Question 13 Marks
If the sides of a triangle are $3\ cm, 4\ cm$ and $6\ cm$ long, determine whether the triangle is a right-angled triangle.
AnswerWe have,
$a = 3cm$
$b = 4cm$
$c = 6cm$
In order to prove that the triangle is a right angled triangle we have to prove that square of the larger side is equal to the sum of the squares of the other two sides.
Here, the larger side is $c = 6cm$
Hence, we have to prove that $a^2+b^2=c^2$
Let solve the left hand side of the above equation.
$ a^2+b^2=3^2+4^2 $
$ =9+16$
$=25$
Now we will solve the right hand side of the equation,
$ c^2=6^2$
$ =36$
Here we can observe that left hand side is not equal to the right hand side.
Therefore, the given triangle is not a right angled triangle.
View full question & answer→Question 23 Marks
In fig. $\triangle\text{ABC}$ is a triangle such that $\frac{\text{AB}}{\text{AC}}=\frac{\text{BD}}{\text{DC}},\angle\text{B}=70^\circ,\angle\text{C}=50^\circ.$ Find the $\angle\text{BAD}.$
Answer
In $\triangle\text{ABC},$
$\frac{\text{AB}}{\text{AC}}=\frac{\text{BD}}{\text{DC}},\angle\text{B}=70^\circ,\angle\text{C}=50^\circ$
We know that,
$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
$\Rightarrow\angle\text{A}+70^\circ+50^\circ=180^\circ$
$\Rightarrow\angle\text{A}+120^\circ=180^\circ$
$\Rightarrow\angle\text{A}=180^\circ-120^\circ$
$\Rightarrow\angle\text{A}=60^\circ$
$\because\frac{\text{AB}}{\text{AC}}=\frac{\text{BD}}{\text{DC}}$
$\therefore AD$ is bisector of $\angle\text{A}$ so,
$\angle1=\angle2$
$\Rightarrow\angle\text{A}=\angle1+\angle2$
$\Rightarrow60^\circ=2\angle1$
$\Rightarrow\angle1=\frac{60^\circ}{2}$
$\Rightarrow\angle1=30^\circ$
$\Rightarrow\angle\text{BAD}=30^\circ$ View full question & answer→Question 33 Marks
In the given figure, $DE || BC$ such that $\text{AE}=\Big(\frac{1}{4}\Big) AC$. If $AB = 6\ cm,$ find $AD.$
AnswerWe have,
$DE || BC$ and $\text{AE}=\frac{1}{4}\text{ AC}, AB = 6\ cm$
In $\triangle\text{ADE}$ and $\triangle\text{ABC},$
$\angle\text{A}=\angle\text{A}$ (Common)
$\triangle\text{ADE}\sim\triangle\text{ABC}$
$\Rightarrow\frac{\text{AD}}{\text{AB}}=\frac{\text{AE}}{\text{AC}}$
$\Rightarrow\frac{\text{AD}}{6}=\frac{1}{4}$
$\Rightarrow\text{AD}=\frac{6}{4}=\frac{3}{2}$
$\Rightarrow\text{AD}=1.5\text{cm}$
View full question & answer→Question 43 Marks
In $\triangle\text{ABC and }\triangle\text{DEF},$ it is being given that: $AB = 5\ cm, BC = 4\ cm$ and $CA = 4.2\ cm; DE = 10\ cm, EF = 8\ cm$ and $FD = 8.4\ cm.$ If $\text{AL}\perp\text{BC}$ and $\text{DM} \perp \text{EF,}$ find $AL : DM.$
Answer
Since, $\frac{\text{AB}}{\text{DE}}=\frac{\text{BC}}{\text{EF}}=\frac{\text{AC}}{\text{DE}}=\frac{1}{2}$
Then, $\triangle\text{ABC}\sim\triangle\text{DEF} [$By $SSS$ similarity$]$
Now, In $\triangle\text{ABL}\sim\triangle\text{DEF}$
$\angle\text{B}=\angle\text{E}$ $\big[\triangle\text{ABC}\sim\triangle\text{DEF}\big]$
$\angle\text{ALB}=\angle\text{DME}[$ Each $90^\circ ]$
Then, $\triangle\text{ABL}\sim\triangle\text{DEM} [$By $AA$ similarity$]$
$\therefore\frac{\text{AB}}{\text{DE}}=\frac{\text{AL}}{\text{DM}}$ [Corresponding parts of similar $\triangle$ are proportional]
$\Rightarrow\frac{5}{10}=\frac{\text{AL}}{\text{DM}}$
$\Rightarrow\frac{1}{2}=\frac{\text{AL}}{\text{DM}}$ View full question & answer→Question 53 Marks
In the given figure, $\angle\text{ABC}=90^\circ$ and $\text{BD}\perp\text{AC}.$ If $BD = 8cm$ and $AD = 4cm,$ find $CD.$
Answer
In right $\triangle\text{ABC},$
$\text{BD}\perp\text{AC}$
$\therefore\triangle\text{ABD}\sim\triangle\text{CBD}$
$\therefore\frac{\text{AD}}{\text{BD}}=\frac{\text{BD}}{\text{CD}}$
$\Rightarrow\frac{4}{8}=\frac{8}{\text{CD}}\Rightarrow\text{CD}=\frac{8\times8}{4}=16$
$\therefore\text{CD}=16\text{cm}$ View full question & answer→Question 63 Marks
A vertical stick of length $6\ m$ casts a shadow $4\ m$ long on the ground and at the same time a tower casts a shadow $28\ m$ long. Find the height of the tower.
Answer
Let $AB$ be a tower
$CD$ be a stick, $CD = 6m$
Shadow of $AB$ is $BE = 28m$
Shadow of $CD$ is $DF = 4m$
At same time light rays from sun will fall on tower and stick same angle.
So, $\angle\text{DCF}=\angle\text{BAE}$
And $\angle\text{DFC}=\angle\text{BEA}$
$\angle\text{CDF}=\angle\text{ABE}$ [tower and stick are vertical to ground)
Therefore $\triangle\text{ABE}\sim\triangle\text{CDF}($ By $AA$ Similarity$)$
So, $\frac{\text{AB}}{\text{CD}}=\frac{\text{BE}}{\text{DF}}$
$\frac{\text{AB}}{6}=\frac{28}{4}$
$\text{AB}=28\times\frac{6}{4}=42\text{m}$
So, height of tower will be $42$ metres. View full question & answer→Question 73 Marks
In an equilateral $\triangle\text{ABC},\ \text{AD}\perp\text{BC}$ prove that $AD^2= 3BD^2$.
AnswerGiven: $\triangle\text{ABC}$ is an equilateral in which $AB = BC = CA.$
$\text{AD}\perp\text{BC}$
To prove: $AD^2= 3BD^2$
Proof: In equilateral $\triangle\text{ABC}.$
$\because\text{AD}\perp\text{BC}$
$\therefore AD$ bisects $BC$ at $D$
$\therefore\text{BD}=\frac{1}{2}\text{BC}$
Now in right $\triangle\text{ABD}$
$ A B^2=A D^2+B D^2 \text { (pythagoras theorem) } $
$ \therefore A D^2=A B^2-B D^2 $
$ A D^2=B C^2-B D^2\{A B=A C=B C \text { given }\} $
$ A D^2=(2 B D)^2-B D^2\{B C=2 B D\} $
$ A D^2=4 B D^2-B D^2 $
$ A D^2=3 B D^2$
Hence proved. View full question & answer→Question 83 Marks
In a $\triangle\text{ABC, D}$ and $E$ are points on the sides $AB$ and $AC$ respectively. For the following cases show that $DE || BC:$
$AB = 10.8cm, BD = 4.5cm, AC = 4.8cm$ and $AE = 2.8cm.$
AnswerIt is given that $D$ and $E$ are point on sides $AB$ and $AC.$
We have to prove that $DE || BC.$
According to thales theorem we have
$\frac{\text{AD}}{\text{DB}}=\frac{\text{AE}}{\text{CE}}$
$AD = AB - DB = 10.8 - 4.5 = 6.3$
And $EC = AC - AE = 4.8 - 2.8 = 2$
Now
$\frac{6.3}{4.5}=\frac{2.8}{2.0}$
Hence, $DE || BC.$
View full question & answer→Question 93 Marks
In $\triangle\text{ABC},\ \angle\text{A}$ is obtuse, $\text{PB}\perp\text{AC}$ and $\text{QC}\perp\text{AB}$ Prove that:
$AB × AQ = AC × AP$
Answer
Then, $\triangle\text{APB}\sim\triangle\text{AQC} [$By $AA$ similarity$]$
$\therefore\frac{\text{AP}}{\text{AQ}}=\frac{\text{AB}}{\text{AC}} [$Corresponding parts of similar $\triangle$ are proportional$]$
$⇒ AP × AC = AQ × AB .....(i)$ View full question & answer→Question 103 Marks
Two poles of heights $6m$ and $11m$ stand on a plane ground. If the distance between their feet is $12m,$ find the distance between their tops.
AnswerLet us draw the diagram from the given information.
Let us draw a perpendicular from $B$ on $CD$ which meets $CD$ at $P.$
It is clear that $BP = 12m$ because it is given that distance between feet of the two poles is $12m.$
After drawing the perpendicular we get a rectangle $BACP$ such that $AB = PC$ and $BP = AC.$
Because of this construction we also obtained a right angled triangle $BPD.$
Now we will use Pythagoras theorem,
$B D^2=B P^2+P D^2$
$\text { Let us substitute the values of } B P \text { and PD we get, }$
$ \mathrm{BD}^2=12^2+5^2 $
$ \therefore \mathrm{BD}^2=144+25 $
$ \mathrm{BD}^2=169$
Taking the square root we get, $BD = 13$
Therefore, distance between the top of the two poles is $13m.$ View full question & answer→Question 113 Marks
In a $\triangle\text{ABC,D}\ \text{and E}$ are points on the sides $AB$ and $AC$ respectively such that $DE || BC.$
If $AD = x, DB = x - 2, AE = x + 2$ and $EC = x - 1,$ find the value of $x.$
AnswerIn the figure,
$AD = x, DB = x - 2, AE = x + 2$ and $EC = x - 1$
$\because DE || BC$
$\therefore\frac{\text{AD}}{\text{DB}}=\frac{\text{AE}}{\text{EC}}\Rightarrow\frac{\text{x}}{\text{x}-2}=\frac{\text{x}+2}{\text{x}-1}$
$ \Rightarrow x(x-1)=(x+2)(x-2) $
$ \Rightarrow x^2-x=x^2-4 $
$ \Rightarrow x^2-x-x^2=-4 \Rightarrow-x=-4$
$ \Rightarrow x=4 $
$ \therefore x=4$
$\therefore x = 4$ View full question & answer→Question 123 Marks
In fig. $AE$ is the bisector of the exterior $\angle\text{CAD}$ meeting $BC$ produced in E. If $AB = 10cm, AC = 6cm$ and $BC = 12cm,$ find $CE.$
AnswerIn $\triangle\text{ABC, AD}$ is the bisector of $\angle\text{A}.$
We know that, the internal bisector of an angle of a triangle divides the opposite side internally in the retio of the sides containing the angle.
$\therefore\frac{\text{BC}}{\text{CE}}=\frac{\text{AB}}{\text{AC}}\Rightarrow\frac{12-\text{x}}{\text{x}}=\frac{10}{6}$
$\Rightarrow6(12+\text{x})=10\text{x}$
$\Rightarrow72+6\text{x}=10\text{x}$
$\Rightarrow4\text{x}-72=0$
$\Rightarrow\text{x}=\frac{72}{4}=18\text{cm}$
$\therefore\text{CE}=\text{x}=18\text{cm}$
View full question & answer→Question 133 Marks
In a $\triangle\text{ABC,D}\ \text{and E}$ are points on the sides $AB$ and $AC$ respectively such that $DE || BC.$
If $AD = 2\ cm, AB = 6\ cm$ and $AC = 9\ cm,$ find $AE.$
Answer
We have,
$AD = 2\ cm, AB = 6\ cm$
$\therefore DB = AB - AD$
$= 6 - 2$
$\Rightarrow DB = 4\ cm$
And, $DE || BC$
Therefore, by basic proportionality theorem, we have,
$\frac{\text{AD}}{\text{DB}}=\frac{\text{AE}}{\text{EC}}$
Taking reciprocal on both sides, we get,
$\frac{\text{DB}}{\text{AD}}=\frac{\text{EC}}{\text{AE}}$
$\frac{4}{2}=\frac{\text{EC}}{\text{AE}}$
Adding $1$ on both sides, we get
$\frac{4}{2}+1=\frac{\text{EC}}{\text{AE}}+1$
$\Rightarrow\frac{4+2}{2}=\frac{\text{EC}+\text{AE}}{\text{AE}}$
$\Rightarrow\frac{6}{2}=\frac{\text{AC}}{\text{AE}}\ \ [\because\text{EC}+\text{AE}=\text{AC}]$
$\Rightarrow\frac{6}{2}=\frac{9}{\text{AE}}\ \ [\because\text{AC}=9\text{cm}]$
$\text{AE}=\frac{9\times2}{6}$
$\Rightarrow\text{AE}=3\text{cm}$ View full question & answer→Question 143 Marks
In a $\triangle\text{ABC,D}\ \text{and E}$ are points on the sides $AB$ and $AC$ respectively such that $DE || BC.$
If $\frac{\text{AD}}{\text{DB}}=\frac{2}{3}$ and $AC = 18\ cm,$ find $AE$.
AnswerGiven, $\frac{\text{AD}}{\text{DB}}=\frac{2}{3},$ and $AC = 18\ cm$
In the figure, $DE || BC$
$\therefore\frac{\text{AD}}{\text{DB}}=\frac{\text{AE}}{\text{EC}}$
$\Rightarrow\frac{2}{3}=\frac{\text{AE}}{\text{EC}}$
$\text{Let AE}=\text{x}$
$\therefore\text{EC}=\text{AC}-\text{AE}=18-\text{x}$
$\therefore\frac{\text{x}}{18-\text{x}}=\frac{2}{3}\Rightarrow3\text{x}=36-2\text{x}$
$\Rightarrow3\text{x}+2\text{x}=36\Rightarrow5\text{x}=36$
$\Rightarrow\text{x}=\frac{36}{5}=7.2$
$\therefore\text{AE}=7.2\text{cm}$ View full question & answer→Question 153 Marks
Prove that in an equilateral triangle, three times the square of a side is equal to four times the square of its altitudes.
AnswerGiven: In an equilateral $\triangle\text{ABC},$ $\text{AD}\perp\text{BC}$
To prove: $3AB^2= 4AD^2$
Proof: The altitude of an equilateral triangle bisects the opposite side
$\because D$ is the mid-point of $BC$
Now in right $\triangle\text{ABD},$
$A B^2=B D^2+A D^2$ (pythagoras Theorem)
$\Rightarrow\text{AB}^2=\Big(\frac{1}{2}\text{BC}\Big)^2+\text{AD}^2=\frac{1}{4}\text{BC}^2+\text{AD}^2$
$\Rightarrow\text{AB}^2=\frac{1}{4}\text{AB}^2+\text{AD}^2$
$\Rightarrow4\text{AB}^2=\text{AB}^2+4\text{AD}^2$
$\Rightarrow4\text{AB}^2-\text{AB}^2=4\text{AD}^2$
$\therefore3\text{AB}^2=4\text{AD}^2$
Hence proved. View full question & answer→Question 163 Marks
If a $\triangle\text{ABC}, AD$ is the bisector of $\angle\text{A},$ Meeting side $BC$ at $D.$
If $AB = 3.5\ cm, AC = 4.2\ cm$ and $DC = 2.8cm,$ find $BD.$
Answer
In $\triangle\text{ABC}, AD$ is the bisector of $\angle\text{A},$
We know that, the internal bisector of an angle of a triangle divides the opposite side internally in the retio of the sides containing the angle.
$\therefore\frac{\text{BD}}{\text{DC}}=\frac{\text{AB}}{\text{AC}}$
$\Rightarrow\frac{\text{BD}}{2.8}=\frac{3.5}{4.2}$
$=\frac{3.5\times2}{3}$
$=\frac{7}{3}=2.33\text{cm}$ View full question & answer→Question 173 Marks
In the given figure, $\triangle\text{ACB}\sim\triangle\text{APQ}.$ If $BC = 10\ cm, PQ = 5\ cm, BA = 6.5\ cm$ and $AP = 2.8\ cm,$ find $CA$ and $AQ.$ Also, find the area$ (\triangle\text{ACB)}$ : area $ (\triangle\text{APQ)}.$
AnswerGiven: $\triangle\text{ACB}$ is similar to $\triangle\text{APQ}$
$BC = 10\ cm, PQ = 5\ cm, BA = 6.5\ cm$ and $AP = 2.8\ cm$
To find:
- $CA$ and $AQ$
- Area of $\triangle\text{ACB}$ : Area of $\triangle\text{APQ}$
- It is given that $\triangle\text{ACB}\sim\triangle\text{APQ}$
We know that for any two similar triangles the sides are proportional.
Hence, $\frac{\text{AB}}{\text{AQ}}=\frac{\text{BC}}{\text{PQ}}=\frac{\text{AC}}{\text{AP}}$
$\frac{\text{AB}}{\text{AQ}}=\frac{\text{BC}}{\text{PQ}}$
$\frac{6.5}{\text{AQ}}=\frac{10}{5}$
$AQ = 3.25\ cm$
Similarly,
$\frac{\text{BC}}{\text{PQ}}=\frac{\text{CA}}{\text{AP}}$
$\frac{\text{CA}}{2.8}=\frac{10}{5}$
$CA = 5.6\ cm$
- We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides.
$\frac{\text{ar}(\triangle\text{ACQ})}{\text{ar}(\triangle\text{APQ})}=\Big(\frac{\text{BC}}{\text{PQ}}\Big)^2=\Big(\frac{10}{5}\Big)^2=\Big(\frac{2}{1}\Big)^2=\frac{4}{1}$ View full question & answer→Question 183 Marks
In $\triangle\text{ABC, PQ}$ is a line segment intersecting $AB$ at $P$ and $AC$ at $Q$ such that $PQ || BC$ and $PQ$ divides $\triangle\text{ABC}$ into two parts equal in area. Find $\frac{\text{BP}}{\text{AB}}.$
Answer
We have,
$PQ || BC$
And ar $(\triangle\text{APQ}) = $ar$($trap. $PQCB)$
$\Rightarrow\text{ar}(\triangle\text{APQ})=\text{ar}(\triangle\text{ABC})-\text{ar}(\triangle\text{APQ})$
$\Rightarrow\text{2ar}(\triangle\text{APQ})=\text{ar}(\triangle\text{ABC})\ \ ...(\text{i})$
In $\triangle\text{APQ and }\triangle\text{ABC}$
$\angle\text{A}=\angle\text{A} [$Common$]$
$\angle\text{APQ}=\angle\text{B}[$ Corresponding angles$]$
Then, $\triangle\text{APQ}\sim\triangle\text{ABC} [$By $AA$ similarity$]$
$\therefore$ By area of similar triangle theorem
$\frac{\text{ar}(\triangle\text{APQ})}{\text{ar}(\triangle\text{ABC})}=\frac{\text{AP}^2}{\text{AB}^2}$
$\Rightarrow\frac{\text{ar}(\triangle\text{APQ})}{\text{ar}(\triangle\text{APQ})}=\frac{\text{AP}^2}{\text{AB}^2}\ \ [\text{By using (i)}]$
$\Rightarrow\frac{1}{2}=\frac{\text{AP}^2}{\text{AB}^2}$
$\Rightarrow\frac{1}{\sqrt{2}}=\frac{\text{AP}}{\text{AB}^2}$
$\Rightarrow\frac{1}{\sqrt{2}}=\frac{\text{AP}}{\text{AB}}$ [Taking square root]
$\Rightarrow\frac{1}{\sqrt{2}}=\frac{\text{AB}-\text{BP}}{\text{AB}}$
$\Rightarrow\frac{1}{\sqrt{2}}=\frac{\text{AB}}{\text{AB}}-\frac{\text{BP}}{\text{AB}}$
$\Rightarrow\frac{1}{\sqrt{2}}=1-\frac{\text{BP}}{\text{AB}}$
$\Rightarrow\frac{\text{BP}}{\text{AB}}=1-\frac{1}{\sqrt{2}}$
$\Rightarrow\frac{\text{BP}}{\text{AB}}=\frac{\sqrt{2}-1}{\sqrt{2}}$ View full question & answer→Question 193 Marks
If a $\triangle\text{ABC},$ AD is the bisector of $\angle\text{A},$ Meeting side $BC$ at $D.$
If $AC = 4.2\ cm, DC = 6\ cm$ and $BC = 10\ cm,$ find $AB$.
Answer$AC = 4.2\ cm, DC = 6\ cm$ and $BC = 10\ cm$
$\because BD = BC - DC = 10 - 6 = 4\ cm$
$\because AD$ is the bisector of $\angle\text{A}\ \text{of}\ \triangle\text{ABC}$
$\therefore\frac{\text{AB}}{\text{AC}}=\frac{\text{BD}}{\text{DC}}$
$\Rightarrow\frac{\text{AB}}{4.2}=\frac{4}{6}$
$\Rightarrow\text{AB}=\frac{4\times4.2}{6}=2.8$
$\therefore\text{AB}=2.8\text{cm}$ View full question & answer→Question 203 Marks
In an isosceles triangle $ABC, AB = AC = 25\ cm, BC = 14\ cm,$ Calculate the altitude from $A$ on $BC.$
AnswerWe know that altitude that is a perpendicular drawn on the unequal side of the isosceles triangle bisects that side.
Therefore, $BD = DC = 7cm.$
Let us use the Pythagoras theorem in right angled triangle $ADB$ we get,
$\mathrm{AB}^2=\mathrm{AD}{ }^2+\mathrm{BD}^2$
Substituting the values we get,
$25^2=A D^2+49$
Subtracting 49 from both the sides we get,
$ 625-49=A D^2 $
$ \therefore A D^2=576$
Let us take the square root we get,
$AD = 24cm$
Therefore, the altitude of the isosceles triangle is $24cm.$ View full question & answer→Question 213 Marks
The areas of two similar triangles are $169 \mathrm{~cm}^2$ and $121 \mathrm{~cm}^2$ respectively. If the longest side of the larger triangle is $26\ cm,$ what is the length of the longest side of the smaller triangle$?$
AnswerLet $\triangle\text{ABC}$ be the larger triangle and $\triangle\text{PQR}$
be the smaller triangle and their longest sides be $BC$ and $QR$ respectively
Area of $\triangle\text{ABC}=169\text{cm}^2$
Area of $\triangle\text{PQR} = 121\text{cm}^2$
$BC = 26\ cm$
Let $QR = x\ cm$
$\triangle\text{ABC}\sim\triangle\text{PQR}$
$\therefore\frac{\text{area}\triangle\text{ABC}}{\text{area}\triangle\text{PQR}}=\frac{\text{BC}^2}{\text{QR}^2}$
$\Rightarrow\frac{169}{121}=\frac{(26)^2}{\text{x}^2}$
$\Rightarrow\frac{(13)^2}{(11)^2}=\frac{(26)^2}{(\text{x})^2}\Rightarrow\frac{13}{11}=\frac{26}{\text{x}}$
$\Rightarrow\text{x}=\frac{26\times11}{13}=22$
$\therefore\text{QR}=22\text{cm}$
View full question & answer→Question 223 Marks
In $\triangle\text{ABC},$ the bisector of $\angle\text{A}$ intersects $BC$ in $D.$ If $AB = 18\ cm, AC = 15\ cm$ and $BC = 22\ cm,$ find $BD.$
Answer
We have to find the value of $BD.$
Given: $AB = 18\ cm, AC = 15\ cm$ and $BC = 22\ cm.$
In $\triangle\text{ABC}, AD$ the bisector of $\angle\text{A}.$
$\frac{\text{AB}}{\text{AC}}=\frac{\text{BD}}{\text{DC}}$
$\frac{\text{AB}}{\text{AC}}=\frac{\text{BD}}{\text{BC}-\text{BD}}$
$\frac{18}{15}=\frac{\text{BD}}{22-\text{BD}}$
On cross multiplication, we get,
$6(22 - BD) = 5 × BD$
$132 - 6BD = 5BD$
$132 = 5BD + 6BD$
$132 = 11BD$
$\text{BD}=\frac{132}{11}$
$BD = 12\ cm$
Hence, the value of $BD$ is $12\ cm.$ View full question & answer→Question 233 Marks
The perimeters of two similar triangles are $25\ cm$ and $15\ cm$ respectively. If one side of first triangle is 9cm, what is the corresponding side of the other triangle$?$
AnswerLet perimeter of $\triangle\text{ABC}=25\text{cm}$
and perimeter of $\triangle\text{DEF}=15\text{cm}$
and side $BC$ of $\triangle\text{ABC}=9\text{cm}$
Now we have to find the side EF of $\triangle\text{DEF}$
$\triangle\text{ABC}\sim\triangle\text{DEF}$ (given)
$\therefore\frac{\text{Perimeter of }\triangle\text{ABC}}{\text{Perimeter of }\triangle\text{DEF}}=\frac{\text{BC}}{\text{EF}}$
$\Rightarrow\frac{25}{15}=\frac{9}{\text{EF}}$
$\Rightarrow\text{EF}=\frac{15\times9}{25}=\frac{27}{5}=5.4\text{cm}$ View full question & answer→Question 243 Marks
In a $\triangle\text{ABC,D}\ \text{and E}$ are points on the sides $AB$ and $AC$ respectively such that $DE || BC.$
If $AD = 4, AE = 8, DB = x - 4,$ and $EC = 3x - 19$, find $x.$
Answer
We have,
$DE || BC$
therefore, by basic proportionally theorem,
We have $\frac{\text{AD}}{\text{DB}}=\frac{\text{AE}}{\text{EC}}$
$\frac{4}{\text{x}-4}=\frac{8}{3\text{x}-19}$
$\Rightarrow4(3\text{x}-19)=8(\text{x}-4)$
$\Rightarrow12\text{x}-76=8\text{x}-32$
$\Rightarrow12\text{x}-8\text{x}=-32+76$
$\Rightarrow4\text{x }=44$
$\Rightarrow\text{x}=\frac{44}{4}=11\text{cm}$
$\therefore\text{x}=11\text{cm}$ View full question & answer→Question 253 Marks
In $\triangle\text{ABC},$ $P$ and $Q $ are points on sides $AB$ and $AC$ respectively such that $PQ || BC.$ If $AP = 3\ cm, PB = 5\ cm$ and $AC = 8\ cm,$ find $AQ.$
AnswerIn $\triangle\text{ABC}, P$ and $Q$ are points on the sides $AB$ and $AC$ such that $PQ || BC$ and $AP = 3\ cm, PQ = 5\ cm, AC = 8\ cm$
Let $AQ = x$
Then $QC = (8 - x)$
$\because$ In $\triangle\text{ABC}, PQ || BC$
$\therefore\frac{\text{AP}}{\text{PB}}=\frac{\text{AQ}}{\text{QC}}\Rightarrow\frac{3}{5}=\frac{\text{x}}{8-\text{x}}$
$\Rightarrow5\text{x}=24-3\text{x}\Rightarrow5\text{x}+3\text{x}=24$
$8x = 24$
$⇒ x = 3$
$AQ = 3\ cm.$ View full question & answer→Question 263 Marks
If the altitude of two similar triangles are in the ratio $2 : 3,$ what is the ratio of their areas$?$
AnswerGiven: Altitudes of two similar triangles are in ratio $2 : 3.$
To find: Ratio of the areas of two similar triangles.
Let first triangle be $\triangle\text{ABC}$ and the second triangle be $\triangle\text{PQR}$
We know that the areas of two similar triangles are in the ratio of the squares of the corresponding altitudes.
$\Rightarrow\frac{\text{Area}(\triangle\text{ABC})}{\text{Area}(\triangle\text{PQR)}}=\frac{2^2}{3^2}$
$\frac{\text{Area}(\triangle\text{ABC})}{\text{Area}(\triangle\text{PQR)}}=\frac{4}{9}$
View full question & answer→Question 273 Marks
In a $\triangle\text{ABC,D}\ \text{and E}$ are points on the sides $AB$ and $AC$ respectively such that $DE || BC.$
If $AD = 8\ cm, AB = 12\ cm$ and $AE = 12\ cm,$ find $CE.$
Answer$AD = 8\ cm, AB = 12\ cm, AE = 12\ cm$
$DB = AB - AD$
$DB = 12 - 8$
$DB = 4\ cm$
by thales theorem,
$\frac{\text{AD}}{\text{DB}}=\frac{\text{AE}}{\text{EC}}$
$\Rightarrow\frac{8}{4}=\frac{12}{\text{EC}}$
$\Rightarrow\text{EC}=\frac{12\times4}{8}$
$\Rightarrow\text{EC}=6\text{cm}$
View full question & answer→Question 283 Marks
$M$ and $N$ are points on the sides $PQ$ and $PR$ respectively of a $\triangle\text{PQR}.$ For the following cases, state whether $MN || QR.$
$PQ = 1.28\ cm, PR = 2.56\ cm, PM = 0.16\ cm, PN = 0.32\ cm.$
AnswerIn $\triangle\text{PQR}, M$ and $N$ are points on $PQ$ and $PR$ respectively and $PQ = 1.28\ cm, PR = 2.56\ cm, PM = 0.16\ cm, PN = 0.32\ cm.$
$QM = PQ - PM = 1.28 - 0.16 = 1.12$ and $RN = PR - PN$
$= 2.56 - 0.32 = 2.24\ cm$
Now $\frac{\text{PM}}{\text{MQ}}=\frac{0.16}{1.12}=\frac{1}{7}$
and $\frac{\text{PN}}{\text{NR}}=\frac{0.32}{2.24}=\frac{1}{7}$
$\because\frac{\text{PM}}{\text{MQ}}=\frac{\text{PN}}{\text{NR}}$
$\therefore\text{MN}||\text{QR}$ View full question & answer→Question 293 Marks
Calculate the height of an equilateral triangle each of whose sides measures $12\ cm.$
Answer
We have,
$\triangle\text{ABC}$ is an equilateral $\triangle$ with side $12\ cm$
Draw $\text{AD}\perp\text{BC}$
In $\triangle\text{ABD}$ and $\triangle\text{ACD}$
$\angle\text{ADB}=\angle\text{ADC}[$ Each $90^\circ ]$
$AB = AC [$Each $12\ cm]$
$AD = AD [$Common$]$
Then, $\triangle\text{ABD}\cong\triangle\text{ACD} [$By RHS condition$]$
$ \therefore A D^2+B D^2=A B^2$
$\Rightarrow A D^2+6^2=12^2$
$\Rightarrow A D^2=144-36=108$
$\Rightarrow\text{AD}=\sqrt{108}=10.39\text{cm}$ View full question & answer→Question 303 Marks
In figure, $PQ || BC$ and $AP : PB = 1 : 2.$
Find: $\frac{\text{area}(\triangle\text{APQ})}{\text{area}(\triangle\text{ABC})}$
AnswerIn $\triangle\text{ABC} PQ || BC$
$\triangle\text{APQ}\sim\triangle\text{ABC}$
But $AP : PB = 1 : 2$
The ratio of the areas of two similar triangles are proportional to the square of their corresponding sides
$\therefore\frac{\text{area}(\triangle\text{APQ})}{\text{area}(\triangle\text{ABC})}=\frac{\text{AP}^2}{\text{AB}^2}=\frac{(1)^2}{(1+2)^2}=\frac{(1)^2}{(3)^2}$
$=\frac{(1)^2}{(3)^2}=\frac{1}{9}$
$\therefore$ Ratio $= 1 : 9.$
View full question & answer→Question 313 Marks
In a $\triangle\text{ABC,} P$ and $Q$ are points on sides $AB$ and $AC$ respectively, such that $PQ || BC.$ If $AP = 2.4\ cm, AQ = 2\ cm, QC = 3\ cm$ and $BC = 6\ cm,$ find the $AB$ and $PQ.$
Answer
$PQ || BC$. If $AP = 2.4\ cm, AQ = 2\ cm, QC = 3\ cm$ and $BC = 6\ cm,$
$\frac{\text{AP}}{\text{AB}}=\frac{\text{AQ}}{\text{AC}}$ (by thales theorem)
$\Rightarrow\frac{\text{AP}}{\text{AB}}=\frac{\text{AQ}}{\text{AQ}+\text{QC}}$
$\Rightarrow\frac{2.4}{\text{AB}}=\frac{2}{2+3}$
$\Rightarrow\text{AB}=\frac{2.4\times5}{2}$
$\Rightarrow\text{AB}=1.2\times5$
$\Rightarrow\text{AB}=6\text{cm}$
and, $\frac{\text{AP}}{\text{AB}}=\frac{\text{PQ}}{\text{BC}}$
$\Rightarrow\frac{2.4}{6}=\frac{\text{PQ}}{6}$
$\Rightarrow\text{PQ}=\frac{2.4\times6}{6}$
$\Rightarrow\text{PQ}=2.4\text{cm}$
Thus, $AB = 6\ cm$ and $PQ = 2.4\ cm.$ View full question & answer→Question 323 Marks
A triangle has sides $5\ cm, 12\ cm$ and $13\ cm.$ Find the length to one decimal place, of the perpendicular from the opposite vertex to the side whose length is $13\ cm.$
Answer
Since $\text{BD}\perp\text{AC}$ we obtained two right angled triangles, $\triangle\text{ABD}$ and $\triangle\text{BDC}.$
In $\triangle\text{ABC}$ and $\triangle\text{ABD}$
$\angle\text{A}=\angle\text{A}$ (Common angle)
$\angle\text{B}=\angle\text{D}$
So, by $AA-$criterion $\triangle\text{ABC}\sim\triangle\text{ADB}$
$\therefore\frac{\text{AB}}{\text{AD}}=\frac{\text{BC}}{\text{BD}}=\frac{\text{AC}}{\text{AB}}$
$\therefore\frac{\text{BC}}{\text{BD}}=\frac{\text{AC}}{\text{AB}}$
Now we will multiply both sides of the equation by $AB × BD$.
$BC × AB = BD × AC .....(1)$
Let us simplify the equation $(1)$ as given below,
$\text{BD}=\frac{\text{BC}\times\text{AB}}{\text{AC}}$
Now we will substitute the values of $BC, AB$ and $AC.$
$\text{BD}=\frac{12\times5}{13}$
$\therefore\text{BD}=\frac{60}{13}$
$\therefore\text{BD}=4.6\text{cm}$
Therefore, the length of the altitude is $4.6\ cm.$ View full question & answer→Question 333 Marks
There is a staircase as shown in the given figure, connecting points $A$ and $B.$ Measurements of steps are marked in the figure. Find the straight line distance between $A$ and $B.$
AnswerThere are $4$ steps in staircase $AB$
Taking first step,
In $\triangle\text{ALP}$
$\text{AP}^2=\text{AL}^2+\text{LP}^2 ($Pythagoras Theorem$)$
$=2^2+1^2=4+1=5$
$\therefore\text{AP}=\sqrt{5}=2.236=2.24$
Similarly
$\text{PQ}^2=2^2+1.6^2=4+2.56=6.56$
$\text{PQ}=\sqrt{6.56}=2.56$
$\text{QR}=\sqrt{2^2+1.6^2}=\sqrt{4+2.56}$
$=\sqrt{6.56}=2.56$
$\text{RB}=\sqrt{2^2+(1.8)^2}=\sqrt{4+3.24}$
$=\sqrt{7.24}=2.69$
$\therefore\text{AB}=2.24+2.56+2.56+269$
$=10.05=10\text{cm (Approx})$ View full question & answer→Question 343 Marks
In FIg. check whether $AD$ is the bisector of $\angle\text{A}$ of $\triangle\text{ABC}$ in the following.
$AB = 4\ cm, AC = 6\ cm, BD = 1.6\ cm$ and $CD = 2.4\ cm.$ Answer$AB = 4\ cm, AC = 6\ cm, BD = 1.6\ cm$ and $CD = 2.4\ cm.$
Now $\frac{\text{AB}}{\text{AC}}=\frac{4}{6}=\frac{2}{3}$
$\frac{\text{BD}}{\text{CD}}=\frac{1.6}{2.4}=\frac{2}{3}$
$\because\frac{\text{AB}}{\text{AC}}=\frac{\text{BD}}{\text{CD}}$
$\therefore AD$ is the bisector of $\angle\text{A}$
View full question & answer→Question 353 Marks
If $D$ is a point on the side $AB$ of $\triangle\text{ABC}$ such that $AD : DB = 3.2$ and $E$ is a point on $BC$ such that $DE || AC.$ Find the ratio of areas of $\triangle\text{ABC}$ and $\triangle\text{BDE}.$
AnswerIn $\triangle\text{ABC}, D$ is a point on $AB$ such that $AD : DB = 3 : 2$
$DE || AC$ is drawn meeting $BC$ is $E$
$\therefore$ $DE || AC$
$\therefore\triangle\text{BED}\sim\triangle\text{ABC}$
$\therefore\frac{\text{area}(\triangle\text{ABC)}}{\text{area}(\triangle\text{BDE})}=\frac{\text{AB}^2}{\text{BD}^2}=\frac{(\text{BD+AD})^2}{\text{BD}^2}$
$=\frac{(2+3)^2}{(2)^2}=\frac{(5)^2}{(2)^2}=\frac{25}{4}$
$\therefore$ Ratio $= 25 : 4$ View full question & answer→Question 363 Marks
Two poles of height $9\ m$ and $14\ m$ stand on a plane ground. If the distance between their feet is $12\ m,$ find the distance between their tops.
Answer
let $AB$ and $CD$ be two poles and distance blw
their feet be $12m$
$AB = 9m, CD = 14m$ and $BC = 12m$
$AB = EC = 9m$
$ED = CD - EC$
$ED = 14 - 9 = 5m$
In $\triangle\text{AED}$
$ A D^2=A E^2+E D^2 $
$ \Rightarrow A D^2=(12)^2+(5)^2 $
$ \Rightarrow A D^2=144+25 $
$ \Rightarrow A D^2=169 $
$ \Rightarrow A D=13 \mathrm{~m}$
thus, distance blw top of poles is $13m.$ View full question & answer→Question 373 Marks
$ABCD$ is a trapezium having $AB || DC.$ Prove that $O,$ the point of intersection of diagonals, divides the two diagonals in the same ratio. Also prove that $\frac{\text{ar}(\triangle\text{OCD})}{\text{ar}(\triangle\text{OAB})}=\frac{1}{9},$ if $AB = 3CD.$
AnswerGiven: $ABCD$ is a trapezium in which $AB || DC$ and diagonals $AC$ and $BD$ intersect each other at $O.$
To prove:
- $O$ divides the diagonals in the same ratio or $\frac{\text{OA}}{\text{OC}}=\frac{\text{OB}}{\text{OD}}$
- $\frac{\text{ar}(\triangle\text{OCD})}{\text{ar}(\triangle\text{OAB})}=\frac{1}{9}$ if $AB = 3CD$
Proof:
- In $\triangle\text{AOB}$ and $\triangle\text{COD}$
$\angle\text{AOB}=\angle\text{COD}$ (Vertically opposite angles)
$\angle\text{OAB}=\angle\text{OCD}$ (Alternate angles)
$\therefore\triangle\text{AOB}\sim\triangle\text{COD}$ (AA criterion)
$\frac{\text{AB}}{\text{DC}}=\frac{\text{OA}}{\text{OC}}=\frac{\text{OB}}{\text{OD}}$
$\Rightarrow\text{OA}\times\text{OD}=\text{OB}\times\text{OC}$
Hence O divides the diagonals in the same ratio
- $\because\triangle\text{AOB}\sim\triangle\text{COD}$ (proved)
$\therefore\frac{\text{area}(\triangle\text{OCD})}{\text{area}(\triangle\text{OAB})}=\frac{\text{CD}^2}{\text{AB}^2}$
But $AB = 3CD$
$\frac{\text{area}(\triangle\text{OCD})}{\text{area}(\triangle\text{OAB})}=\frac{\text{CD}^2}{(3\text{CD})^2}=\frac{\text{CD}^2}{9\text{CD}^2}=\frac{1}{9}$
Hence proved. View full question & answer→Question 383 Marks
The areas of two similar triangles $ABC$ and $PQR$ are in the ratio $9 : 16.$ If $BC = 4.5\ cm$, find the length of $QR.$
AnswerGiven: The areas of two similar triangles $ABC$ and $PQR$ are in the ratio $9 : 16, BC = 4.5\ cm.$
To find: length of $QR$
We know that the ratio of areas of two similar triangle is equal to the ratio of their corresponding sides.
$\frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{PQR})}=\Big(\frac{\text{BC}}{\text{QR}}\Big)^2$
$\frac{9}{16}=\Big(\frac{4.5}{\text{QR}}\Big)^2$
$\frac{3}{4}=\frac{4.5}{\text{QR}}$
$\text{QR}=\frac{4\times4.5}{3}$
$\text{QR}=6\text{cm}$
View full question & answer→Question 393 Marks
The foot of a ladder is $6m$ away from a wall and its top reaches a window $8m$ above the ground. If the ladder is shifted in such a way that its foot is $8m$ away from the wall, to what height does its tip reach$?$
AnswerIn first case,
The foot of the ladder are $6m$ away from the wall and its top reaches window $8m$ high
Let $AC$ be ladder and $BC = 6m, AB = 8m$
Now in right $\triangle\text{ABC}$
Using Pythagoras Theorem
$\mathrm{ A C^2=B C^2+A B^2=(6)^2+(8)^2=36+64=100=(10)^2}$
$ \mathrm{A C=10 \mathrm{~m}}$
$\text { In second case, }$
$ \mathrm{ED}=\mathrm{AC}=10 \mathrm{~m} $
$ \mathrm{BD}=8 \mathrm{~m}, \text { let } \mathrm{ED}=\mathrm{x} $
$ \mathrm{ED}^2=\mathrm{BD}^2+\mathrm{EB}^2 $
$ \Rightarrow(10)^2=(8)^2+\mathrm{x}^2 $
$ \Rightarrow 100=64+\mathrm{x}^2 $
$ \Rightarrow \mathrm{x}^2=100-64=36=(6)^2 $
$ x=6$
Height of the ladder on the wall $= 6m$
View full question & answer→Question 403 Marks
In $\triangle\text{ABC}, D$ and $E$ are points on sides $AB$ and $AC$ respectively such that $AD × EC = AE × DB$. Prove that $DE || BC.$
AnswerGiven: In $\triangle\text{ABC},$ and $E$ are points on sides $AB$ and $AC$ such that $AD × EC = AE × DB$
To Prove: $DE || BC$
Proof:
Since $AD × EC = AE × DB$
$\Rightarrow\frac{\text{DB}}{\text{AD}}=\frac{\text{EC}}{\text{AE}}$
$\Rightarrow\frac{\text{DB}}{\text{AD}}+1=\frac{\text{EC}}{\text{AE}}+1$
$\Rightarrow\frac{\text{DB}+\text{AD}}{\text{AD}}=\frac{\text{EC}+\text{AE}}{\text{AE}}$
$\Rightarrow\frac{\text{AB}}{\text{AD}}=\frac{\text{AC}}{\text{AE}}$
$\therefore\text{DE}||\text{BC} ($Converse of basic proportionality theorem$)$ View full question & answer→Question 413 Marks
In the given figure, $\angle\text{A}=\angle\text{CED},$ prove that $\triangle\text{CAB}\sim\triangle\text{CED}.$ Also, find the value of $x.$
AnswerWe have,
$\angle\text{A}=\angle\text{CED}$
In $\triangle\text{CAB}$ and $\triangle\text{CED}$
$\angle\text{A}=\angle\text{CED}$
$\angle\text{C}=\angle\text{C}$
$\triangle\text{CAB}\sim\triangle\text{CED} ($By $AA$ criteria$)$
Now, $\frac{\text{CA}}{\text{CE}}=\frac{\text{AB}}{\text{ED}}$
$\Rightarrow\frac{7+8}{10}=\frac{9}{\text{x}}$
$\Rightarrow\text{x}=\frac{9\times10}{15}$
$\Rightarrow\text{x}=\frac{90}{15}$
$\Rightarrow\text{x}=6\text{cm}.$
View full question & answer→Question 423 Marks
$\triangle\text{ABD}$ is a right triangle right-angled at $A$ and $\text{AC}\perp\text{BD}.$ Show that
$AC^2= BC × DC$
Answer
Let $\angle\text{CAB}=\text{x}$
In $\triangle\text{CBA}$
$\angle\text{CBA}=180^\circ-90^\circ-\text{x}$
$\angle\text{CBA}=90^\circ-\text{x}$
Similarly in $\triangle\text{CAD}$
$\angle\text{CAD}=90^\circ-\angle\text{CAD}=90^\circ-\text{x}$
$\angle\text{CDA}=90^\circ-\angle\text{CAB}$
$=90^\circ-\text{x}$
$\angle\text{CDA}=180^\circ-90^\circ-(90^\circ-\text{x})$
$\angle\text{CDA}=\text{x}$
Now in $\triangle\text{CBA}$ and $\triangle\text{CAD}$ we may observe that,
$\angle\text{CBA}=\angle\text{CAD}$
$\angle\text{CAB}=\angle\text{CDA}$
$\angle\text{ACB}=\angle\text{DCA}=90^\circ$
Therefore $\triangle\text{CBA}\sim\triangle\text{CAD} ($by $AAA$ rule$)$
Therefore $\frac{\text{AC}}{\text{DC}}=\frac{\text{BC}}{\text{AC}}$
$\Rightarrow\text{AC}^2=\text{DC}\times\text{BC}$ View full question & answer→Question 433 Marks
If $\triangle\text{ABC}$ and $\triangle\text{DEF}$ are similar triangles such that $AB = 3\ cm, BC = 2\ cm, CA = 2.5\ cm$ and $EF = 4\ cm,$ write the perimeter of $\triangle\text{DEF}.$
Answer$\because\triangle\text{ABC}\sim\triangle\text{DEF}$
$\therefore\frac{\text{AB}}{\text{DE}}=\frac{\text{BC}}{\text{EF}}=\frac{\text{CA}}{\text{FD}}$
But $AB = 3\ cm, BC = 2\ cm, CA = 2.5\ cm$ and $EF = 4\ cm.$
$\therefore\frac{3}{\text{DE}}=\frac{2}{4}=\frac{2.5}{\text{FD}}$
Now $\frac{3}{\text{DE}}=\frac{2}{4}\Rightarrow\text{DE}=\frac{3\times4}{2}=6\text{cm}$
and $\frac{\text{CA}}{\text{FD}}=\frac{2}{4}\Rightarrow\frac{2.5}{\text{FD}}=\frac{2}{4}$
$\text{FD}=\frac{2.5\times4}{2}=5\text{cm}$
$\therefore$ Perimeter of $\triangle\text{DEF}=6+4+5=15\text{cm}$ View full question & answer→Question 443 Marks
In a quadrilateral $ABCD, \angle\text{B}=90^\circ,$ $A D^2=A B^2+B C^2+C D^2$, prove that $\angle\text{ACD}=90^\circ.$
Answer
In order to prove angle $\angle A C D=90^{\circ}$ it is enough to prove that $A D^2=A C^2+C D^2$
Given: $A D^2=A B^2+B C^2+C D^2$
$A D^2-C D^2=A B^2+B C^2 \ldots \ldots .(1)$
Since $\angle B=90^{\circ}$, so applying pythagoras theorem in the right angled triangle $A B C$, we get, $A C^2=A B^2+B C^2 \ldots(2)$
From $(1)$ and $(2),$ we get
$ A C^2=A D^2-C D^2 $
$ A C^2+C D^2=A D^2$
Therefore, angle $\triangle \mathrm{ACD}=90^{\circ}. ($Converse of pythagoras theorem$)$ View full question & answer→Question 453 Marks
In $\triangle\text{ABC},$ given that $AB = AC$ and $\text{BD}\perp\text{AC}.$ Prove that $BC^2= 2AC × CD.$
AnswerGiven: In $\triangle\text{ABC},$ $\text{AB} = \text{AC, } \text{BD}\perp\text{AC}$
To prove: $B C^2=2 A C \times C D$
$ \therefore A B^2=B D^2+A D^2 \text { (Pythagoras Theorem) } $
$ \Rightarrow B D^2=A B^2-A D^2 \ldots . . .(i)$
Similarly in right $\triangle \mathrm{BDC}$,
$ B C^2=B D^2+D C^2 $
$ =A B^2-A D^2+D C^2[\text { From (i) }] $
$ =A B^2-(A C-C D)^2+C D^2 $
$ =A B^2-\left(A C^2+C D^2-2 A C \times C D\right)+C D^2 $
$ =A C^2-A C^2-C D^2+2 A C \times C D+C D^2(\because A B=A C) $
$ =2 A C \times C D \text { (given) }$
Hence $B C^2=2 A C \times C D$ View full question & answer→Question 463 Marks
In $\triangle\text{ABC}$ (Fig.), if $\angle1=\angle2,$ prove that $\frac{\text{AB}}{\text{AC}}=\frac{\text{BD}}{\text{DC}}.$
Answer
Given: A $\triangle\text{ABC}$ in which $\angle1=\angle2$
To prove: $\frac{\text{AB}}{\text{AC}}=\frac{\text{BD}}{\text{DC}}$
Construction: Draw $CE || DA$ to meet $BA$ produced in $E.$
$$Proof: since, $CE || DA$ and $AC$ cuts them.
$\therefore\angle2=\angle3\ ....(\text{i}) [$Alternate angles$]$
And, $\angle1=\angle4\ ...(\text{ii})[$ Corresponding angles$]$
But, $\angle1=\angle2 [$Given$]$
Form $(i)$ and $(ii),$ we get
$\angle3=\angle4$
Thus, in $\triangle\text{ACE},$ we have
$\angle3=\angle4$
$\Rightarrow\text{AE}=\text{AC}\ ...(\text{iii})$ [Sides opposite to equal angles are equal]
Now, In $\triangle\text{BCE},$ we have
$DA || CE$
$\Rightarrow\frac{\text{BD}}{\text{DC}}=\frac{\text{BA}}{\text{AE}}$ [Using basic proportionality theorem]
$\Rightarrow\frac{\text{BD}}{\text{DC}}=\frac{\text{AB}}{\text{AC}} [ \because BA - AB$ and $AE - AC$ from $(iii)]$
Hence, $\frac{\text{AB}}{\text{AC}}=\frac{\text{BD}}{\text{DC}}$ View full question & answer→Question 473 Marks
In a triangle $ABC, N$ is a point on $AC$ such that $\text{BN}\perp\text{AC}.$ If $BN^2= AN × NC,$ prove that $\angle\text{B}=90^\circ.$
AnswerIn $\triangle\text{ABC},\ \text{BN}\perp\text{AC}.$
Also, $BN^2= AN × NC$
We have to prove that $\angle\text{B}=90^\circ.$
In triangles $ABN$ and $BNC,$ we have
$ \mathrm{AB}^2=\mathrm{AN}^2+\mathrm{BN}^2 $
$ \mathrm{BC}^2=\mathrm{BN}^2+\mathrm{CN}^2$
Adding above two equations, we get
$\mathrm{AB}^2+\mathrm{BC}^2=\mathrm{AN}^2+\mathrm{CN}^2+2 \mathrm{BN}^2$
Since $B N^2=A N \times N C$
So,
$ A B^2+B C^2=A N^2+C N^2+2 A N \times N C $
$ A B^2+B C^2=(A N+N C)^2 $
$ A B^2+B C^2=A C^2$
Hence $\angle\text{B}=90^\circ$ View full question & answer→Question 483 Marks
$ABCD$ is a trapezium in which $AB || DC. P$ and $Q$ are points on sides $AD$ and $BC$ such that $PQ || AB.$ If $PD = 18, BQ = 35$ and $QC = 15$, find $AD.$
AnswerIn trapezium $ABCD, AB || DC$
$P$ and $Q$ are points on $AD$ and $BC$ respectively such that
$PQ || AB PD = 18, BQ = 35, QC = 15$
Let $AP = x$
$\because DC || AB || PQ$
$\therefore\frac{\text{DP}}{\text{PA}}=\frac{\text{CQ}}{\text{QB}}$
$\Rightarrow\frac{18}{\text{x}}=\frac{15}{35}\Rightarrow\text{x}=\frac{18\times35}{15}=42$
$\therefore\text{AD}=\text{AP}+\text{PD}=42+18=60$ View full question & answer→Question 493 Marks
In $\triangle\text{ABC}, P$ divides the side $AB$ such that $AP : PB = 1 : 2. Q$ is a point in $AC$ such that $PQ || BC.$ Find the ratio of the areas of $\triangle\text{APQ},$ and trapezium $BPQC.$
AnswerIn $\triangle\text{ABC} P $ is a point on $AB$ such that $AP : PQ = 1 : 2$
$PQ || BC$
Now we have to find the ratio between area $\triangle\text{APQ}$ and area trap $BPQC$
$\because\text{PQ}||\text{BC}$
$\therefore\triangle\text{APQ}\sim\triangle\text{ABC}$
$\frac{\text{area}(\triangle\text{APQ)}}{\text{area}(\triangle\text{ABC})}=\frac{(\text{AP)}^2}{(\text{AB})^2}$.
$=\frac{(1)^2}{(1+2)^2}=\frac{(1)^2}{(3)^2}=\frac{1}{9}$
$\therefore9\text{ area}(\triangle\text{APQ})=\text{area}(\triangle\text{ABC})$
$\Rightarrow9\text{ area}(\triangle\text{APQ})=\text{area}(\triangle\text{APQ})+\text{area (trap BPQC})$
$\Rightarrow9\text{ area}(\triangle\text{APQ})-\text{are}(\triangle\text{APQ})=\text{are trap BPQC}$
$\Rightarrow8\text{ and}(\triangle\text{APQ})=\text{area (trap BPQC)}$
$\frac{\text{area}(\triangle\text{APQ)}}{\text{area}(\text{trap. }\text{BPQC})}=\frac{1}{8}$
$\therefore\text{Ratio} = 1 : 8$ View full question & answer→Question 503 Marks
If $\triangle\text{ABC}$ and $\triangle\text{DEF}$ are two triangles such that $\frac{\text{AB}}{\text{DE}}=\frac{\text{BC}}{\text{EF}}=\frac{\text{CA}}{\text{FD}}=\frac{3}{4},$ then write $\text{Area}(\triangle\text{ABC}):\text{Area}(\triangle\text{DEF.})$
AnswerIn two $\triangle\text{s} ABC$ and $DEF$
$\frac{\text{AB}}{\text{DE}}=\frac{\text{BC}}{\text{EF}}=\frac{\text{CA}}{\text{FD}}=\frac{3}{4}$
$\therefore\triangle\text{ABC}\sim\triangle\text{DEF}$ (Sides of two similar triangles are proportional)
$\therefore\frac{\text{area}(\triangle\text{ABC})}{\text{area}(\triangle\text{DEF})}=\frac{\text{AB}^2}{\text{DE}^2}=\frac{(3)^2}{(4)^2}=\frac{9}{16}$
$\text{area}(\triangle\text{ABC}):\text{area}(\triangle\text{DEF})=9:16$
View full question & answer→