State the converse of Pythagpras' theorem.
AnswerIn a triangle, if the square of one side is equal to the sum of the squeares of the other two sides then the angle opposite to the first side is a right angle.
View full question & answer→Question 274 Marks
In the given figure, $\triangle\text{OAB}\sim\triangle\text{OCD}.$ If $AB = 8\ cm, BO = 6.4\ cm$, $OC = 3.5\ cm$ and $CD = 5\ cm$, find.
- $OA$
- $DO$
Answer
- Let OA be $x \ cm.$
$\therefore\triangle\text{OAB}\sim\triangle\text{OCD}$
$\therefore\frac{\text{OA}}{\text{OC}}=\frac{\text{AB}}{\text{CD}}$
$\Rightarrow\frac{\text{x}}{\text{3.5}}=\frac{\text{8}}{\text{5}}$ and
$\Rightarrow\text{x}=\frac{8\times3.5}{5}=5.6$
hence, $OA = 5.6\ cm$
- Let $OD$ be $y \ cm$
$\therefore\triangle\text{OAB}\sim\triangle\text{OCD}$
$\therefore\frac{\text{AB}}{\text{CD}}=\frac{\text{OB}}{\text{OD}}$
$\Rightarrow\frac{\text{8}}{\text{5}}=\frac{\text{6.4}}{\text{y}}$
$\Rightarrow\text{y}=\frac{6.4\times5}{8}=4$ View full question & answer→Question 284 Marks
In a circle, two chords $AB$ and $CD$ intersect a point $P$ inside the circle. Prove that
- $\triangle\text{PAC}\sim\triangle\text{PDB}$
- $\text{PA}.\text{PB}=\text{PC}.\text{PD}$
AnswerIn $\triangle\text{PAC}$ and $\triangle\text{BPD},$
$\angle\text{APC}=\angle\text{BPD}$ .....(Vertically opposite angles)
$\angle\text{CAP}=\angle\text{BDP}$ ......(Angels inscribed in the same are equal)
$\therefore\triangle\text{PAC}\sim\triangle\text{PDB}$ .....(AA similarity criterion)
$\Rightarrow\frac{\text{PA}}{\text{PD}}=\frac{\text{PC}}{\text{PB}}$
$\Rightarrow\text{PA}.\text{PB}=\text{PC}.\text{PD}$
View full question & answer→Question 294 Marks
Naman is diing fly-fishing in a stream. The tip of his fishing rod is $1.8\ m$ above the surface of the water and the fly at the eand of the string rests on the water $3.6\ m$ away from him and $2.4\ m$ from the point directly under the tip of the rod Assuming that the string (from tip of his rod to the fly) is taut, how much string does he have out (see the adjoining figure)? If he pulls in the string at the rate of $5\ cm$ per second, what will be the horizontal distance of the fly from him after $12$ seconds?
Answer
Given that Naman pulls is the string at the rate of $5\ cm$ per second.
Hence, after $12$ second the lenght of the string he pulls
$= 12 × 5 = 60cm = 0.6m$
In right $\triangle\text{BMC},$
By Pythagoras theorem,
$BC^2 = CM^2 + MB^2$
$⇒ BC^2 = (2.4)^2 + (1.8)^2$
$⇒ BC^2 = 5.76 + 3.24$
$⇒ BC^2 = 9$
$⇒ BC^2 = 3m$
So,$ BC' = BC -$ lenght of the string he pulled after $12$ seconds
$⇒ BC' = BC - 0.6$
$⇒ BC' = 3 - 0.6$
$⇒ BC' = 2.4m$
In right $\triangle\text{ABC}'\text{M},$
By Pythagoras theorem,
$C'M^2 = BC'^2 - MB^2$
$⇒ C'M^2 = (2.4)^2 - (1.8)^2$
$⇒ C'M^2 = 5.76 - 3.24$
$⇒ C'M^2 = 2.52$
$⇒ C'M^2 = 1.59m ....$(approximately)
The horizontal distance of the fly from him after $12$ second
$= C' A$
$= C' M + MA$
$= 1.59 + 1.2$
$= 2.79m$ approximately
$= 2.8m$ approximately View full question & answer→Question 304 Marks
In triangles BMP and $CNR$ it is given that $PB = 5\ cm, MP = 6\ cm, BM = 9\ cm$ and $NR = 9\ cm$. If $\triangle\text{BMP}\sim\triangle\text{CNR}$ then find the perimeter of $\triangle\text{CNR}.$
Answer$\triangle\text{BMP}\sim\triangle\text{CNR}$
$\Rightarrow\frac{\text{Perimeters of }\triangle\text{BMP}}{\text{Perimeters of }\triangle\text{CNR}}=\frac{\text{PB}}{\text{RC}}=\frac{\text{MP}}{\text{NR}}=\frac{\text{MB}}{\text{NC}}$
$\Rightarrow\frac{\text{BM+MP+PB }}{\text{Perimeters of }\triangle\text{CNR}}=\frac{\text{PB}}{\text{RC}}=\frac{\text{MP}}{\text{NR}}=\frac{\text{MB}}{\text{NC}}$
$\Rightarrow\frac{\text{9+6+5 }}{\text{Perimeters of }\triangle\text{CNR}}=\frac{\text{5}}{\text{RC}}=\frac{\text{6}}{\text{9}}=\frac{\text{9}}{\text{NC}}$
$\Rightarrow\frac{\text{20}}{\text{Perimeters of }\triangle\text{CNR}}=\frac{\text{6}}{9}$
$\Rightarrow\text{Perimeters of }\triangle\text{CNR}=\frac{\text{20}\times9}{9}$
$\Rightarrow\text{Perimeters of }\triangle\text{CNR}=30\text{cm}$
View full question & answer→Question 314 Marks
If $D, E$ and $F$ are respectively the midpoint of sides $AB, BC$ and $CA$ of $\triangle\text{ABC}$ then what is the ratio of the areas of $\triangle\text{DEF}$ and $\triangle\text{ABC}?$
Answer
Given: A $\triangle\text{ABC}$ in which $D, E$ and $F$ are respectively the midpoints of sides $AB, BC$ and $CA$
To find: $\text{ar}(\triangle\text{DEF}):\text{ar}(\triangle\text{ABC})$
Since E and F are the mid-points of $BC$ and $CA$ respectively,
so $EF || AB$ and $\text{EF}=\frac{1}{2}\text{AB}$
...(By the converse of Basic Proportionality theorem)
In particular, $EF || BD.$
$\therefore$ $BDFE$ is a || gm.
Similarly, $EDAF$ is a || gm.
Now, in $\triangle\text{DEF}$ and $\triangle\text{ABC},$
$\angle\text{E}=\angle\text{A}$ ...(opposite angles of a || gm are equal)
$\angle\text{F}=\angle\text{B}$ ...(opposite angles of a || gm are equal)
$\therefore\triangle\text{DEF}\sim\triangle\text{CAB}$ ...(AA criterion for similarity)
We know that, the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.
$\therefore\frac{\text{ar}(\triangle\text{DEF})}{\text{ar}(\triangle\text{CAB})}=\frac{\text{EF}^2}{\text{AB}^2}=\frac{\Big(\frac{1}{2}\text{AB}\Big)^2}{\text{AB}^2}=\frac{1}{4}$
Hence, $\text{ar}(\triangle\text{DEF}):\text{ar}(\triangle\text{ABC})\text{ is }1:4.$ View full question & answer→Question 324 Marks
A man goes $80\ m$ due east and then $150\ m$ due north. How far is he from the starting point?
Answerstarting from $A$, let the man goas from $A$ to $B$ and from $B$ to $C$, as shows in the figure.
Then,
$\text{AB}=80\text{m,}\text { BC}=150\text{m}$ and $\angle\text{ABC}=90^\circ$
From right $\triangle\text{ABC},$ we have
$\text{AC}^2=\big(\text{AB}^2+\text{BC}^2\big)\text{m}^2$
$=\big[(80)^2+(150)^2\big]\text{m}^2$
$=(6400+22500)\text{m}^2$
$=28900\text{m}^2$
$\therefore\text{AC}=\sqrt{28900}\text{m}=170\text{m}$
Hence, the man is 170m north-east from the starting point. View full question & answer→Question 334 Marks
In the given figure, $\text{DB}\perp\text{BC},\text{DE}\perp\text{AB}$ and $\text{AC}\perp\text{BC}.$
Prove that $\frac{\text{BE}}{\text{DE}}=\frac{\text{AC}}{\text{BC}}.$
Answer
In the given figure : $\text{DB}\perp\text{AB},\text{AC}\perp\text{BC}$ and $DB || AC$
$\therefore\angle\text{DBC}=\angle\text{ACB}$
AB is the transversal
$\therefore\angle\text{DBE}=\angle\text{BAC}$ $\big[\text{Alternate }\angle\text{s}\big]$
In $\triangle\text{DBE}$ and $\triangle\text{ABC}$
$\angle\text{DEB}=\angle\text{ACB}=90^\circ$
$\angle\text{DBE}=\angle\text{BAC}$
$\triangle\text{DBE}\sim\triangle\text{ABC}$ [By AA similarity]
$\Rightarrow\frac{\text{BE}}{\text{DE}}=\frac{\text{AC}}{\text{BC}}$
Hence proved. View full question & answer→Question 344 Marks
For the following statments state whether true $(T)$ or false$(F):$
The sum of the squares on the sides of a rhombus is equal to the sum of the squares on its diagonals.
AnswerTrue.
Solution:
Given: A rhombus $ABCD$ whose diagonals $AC$ and $BD$ intersect at $O.$
To prove: $\left(A B^2+B C^2+C D^2+A D^2\right)=\left(A C^2+B D^2\right)$
Proof:
We know that the diagonals of a rhombus are perpendicular bisectors of each other,
$\therefore\angle\text{AOB}=\angle\text{BOC}=\angle\text{COD}=\angle\text{DOA}=90^\circ$
$\text{OA}=\frac{1}{2}\text{AC}$ and $\text{OB}=\frac{1}{2}\text{BD}$
From right $\triangle\text{AOB},$
$\text{AB}^2=\text{OA}^2+\text{OB}^2$
$\Rightarrow\text{AB}^2=\Big(\frac{1}{2}\text{AC}\Big)^2+\Big(\frac{1}{2}\text{BD}\Big)^2$
$\Rightarrow\text{AB}^2=\frac{1}{4}(\text{AC}^2+\text{BD}^2)$
$\Rightarrow4\text{AB}^2=\text{AC}^2+\text{BD}^2\dots(\text{i})$
Similarly, we have
$4\text{BC}^2-\text{AC}^2+\text{BD}^2\dots(\text{ii})$
$4\text{CD}^2=\text{AC}^2+\text{BD}^2\dots(\text{iii})$
$4\text{DA}^2=\text{AC}^2+\text{BD}^2\dots(\text{iv})$
Addind (i), (ii), (iii) and (iv), we get
$\text{AB}^2+\text{BC}^2+\text{CD}^2+\text{DA}^2=\text{AC}^2+\text{BD}^2$ View full question & answer→Question 354 Marks
QUESIION $D$ and $E$ are points on the sides $A B$ and $A C$ respectively of a $\triangle A B C$ such that $D E \| B C$ : If $\frac{ AD }{ DB }=\frac{4}{7}$ and $AC =6.6 cm$, find $AE .$
AnswerIn $\triangle\text{ABC},$ it is given that $DE || BC.$
Applying Thales' theorem, we get:
$\frac{\text{AD}}{\text{DB}}=\frac{\text{AE}}{\text{EC}}$
$\Rightarrow\frac{\text{4}}{\text{7}}=\frac{\text{AE}}{\text{EC}}$
Adding $1$ to both sides, we get:
$\Rightarrow\frac{\text{11}}{\text{7}}=\frac{\text{AE}}{\text{EC}}$
$\text{EC}=\frac{6.6\times7}{11}=4.2\text{cm}$
Therefore, $\text{AE}=\text{AC}-\text{EC}=6.6-4.2=2.4\text{cm}$
View full question & answer→Question 364 Marks
State the basic proportionality theorem.
AnswerIf a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct point, then the other sides are divided in the same ratio.
View full question & answer→Question 374 Marks
For the following statments state whether true $(T)$ or false$(F):$
The ratio of the perimeters of two similar triangles is the same as the ratio of their corresponding medians.
AnswerTrue.
Solution:
Given $\triangle\text{ABC}\sim\triangle\text{DEF}$
$\Rightarrow\frac{\text{AB}}{\text{DE}}=\frac{\text{BC}}{\text{EF}}=\frac{\text{AC}}{\text{DF}}$
$\Rightarrow\frac{\text{AB}}{\text{DE}}=\frac{2\text{BP}}{2\text{EQ}}=\frac{\text{BP}}{\text{EQ}}$
and $\angle\text{B}=\angle\text{E}$
Now, in $\triangle\text{APB}$ and $\triangle\text{DQE},$
$\angle\text{B}=\angle\text{E}$
$\frac{\text{AB}}{\text{DE}}=\frac{\text{BP}}{\text{EQ}}$
$\Rightarrow\triangle\text{APB}\sim\triangle\text{DQE}$ .....(SAS criterion for similarity)
$\therefore\frac{\text{AB}}{\text{DE}}=\frac{\text{AP}}{\text{DQ}}$
So, $\frac{\text{AB}}{\text{DE}}=\frac{\text{BC}}{\text{EF}}=\frac{\text{AC}}{\text{DF}}=\frac{\text{AP}}{\text{DQ}}$
We now that,
The ratio of the perimeters of two similar triangles is the same as the ratio of their corresponding sides.
That is, $\frac{\text{AB}+\text{BC}+\text{AC}}{\text{DE+EF+DF}}=\frac{\text{AP}}{\text{DQ}}$ View full question & answer→Question 384 Marks
In the given figure, each one of $P A, Q B$ and $R C$ is perpendicular to $A C$. If $A P=x, Q B=z, R C=y, A B=a$ and $B C=b$, show that $\frac{1}{x}+\frac{1}{y}=\frac{1}{z}$.
AnswerIn $\triangle\text{PAC}$ and $\triangle\text{QBC},$ we have:
$\angle\text{A}=\angle\text{B}$ (Both angles are $90^\circ )$
$\angle\text{P}=\angle\text{Q}$ (Corresponding angle)
and
Therefore, $\triangle\text{PAC}\sim\triangle\text{QBC}$
$\frac{\text{AP}}{\text{BQ}}=\frac{\text{AC}}{\text{BC}}$
$\Rightarrow\frac{\text{x}}{\text{z}}=\frac{\text{a+b}}{\text{b}}$
$\Rightarrow\text{a}+\text{b}=\frac{\text{bx}}{\text{z}}\dots(1)$
In $\triangle\text{RCA}$ and $\triangle\text{QBA},$ we have:
$\angle\text{C}=\angle\text{B}$ (Both angles are $90^\circ )$
$\angle\text{R}=\angle\text{Q}$ (Corresponding angles)
and
$\angle\text{A}=\angle\text{A}$ (common angles)
Therefore, $\triangle\text{RCA}\sim\triangle\text{QBA}$
$\frac{\text{RC}}{\text{BQ}}=\frac{\text{AC}}{\text{AB}}$
$\Rightarrow\frac{\text{y}}{\text{z}}=\frac{\text{a+b}}{\text{a}}$
$\Rightarrow\text{a}+\text{b}=\frac{\text{ay}}{\text{z}}\dots(2)$
From equation $(1)$ and $(2)$, we have:
$\frac{\text{bx}}{\text{z}}=\frac{\text{ay}}{\text{z}}$
$\Rightarrow\text{bx}=\text{ay}$
$\frac{\text{a}}{\text{b}}=\frac{\text{x}}{\text{y}}\dots(3)$
Also,
$\Rightarrow\frac{\text{x}}{\text{z}}=\frac{\text{a+b}}{\text{b}}$
$\Rightarrow\frac{\text{x}}{\text{z}}=\frac{\text{a}}{\text{b}}+1$
Using the value of $\frac{\text{a}}{\text{b}}$ from equation $(3)$, we have:
$\Rightarrow\frac{\text{x}}{\text{z}}=\frac{\text{x}}{\text{y}}+1$
Dividing both side by x, we get:
$\frac{\text{1}}{\text{z}}=\frac{\text{1}}{\text{y}}+\frac{\text{1}}{\text{z}}$
$\therefore\frac{\text{1}}{\text{x}}+\frac{\text{1}}{\text{y}}=\frac{\text{1}}{\text{z}}$
This completes the proof.
View full question & answer→Question 394 Marks
Find the lenght of a diagonal of a rectangle whose adjacent side are $30\ cm$ and $16\ cm.$
AnswerLet $ABCD$ is the given rectangle let $BD$ is a diagonal making a $\triangle\text{ADB}.$
$\Rightarrow\angle\text{BAD}=90^\circ$
Using Pythagoras theorem:
$(\text{DB})^2=\text{AB}^2+\text{AD}^2$
$\text{DB}^2=\big(16^2+30^2\big)\text{cm}^2$
$\text{DB}=\sqrt{16^2+30^2}\text{cm}$
$=\sqrt{256+900}$
$=34\text{cm}$
Hence, lenght of diagonal $DB$ is $34\ cm.$ View full question & answer→Question 404 Marks
$\triangle\text{ABC}$ is an equilateral triangle of side $2a$ units. Find each of its altitudes.
AnswerIn an equilateral triangle all sides are equal.
Then, $A B=B C=A C=2 a$ uints
Const: Draw an altitude $AD \perp BC$
Given $B C=2 a$ Then, $BD = a$
In $\triangle\text{ABD},$
$\angle\text{ADB}=90^\circ$
$(\text{AB})^2=\text{(AD)}^2+\text{(BD})^2$
(by pythagoras theorem)
$\text{(AD})^2=\big(\text{AB}^2-\text{BD}^2\big)$
$=\Big[(2\text{a})^2-(\text{a})^2\Big]\text{sq}.\text{units}$
$=\big(4\text{a}^2-\text{a}^2\big)\text{sq}.\text{unit}=3\text{a}^2\text{sq}.\text{unit}$
$\text{AD}=\sqrt{3\text{a}^2}\text{unit}=\text{a}\sqrt{3}\text{ unit}$
Hence, lenght of each altitude is $\text{a}\sqrt{3}\text{ units}$ View full question & answer→Question 414 Marks
The areas of two similar triangles are $81cm^2$ and $49cm^2$ respectively. If the altitudes of the first triangle is $6.3\ cm$, find the corresponding altitude of the other.
AnswerIt is given that the triangles are similar.
Therefore, the ratio of the areas of these triangles is will be equal to the ratio of squares of their corresponding sides.
Also, the ratio of areas of two similar triangles is equal to the ratio of square of their correspondind altitudes.
Let the two triangles be $ABC$ and $DEF$ with altitudes $AP$ and $DQ$, respectively.
$\frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{PQR})}=\frac{\text{AP}^2}{\text{DQ}^2}$
$\Rightarrow\frac{81}{49}=\frac{\text{6.3}^2}{\text{DQ}^2}$
$\Rightarrow\text{DQ}^2=\frac{49}{81}\times6.3^2$
$\Rightarrow\text{DQ}^2=\sqrt{\frac{49}{81}\times6.3\times6.3}$
$=4.9\text{cm}$
Hence, the altitude of the other triangle is $4.9\ cm.$ View full question & answer→Question 424 Marks
The areas of two similar triangles are $100 \mathrm{~cm}^2$ and $64 \mathrm{~cm}^2$ respectively. If a median of the smaller triangle is $5.6\ cm$, find the correspondin median of the other.
AnswerLet the two triangles be $ABC$ and $PQR$ with median $AM$ and $PN$, respectively.
Therefore, the ratio of areas of two similar triangles will be equal to the ratio of squares of their corrrsponding medians.
$\therefore\frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{PQR})}=\frac{\text{AM}^2}{\text{PN}^2}$
$\Rightarrow\frac{64}{100}=\frac{\text{5.6}^2}{\text{PN}^2}$
$\Rightarrow\text{PN}^2=\frac{64}{100}\times5.6^2$
$\Rightarrow\text{PN}^2=\sqrt{\frac{100}{64}\times5.6\times5.6}$
$=7\text{cm}$
Hence, the median of the larger triangle is $7\ cm.$ View full question & answer→Question 434 Marks
Prove that the internal bisector of an angle of a triangle divides the opposite side internally in the ratio of the sides containing the angle.
AnswerLet the triangle be $ABC$ with $AD$ as the bisector of $\angle\text{A}$ which meets $BC$ at $D.$
We have to prove:
$\frac{\text{BD}}{\text{DC}}=\frac{\text{AB}}{\text{AC}}$
Draw $CE || DA$, meeting $BA$ produced at $E$
$CE || DA$
Therefore,
$\angle2=\angle3$ (Alternate angles)
and $\angle1=\angle4$ (Corresponding angles)
But,
$\angle1=\angle2$
Therefore,
$\angle3=\angle4$
$\Rightarrow\text{AE}=\text{AC}$
In $\triangle\text{BCE},\text{DA }||\text{ CE}.$
Applying Thales' theorem, we have:
$\frac{\text{BD}}{\text{DC}}=\frac{\text{AB}}{\text{AE}}$
$\Rightarrow\frac{\text{BD}}{\text{DC}}=\frac{\text{AB}}{\text{AC}}$
This completes the proof. View full question & answer→Question 444 Marks
State the $SAS-$similarity criterion.
AnswerIf one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional then the two triangles are similar.
View full question & answer→Question 454 Marks
Two chords $AB$ and $CD$ of a circle intersect at a point $P$ outside the ciecle. Prove that.
- $\triangle\text{PAC}\sim\triangle\text{PDB}$
- $\text{PA}.\text{PB}=\text{PC}.\text{PD}$
AnswerIn $\triangle\text{PAC}$ and $\triangle\text{PDB},$
$\angle\text{APC}=\angle\text{BPD}$ .....(common angle)
Since $ABCD$ is a cyclic quadrilateral,
$\angle\text{PAC}=\angle\text{BDP}$
$\therefore\triangle\text{PAC}\sim\triangle\text{PDB}$ .....($AA$ similarity criterion)
$\Rightarrow\frac{\text{PA}}{\text{PD}}=\frac{\text{PC}}{\text{PB}}$
$\Rightarrow\text{PA}.\text{PB}=\text{PC}.\text{PD}$
View full question & answer→Question 464 Marks
In the given figure, $\angle1=\angle2$ and $\frac{\text{AC}}{\text{BD}}=\frac{\text{CB}}{\text{CE}}$
Prove that $\triangle\text{ACB}\sim\triangle\text{DCE}.$
Answer
$\angle1=\angle2$ (given)
$\frac{\text{AC}}{\text{BD}}=\frac{\text{CB}}{\text{CE}}\Rightarrow\frac{\text{AC}}{\text{CB}}=\frac{\text{BD}}{\text{CE}}$ (given)
Also, $\angle2=\angle1$ $\big[\therefore\text{BD}=\text{DC}\big]$
Thus, $\frac{\text{AC}}{\text{CB}}=\frac{\text{BD}}{\text{CE}}$
and $\angle2=\angle1$
Therefore, by $SAS$ similarity criterion $\triangle\text{ACB}\sim\triangle\text{DCE}$ View full question & answer→Question 474 Marks
A man goes $12\ m$ due south and then $35\ m$ due west. How far is he from the starting point?
Answer
Let the starting point be $B$.
Since $\triangle\text{OAB}$ forms a right-angled triangle,
By Pythagoras theorem,
$ O B^2=O A^2+A B^2 $
$ \Rightarrow O B^2=35^2+12^2 $
$ \Rightarrow O B^2=1225+144 $
$ \Rightarrow O B^2=1369 $
$ \Rightarrow O B=37 \mathrm{~m}$
So, he is $37\ m$ away from the starting point. View full question & answer→Question 484 Marks
In the given figure, $D$ is the midpoint of side $B C$ and $A E \perp B C$. If $B C=a, A C=b, A B=c, E D=x, A D=p$ and $A E=h$, prove that..
$(\text{b}^2+\text{c}^2)=2\text{p}^2+\frac{1}{2}\text{a}^2$ AnswerGiven: D is the midpoint of side $\text{BC},\text{AE}\perp\text{BC}$
$\text{BC}=\text{a},\text{AC}=\text{b},\text{AB}=\text{c},\text{ED}=\text{x},\text{AD}=\text{p}$ and $\text{AE}=\text{h}$
In $\triangle\text{AEC},\angle\text{AEC}=90^\circ$
$\text{AD}^2=\text{AE}^2+\text{ED}^2$ (by pythagoras theorem)
$\Rightarrow\text{p}^2=\text{h}^2+\text{x}^2$
Adding $(1)$ and $(2)$,, we get
$\text{b}^2+\text{c} ^2=\text{p}^2+\text{ax}+\frac{\text{a}^2}{4}+\text{p}^2-\text{ax}+\frac{\text{a}^2}{4}$
$\big(\text{b}^2+\text{c}^2\big)=2\text{p}^2+\frac{1}{2}\text{a}^2$
View full question & answer→Question 494 Marks
For the following statments state whether true $(T)$ or false$(F):$
The polygon formed by joining the midpoit of the sides of a quadrilateral is a rhombus.
AnswerFalse.
Solution:
The line segments joining the midpoint of the adjacent sides of a quadrilateral form a parallelogram as shown.
It may or may not be a rhombus.
View full question & answer→Question 504 Marks
In $\triangle\text{ABC},\text{D}$ and $\text{E}$ are the midpoint of $AB$ and $AC$ respectively. Find the ratio of the areas of $\triangle\text{ADE}$ and $\triangle\text{ABC}.$
AnswerIt is given that $D$ and $E$ are midpoint of $AB$ and $AC.$
Applying midpoint theorem, we can conclude that $DE || BC.$
Hence, by $B.P.T$ we get:
$\frac{\text{AD}}{\text{AB}}=\frac{\text{AE}}{\text{AC}}$
Also, $\angle\text{A}=\angle\text{A}$
Applying $SAS$ similarity theorem, we can conclude that $\triangle\text{ADE}\sim\triangle\text{ABC}.$
Therefore, the ratio of the areas of these triangles will be equal to the ratio of squares of their corresponding sides.
$\therefore\frac{\text{ar}(\triangle\text{ADE})}{\text{ar}(\triangle\text{ABC})}=\frac{\text{DE}^2}{\text{BC}^2}$
$=\frac{\big(\frac{1}{2}\text{BC}\big)^2}{\text{BC}^2}$
$=\frac{1}{4}$
View full question & answer→