5 Marks Questions

Question 15 Marks

In this given $\triangle A B C, P Q \| A C$ and it divides the $\triangle$ into two equal parts in area. Then find the ratio of $\frac{A P}{A B}$.

Answer

Now, Given: In
$\begin{array}{l}\triangle A B C, P Q \| A C \text { and } \triangle P B C \sim \triangle A B C \\\text { and } \operatorname{ar}(P B C)=\operatorname{ar}(A P Q C)\end{array}$

To find: ratio of $\frac{A P}{A B}$
Solution: $\operatorname{ar}(P B C)=\operatorname{ar}(A P Q C)\quad \quad \ldots \ldots\text{(i)}$
As $\triangle P B C \sim \triangle A B C$
So,
$\frac{\operatorname{ar}(P B C)}{\operatorname{ar}(A B C)}=\left(\frac{P B}{A B}\right)^2 \times\left(\frac{B Q}{BC}\right)^2=\left(\frac{P Q}{A C}\right)^2$
Taking $\frac{\operatorname{ar}(P B C)}{\operatorname{ar}(\triangle A B C)}=\left(\frac{P B}{A B}\right)^2$
$\frac{\operatorname{ar}(\triangle P B C)}{\operatorname{ar} \triangle P B C+a r(A P Q C)}=\left(\frac{P B}{A B}\right)^2$
$\Rightarrow \frac{\operatorname{ar}(\triangle P B C)}{2 a r(\triangle P B C)}=\left(\frac{P B}{A B}\right)^2 \ldots \text {from (ii)}$
$\Rightarrow \frac{1}{2}=\left(\frac{P B}{A B}\right)^2$
Taking square root of both the sides, we get
$\Rightarrow \frac{1}{\sqrt{2}}=\frac{P B}{A B}$
$\Rightarrow 1-\frac{1}{\sqrt{2}}=1-\frac{P B}{A B}$
$\Rightarrow \frac{\sqrt{2}-1}{\sqrt{2}}=\frac{A B-P B}{A B}$
$\Rightarrow \frac{\sqrt{2}-1}{\sqrt{2}}=\frac{A P}{A B}$
After rationalization of the denominator, we get
$\Rightarrow \frac{A P}{A B}=\frac{2-\sqrt{2}}{2}$
So, the ratio of is
$\Rightarrow \frac{A P}{A B} \text { is } \frac{2-\sqrt{2}}{2} .$

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Question 25 Marks

In $\triangle A B C$, If $\angle A D E=\angle B$, then prove that $\triangle A D E \sim \triangle A B C$.
Also if $A D=7.6 cm, A E=7.2 cm, B E=4.2 cm$ and $B C=5.4 cm$, then find $D E$.

Answer

In $\triangle A D E$ and $\triangle A B C$
Given,
$
\begin{array}{ll}
\angle A D E=\angle B & (\text { Given }) \\
\angle A=\angle A & \text { (Common) } \\
\angle C=\angle E & \left(\text { each } 90^{\circ}\right)
\end{array}
$
Hence, $\triangle A D E \sim \triangle A B C$ by R.H.S criterion.
Now $\triangle A D E \sim \triangle A B C$
So,
$
\begin{array}{l}
\frac{A C}{A E}=\frac{A B}{A D}=\frac{B C}{D E} \\
A B=B E+A E \\
\Rightarrow A B=4.2+7.2=11.4 cm
\end{array}
$
Now, $\frac{A C}{A E}=\frac{A B}{A D}=\frac{B C}{D E}$
$
\Rightarrow \frac{A C}{7.2}=\frac{11.4}{7.6}=\frac{5.4}{D E}
$
Now, $\Rightarrow \frac{11.4}{7.6}=\frac{5.4}{D E}$
$
\begin{array}{l}
\Rightarrow 11.4 \times D E=5.4 \times 7.6 \\
\Rightarrow D E=\frac{7.6 \times 5.4}{11.4}
\end{array}
$
Hence $D E=3.6 cm$.

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Question 35 Marks

In the given figure $\triangle A B C$ is an equilateral. D is a point on $B C$ such that $B D=\frac{1}{3} B C$ Prove that $9 A D^2=7 A B^2$

Answer

It is given that in an equilateral triangle $A B C$, the side $B C$ is trisected at $D$ such that $B D=\frac{1}{3} B C$.
To prove: $9 A D^2=7 A B^2$
Construction: Draw $A E \perp B C$

In $\triangle A B E$ and $\triangle A C E$.
$A B=A C$
$A E=A E \because$ Common sides are equal.
$\angle A E B=\angle A E C=90^{\circ}$
$\therefore \triangle A E B \cong \triangle A E C$
Also, $B E=E C \quad$ By C.P.C.T
$B E=E C=\frac{B C}{2}$
In a right angled triangle $A D E$.
$A D^2=A E^2+D E^2\quad \quad \ldots \ldots\text{(i)}$ $\qquad$
In a right angled triangle $A B E$.
$A B^2=A E^2+B E^2\quad \quad \ldots \ldots\text{(ii)}$ $\qquad$
Subtract equation (ii) from equation (i) we get,
$\begin{array}{l}\Rightarrow A D^2-A B^2=D E^2-B E^2 \\\Rightarrow A D^2-A B^2=(B E-B D)^2-B E^2 \\\Rightarrow A D^2-A B^2=\left(\frac{B C}{2}-\frac{B C}{3}\right)^2-\left(\frac{B C}{2}\right)^2 \\\Rightarrow A D^2-A B^2=\left(\frac{(3 B C-2 B C)}{6}\right)^2-\left(\frac{B C}{2}\right)^2 \\\Rightarrow A D^2-A B^2=\frac{B C^2}{36}-\frac{B C^2}{4} \\\Rightarrow A D^2=\frac{\left(36 A B^2+A B^2-9 A B^2\right)}{36} \\\Rightarrow A D^2=\frac{\left(28 A B^2\right)}{36} \\\Rightarrow A D^2=\frac{\left(7 A B^2\right)}{9} \\\Rightarrow 9 A D^2=7 A B^2\end{array}$
Hence Proved.

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Question 45 Marks

In the given figure, $A B C$ and $D B C$ are two triangles on the same base $B C$. If $A D$ intersects $B C$ at $O$, show that $\frac{\operatorname{ar}(\triangle A B C)}{\operatorname{ar}(\triangle D B C)}=\frac{A O}{D O}$

Answer

Given: $B C$ is the common base of $\triangle A B C$ and $(\triangle D B C)$.
To Prove: $\frac{\operatorname{ar}(\triangle A B C)}{\operatorname{ar}(\triangle D B C)}=\frac{A O}{D O}$
Draw,
$A E \perp B C$ and $D F \perp B C$ in $\triangle A B C$ and $\triangle D B C$ respectively.

Hence, $\operatorname{ar}(\triangle A B C)=\frac{1}{2} \times B C \times A E\quad \quad \ldots \ldots\text{(i)}$
And, $\operatorname{ar}(\triangle D B C)=\frac{1}{2} \times B C \times D F\quad \quad \ldots \ldots\text{(ii)}$
Taking ratio of equation (i) and equation (ii), we get,
$\frac{\operatorname{ar}(\triangle A B C)}{\operatorname{ar}(\triangle D B C)}=\frac{A E}{D F}$
But we need to prove that,
$\frac{\operatorname{ar}(\triangle A B C)}{\operatorname{ar}(\triangle D B C)}=\frac{A O}{D O}$
Hence, we need to prove that, $\frac{A O}{D O}=\frac{A E}{D F}$
In $\triangle A O E$ and $\triangle D O F$,
$\angle A E O=\angle D F O=90^{\circ}$
$\angle A O E=\angle D O F \quad$ (Vertically opposite angles)
Hence, by $A A$ similarity criterion
$\triangle A O E=\triangle D O F$
We know that, if two triangles are similar, their corresponding sides are in the same ratio
Therefore, $\frac{A E}{D F}=\frac{A O}{D O}\quad \quad \ldots \ldots\text{(iii)}$
Now, we have,
$\frac{\operatorname{ar}(\triangle A B C)}{\operatorname{ar}(\triangle D B C)}=\frac{A E}{D F}\quad \quad \ldots \ldots\text{(iv)}$
On comparing equation (iii) and equation (iv), we get,
$\frac{a r(\triangle A B C)}{a r(\triangle D B C)}=\frac{A O}{D O}$
Hence proved.

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Question 55 Marks

Two poles of height $p$ and $q$ metres are standing vertically on a level ground, a metres apart. Prove that the height of the point of intersection of the lines joining the top of each pole to the foot of the opposite pole is given by $\frac{p q}{p+q}$ metres.

Answer

Let height of the point of intersection of the lines joining the top of each pole to the foot of the opposite pole be $h$.

Consider $\triangle A B C$ and $\triangle E F C$
$\begin{array}{l}\angle A B C=\angle E F C=90^{\circ} \\A C B=\angle E C F\end{array}$
$\Rightarrow \triangle A B C \cong \triangle E F C \text { (AAA similarity criterion) }$
$\begin{array}{l}\Rightarrow \frac{A B}{E F}=\frac{B C}{F C} \Rightarrow \frac{p}{h}=\frac{a}{F C} \\\Rightarrow F C=\frac{a h}{p}\quad \quad \ldots \ldots\text{(i)}\end{array}$
Consider $\triangle D C B$ and $\triangle E F B$
$\begin{array}{l}\angle D C B=\angle E F B=90^{\circ} \\\angle D B C=\angle E B F \\\Rightarrow \triangle D C B \cong \triangle E F B \text { (AA similarity criterion) } \\\Rightarrow \frac{D C}{E F}=\frac{B C}{F B} \Rightarrow \frac{q}{h}=\frac{a}{F B} \\\Rightarrow F B=\frac{a h}{q}\quad \quad \ldots \ldots\text{(ii)}\end{array}$
Add equations (i) and (ii),
$\begin{array}{l}\Rightarrow F C+F B=\frac{a h}{p}+\frac{a h}{q} \\\Rightarrow a=a h\left(\frac{1}{p}+\frac{1}{q}\right) \Rightarrow h=\frac{1}{\frac{1}{p}+\frac{1}{q}}=\frac{1}{\frac{q+p}{p q}} \\\Rightarrow h=\frac{p q}{p+q}\end{array}$
Hence proved.

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Question 65 Marks

In the given figure, in $\triangle A B C, X Y| | A C$ and $X Y$ divides the $\triangle A B C$ into two regions such that $\operatorname{ar}(\triangle B X Y)=) 2 \operatorname{ar}(A C Y X)$. Determine $\frac{A X}{A B}$.

Answer

$\begin{array}{l}\operatorname{ar}(\triangle B X Y)=2 \operatorname{ar}(A C Y X) . \\\operatorname{ar}(\triangle B X Y)=2 \operatorname{ar}(\triangle A B C-\triangle B X Y) \\\operatorname{3ar}(\triangle B X Y)=2 \operatorname{ar}(\triangle A B C) \\\Rightarrow \frac{\operatorname{ar}(\triangle B X Y)}{\operatorname{ar}(\triangle A B C)}=\frac{2}{3} \quad \ldots \text { (i) }\end{array}$
Consider $\triangle B X Y$ and $\triangle A B C$,
$\angle B X Y=\angle B A C$ (Corresponding angles as $X Y \| A C)$
$\angle B Y X=\angle B C A$ (Corresponding angles as $X Y \| A C)$
$\begin{array}{l}\therefore \triangle B Y X \cong \triangle B C A \text { (AA similarity criterion) } \\\Rightarrow \frac{\operatorname{ar}(\triangle B X Y)}{\operatorname{ar}(\triangle A B C)}=\left(\frac{B X}{A B}\right)^2\quad \quad \ldots\ldots\text{(ii)}\end{array}$
From (i) and (ii),
$\left(\frac{B X}{A B}\right)^2=\frac{2}{3}$
Now, $B X=A B-A X$
$\begin{array}{l}\left(\frac{A B-A X}{A B}\right)^2=\frac{2}{3} \\\Rightarrow\left(1-\frac{A X}{A B}\right)^2=\frac{2}{3}\end{array}$
Taking square root,
$\begin{array}{l}1-\frac{A X}{A B}=\sqrt{\frac{2}{3}} \\\Rightarrow \frac{A X}{A B}=1-\sqrt{\frac{2}{3}}\end{array}$

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Question 75 Marks

Prove that, in a right triangle, the square on the hypotenuse is equal to the sum of the squares on the other two sides.

Answer

Consider a right triangle ABC right angled at B .
Draw $B D \perp A C$
Consider $\triangle A D B$ and $\triangle A B C$,
$\begin{array}{l}\angle A D B=\angle A B C=90^{\circ} \\ \angle B A D=\angle C A B \text { (Common) }\end{array}$

$\triangle A D B \sim \triangle A B C$ (Using AA similarity criterion)
$\frac{A D}{A B}=\frac{A B}{A C}$ (Corresponding sides of similar triangles are proportional)
$\Rightarrow A B^2=A D \times A C\quad \quad \ldots \ldots\text{(i)}$
Consider $\triangle B D C$ and $\triangle A B C$,
$\angle B D C=\angle A B C=90^{\circ}$
$\angle B C D=\angle A C B$ (Common)
$\triangle B D C \sim \triangle A B C$ (Using AA similarity criterion)
$\frac{D C}{B C}=\frac{B C}{A C}$ (Corresponding sides of similar
triangles are proportional)
$\Rightarrow B C^2=C D \times A C\quad \quad \ldots \ldots\text{(ii)}$
Adding (i) and (ii), we get
$\begin{array}{l}A B^2+B C^2=(A D \times A C)+(C D \times A C) \\\Rightarrow A B^2+B C^2=A C \times(A D+C D) \\\Rightarrow A B^2+B C^2=A C^2\end{array}$
Hence proved.

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Question 85 Marks

In below Fig., in an equilateral triangle ABC, $A D \perp B C, B E \perp A C$ and $C F \perp A B$. Prove that $4\left( AD ^2+ BE ^2+ CF ^2\right)=9 AB ^2.$

Answer

To Prove :
$4\left(AD^2+BE^2+CF^2\right)=9 AB^2$
In $\triangle ABD$,
$AB ^2= AD ^2+ BD ^2\left[\because BD =\frac{1}{2} BC \right]$
$\Rightarrow AB ^2= AD ^2+\frac{1}{4} BC ^2 \quad[\because BC = AB ]$
$\begin{array}{l}\Rightarrow 4 AB ^2=4 AD ^2+ AB ^2 \\ \Rightarrow 3 AB ^2=4 AD ^2 \ldots \text { (i) }\end{array}$
Similarly, $3 AB ^2=4 BE ^2\ldots\ldots \text {(ii)}$
$3 AB ^2=4 CF ^2 \ldots\ldots\text {(iii)}$
By adding (i), (ii) and (iii), we get
$9 AB^2=4\left(AD^2+BE^2+CF^2\right)$
Hence proved

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Question 95 Marks

If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, prove that the other two sides are divided in the same ratio.

Answer

Given : $A \triangle ABC$ in which $DE |\mid BC$ and DE intersects AB at D and AC at E
To prove : $\frac{A D}{B D}=\frac{A E}{E C}$
Cons: Join BE and $CD , EL \perp AB , DM \perp AC$
$\begin{array}{l}\text { Proof: } \operatorname{ar}(ADE)=\frac{1}{2} \times AD \times EL \\\text { and are }(DBE)=\frac{1}{2} \times BD \times EL \\\begin{aligned}\frac{\operatorname{ar}(A D E)}{\operatorname{ar}(B D E)}= & \frac{\frac{1}{2} \times A D\times E L}{\frac{1}{2} \times B D \times E L} \\= & \frac{AD}{BD}\quad \quad \ldots \ldots\text{(1)}\end{aligned}\end{array}$
$\operatorname{ar}( ADE )=\operatorname{ar}( AED )=\frac{1}{2} \times AE \times DM$ and $\operatorname{ar}( ECD )=\frac{1}{2} \times EC \times DM$ $\begin{aligned} \frac{\operatorname{ar}( ADE )}{\operatorname{ar}( ECD )} & =\frac{\frac{1}{2} \times A E \times D M}{\frac{1}{2} \times E C \times D M} \\ & =\frac{A E}{E C}\quad \quad \ldots \ldots\text{(2)}\end{aligned}$
$\operatorname{ar}( DBE )= ar ( ECD )$ (on same base DE and between same parallels)
From (1) and (2),
$\frac{A D}{B D}=\frac{A E}{E C}$ Hence proved.

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Question 105 Marks

Construct a triangle with sides 5 cm 6 cm and 7 cm and then another triangle whose sides are $\frac{3}{5}$ of the corresponding sides of the first triangle.

Answer

1. Draw a line regiment $BC =6 cm$
2. With $B$ as a centre and radius $=A B=5 cm$, draw an arc.
3. With C as centre and radius $= AC =7 cm$, draw another are, intersecting the are drawn in step 2 at the point A .
4. Join $A B$ and $A C$ to obtain $\triangle A B C$.
5. Below BC, mark an acute angle $\angle CBX$.
6. Along $B X$ mark off five points $B _1, B_2, B_3, B_4, B_5$ such that $BB _1= B _1 B_2= B _2 B_3= B _3 B_4= B _4 B_5$
7. Join $B_5 C$.
8. From $B _3$, draw $B _3 C ^{\prime}| | B _5 C$.
9. From C , draw $C ^{\prime} A ^{\prime}$ I I CA , meeting BA at the point $A^{\prime}$

Then $A ^{\prime} BC ^{\prime}$ is the required triangle.

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Question 115 Marks

In $\triangle ABC$ (Figure), $AD \perp BC$. Prove that $AC ^2= AB ^2+ BC ^2-2 BC \times BD$

Answer

Given: ABC is a triangle where
$\angle ABC <90^{\circ}$ and $AD \perp BC$

To prove: $AC ^2= AB ^2+ BC ^2-2 BC \times BD$
Proof: In $\triangle ADB$, by Pythagoras theorem,
$AB^2=AD^2+BD^2\ldots\ldots \text {(i)}$
In $\triangle ADC$, by Pythagoras theorem
$AC^2=AD^2+DC^2$
$\begin{array}{l}\Rightarrow AC ^2= AD ^2+( BC - BD )^2[\because DC = BC - BD ] \\ \Rightarrow AC ^2= AD ^2+ BC ^2+ BD ^2-2 BC \times BD \\ \Rightarrow AC ^2=\left( AD ^2+ BD ^2\right)+ BC ^2-2 BC \times BD \\ \Rightarrow AC ^2= AB ^2+ BC ^2-2 BC \times BD \\ \text { [from eqn. (i)] }\end{array}$

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