[4 marks sum] - MATHEMATICS STD 9 Questions - Vidyadip

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Question 14 Marks

In a quadrilateral $\text{ABCD}, \angle B=90^{\circ}$ and $\angle D=90^{\circ}$.Prove that: $2 A C^2-A B^2=B C^2+C D^2+D A^2$

Answer

In quadrilateral $ABCD , \angle B =90^{\circ}$ and $\angle D =90^{\circ}$.
So, $\triangle ABC$ and $\triangle ADC$ are right$-$angled triangles.
In $\triangle ABC$, using Pythagoras theorem,
$A C^2=A B^2+B C^2$
$\Rightarrow A B^2=A C^2-B C^2 \ldots...(i)$
In $\triangle A D C$, using Pythagoras theorem,
$A C^2=A D^2+D C^2 ...(ii)$
$LHS =2 AC ^2- AB ^2$
$=2 A C^2-\left(A C^2-B C^2\right) .....[$From$(i) ]$
$=2 A C^2-A C^2+B C^2$
$= AC ^2+ BC ^2$
$ = AD ^2+ DC ^2+ BC ^2 \ldots .[$From$(ii)]$
$ =\text{RHS}$

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Question 24 Marks

In $\triangle ABC,\angle B = 90^o$ and $D$ is the mid$-$point of $BC.$Prove that: $AC^2= AD^2+ 3CD^2.$

Answer

Pythagoras theorem states that in a right$-$angled triangle,
the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.
In $\triangle ABC C_t \angle B =90^{\circ}$ and $D$ is the mid$-$point of $BC$.
Join $AD$.
Therefore, $BD = DC$
First, we consider the $\triangle A D B$,
and applying Pythagoras theorem we get,
$A D^2=A B^2+B D^2$
$A B^2=A D^2-B D^2\ldots...(i)$
Similarly, we get from $rt.$ angle triangles $ABC$ we get,
$A C^2=A B^2+B C^2$
$A B^2=A C^2-B C^2 \dots.....(ii)$
From $(i)$ and $(ii)$
$A C^2-B C^2=A D^2-B D^2$
$A C^2=A D^2-B D^2+B C^2$
$A C^2=A D^2-C D^2+4 C D^2 \quad \ldots .\left[B D=C D=\frac{1}{2} B C\right]$
$A C^2=A D^2+3 C D^2$
Hence proved.

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Question 34 Marks

In the figure, given below, $A D \perp B C$.Prove that: $c^2=a^2+b^2-2 a x$.

Answer

Pythagoras theorem states that in a right$-$angled triangle,
the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.
First, we consider the $\triangle ABD$ and applying Pythagoras theorem we get,
$AB^2 = AD^2 + BD^2$
$c^2 = h^2 + ( a - x )^2$
$h^2 = c^2 - ( a - x )^2 ......(i)$
First, we consider the $\triangle ACD$ and applying Pythagoras theorem we get,
$AC^2 = AD^2 + CD^2$
$b^2 = h^2 + x^2$
$h^2 = b^2 - x^2 ......(ii)$
From $(i)$ and $(ii)$ we get,
$c^2 - ( a - x )^2 = b^2 - x^2$
$c^2 - a^2- x^2 + 2ax = b^2 - x^2$
$c^2 = a^2 + b^2 - 2ax$
Hence Proved.

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Question 44 Marks

In $\triangle ABC, AB = AC$ and $BD$ is perpendicular to $AC.$Prove that: $BD^2- CD^2= 2CD \times AD.$

Answer

In right-angled $\triangle ADB$,
$A B^2=A D^2+B D^2$
$\Rightarrow A D^2=A B^2-B D^2 .....(i)$
$ A C=A D+D C$
$ \Rightarrow A C^2=(A D+D C)^2$
$ \Rightarrow A C^2=A D^2+D C^2+2 A D \times D C$
$ \Rightarrow AC ^2= AB ^2- BD ^2+ DC ^2+2 AD \times DC \ldots[$ From $(i)]$
$ \Rightarrow A C^2=A C^2-B D^2+D C^2+2 A D \times D C \ldots[A B=A C]$
$ \Rightarrow BD ^2- DC ^2=2 AD \times DC.$

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Question 54 Marks

Two poles of heights $6\ m$ and $11\ m$ stand vertically on a plane ground. If the distance between their feet is $12\ m;$find the distance between their tips.

Answer

The diagram of the given problem is given below,

We have Pythagoras theorem which states that in a right$-$angled triangle,
the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.
Here, $11-6=5 \ m \dots... ($ Since $DC$ is perpendicular to $BC)$
base $=12 \ cm$
Applying Pythagoras theorem we get, hypotenuse ${ }^2=5^2+12^2$
$h^2=25+144$
$ h^2=169$
$ h=13$
Therefore, the distance between the tips will be $13\ m$.

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Question 64 Marks

The given figure shows a quadrilateral $\text{ABCD}$ in which $A D=13 \ cm$, $D C=12 \ cm , B C=3 \ cm$ and $\angle A B D=\angle B C D=90^{\circ}$. Calculate the length of $A B$.

Answer

Pythagoras theorem states that in a right$-$angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.
First, we consider the $\triangle BDC$ and applying Pythagoras theorem we get,
$DB^2 = DC^2 + BC^2$
$DB^2 = 12^2 + 3^2$
$DB^2 = 144 + 9$
$DB^2 = 153$
Now, we consider the $\triangle ABD$ and applying Pythagoras theorem we get,
$DA^2 = DB^2 + BA^2$
$13^2 = 153 + BA^2$
$BA^2 = 169 - 153$
$BA = 4$
The length of $AB$ is $4 \ cm.$

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Question 74 Marks

In the figure: $\angle PSQ =90^{\circ}, PQ =10 \ cm , QS =6 \ cm$ and $RQ =9 \ cm$. Calculate the length of $PR.$

Answer

Pythagoras theorem states that in a right$-$angled triangle,
the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.
First, we consider the $\triangle PQS$ and applying Pythagoras theorem we get,
$PQ^2 = PS^2 + QS^2$
$10^2= PS^2 + 6^2$
$PS^2 = 100 - 36$
$PR = 8$
Now, we consider the $\triangle PRS$ and applying Pythagoras theorem we get,
$PR^2 = RS^2 + PS^2$
$PR^2 = 15^2 + 8^2$
$PR = 17$
The length of $\text{PR}\ 17 \ cm.$

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Question 84 Marks

A man goes $40\ m$ due north and then $50\ m$ due west. Find his distance from the starting point.

Answer

Here, we need to measure the distance $A B$ as shown in the figure below,

Pythagoras theorem states that in a right$-$angled triangle,
the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.
Therefore, in this case
$A B^2=B C^2+C A^2$
$ A B^2=50^2+40^2$
$ A B^2=2500+1600$
$ A B^2=4100$
$ A B=64.03$
Therefore the required distance is $64.03\ m$.

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Question 94 Marks

A ladder $13\ m$ long rests against a vertical wall. If the foot of the ladder is $5\ m$ from the foot of the wall, find the distance of the other end of the ladder from the ground.

Answer

The pictorial representation of the given problem is given below,

Pythagoras theorem states that in a right$-$angled triangle,
the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.
Here, $A B$ is the hypotenuse.
Therefore applying the Pythagoras theorem we get,
$ A B^2=B C^2+C A^2$
$ 13^2=5^2+C A^2$
$ C A^2=13^2-5^2$
$ C A^2=144$
$ C A=12\ m$
Therefore, the distance of the other end of the ladder from the ground is $12\ m$.

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Question 104 Marks

In the following figure, $O P, O Q$ and $O R$ are drawn perpendiculars to the sides $B C, C A$ and $A B$ respectively of $\triangle A B C$.Prove that: $AR ^2+ BP ^2+ CQ ^2= AQ ^2+ CP ^2+ BR ^2$

Answer

Here, we first need to join $OA , OB$, and $OC$ after which the figure becomes as follows,

Pythagoras theorem states that in a right$-$angled triangle,
the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.
First, we consider the $\triangle A R O$ and applying Pythagoras theorem we get,
$A Q^2=A R^2+O R^2$
$A R^2=A O^2-O R^2 \dots....(i)$
Similarly, from triangles, $\text{BPO} , \text{COQ} , \text{AOQ}, \text{CPO}$, and $\text{BRO}$ we get the following results,
$BP ^2= BO ^2-O P^2 \dots....(ii)$
$C Q^2=O C^2-O Q^2\dots ....(iii)$
$AQ ^2= AO ^2- OQ ^2\dots ....(iv)$
$C P^2=O C^2-O P^2\dots ....(v)$
$BR ^2=O B^2-O R^2\dots ....(vi)$
Adding $(i), (ii)$ and $(iii),$ we get
$A R^2+B P^2+C Q^2=A O^2-O R^2+B O^2-O P^2+O C^2-O Q^2\dots....(vii)$
Adding $(iv), (v)$ and $(vi),$ we get,
$A Q^2+C P^2+B R^2=A O^2-O R^2+B O^2-O P^2+O C^2-O Q^2 \dots....(viiii)$
From $(vii)$ and $(viii)$, we get,
$A R^2+B^2+C Q^2=A Q^2+C P^2+B R^2$
Hence proved.

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Question 114 Marks

$O$ is any point inside a rectangle $\text{ABCD}$.Prove that: $OB^2+ OD^2= OC^2+ OA^2$.

Answer

Draw rectangle $\text{ABCD}$ with arbitrary point $O$ within it,
and then draw lines $O A, O B, O C, O D$.
Then draw lines from point $O$ perpendicular to the sides: $OE , OF , OG , OH$.
Pythagoras theorem states that in a right$-$angled triangle,
the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.
Using Pythagorean theorem we have from the above diagram:
$ O A^2=A H^2+O H^2=A H^2+A^2$
$ O C^2=C G^2+O G^2= EB ^2+ HD ^2$
$ O B^2=E O^2+ BE ^2=A H^2+ BE ^2$
$ O D^2= HD ^2+ OH ^2= HD ^2+ AE ^2$
Adding these equalities we get:
$O A^2+O C^2=A H^2+H D^2+A E^2+E B^2$
$ O B^2+O D^2=A H^2+H D^2+A E^2+E B^2$
From which we prove that for any point within the rectangle there is the relation
$O A^2+O C^2=O B^2+O D^2$
Hence Proved.

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Question 124 Marks

In a rectangle $\text{ABCD},$prove that : $AC^2+ BD^2= AB^2+ BC^2+ CD^2+ DA^2.$

Answer

Pythagoras theorem states that in a right$-$angled triangle,
the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.
Since, $\text{ABCD}$ is a rectangle angles $A, B, C$ and $D$ are $rt.$ angles.
First, we consider the $\triangle A C D$,
and applying Pythagoras theorem we get,
$AC ^2= DA ^2+ CD ^2 \dots....(i)$
Similarly, we get from $rt.$ angle $\triangle BDC$ we get,
$BD ^2= BC ^2+ CD ^2$
$=B C^2+A B^2\ldots .[$ In a rectangle, opposite sides are equal, $\therefore C D=AB ] \ldots (ii)$
Adding $(i)$ and $(ii)$
$A C^2+B D^2=A B^2+B C^2+C D^2+D A^2$
Hence proved.

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Question 134 Marks

$M$ and $N$ are the mid$-$points of the sides $QR$ and $PQ$ respectively of a $\text{PQR},$ right$-$angled at $Q$.Prove that :$(i) \ PM^2+ RN^2= 5 MN^2(ii) \ 4 PM^2= 4 PQ^2+ QR^2$$(iii)\ 4 RN^2= PQ^2+ 4 QR^2(iv) \ 4 (PM^2+ RN^2) = 5 PR^2$

Answer

We draw$, PM, MN, NR$
Pythagoras theorem states that in a right$-$angled triangle,
the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.
Since $M$ and $N$ are the mid$-$points of the sides $Q R$ and $P Q$ respectively,
therefore, $PN = NQ , QM = RM$
$(i)$ First, we consider the $\triangle PQM$,
and applying Pythagoras theorem we get,
$ PM ^2= PQ ^2+ MQ ^2$
$ =( PN + NQ )^2+ MQ ^2$
$ = PN + NQ ^2+2 PN \cdot NQ + MQ ^2$
$= MN ^2+ PN ^2+2 PN . NQ \ldots\left[\right.$ From $\left., \Delta MNQ , MN ^2= NQ ^2+ MQ ^2\right] \ldots \ldots .. (i)$
Now, we consider the $\triangle RNQ$,
and applying Pythagoras theorem we get,
$ R^2=N Q^2+R Q^2$
$ =N Q^2+(Q M+R M)^2$
$ =N Q^2+Q^2+R M^2+2 Q M \cdot R M$
$= MN ^2+ RM ^2+2 QM \cdot RM \dots.......(ii)$
Adding $(i)$ and $(ii)$ we get,
$ PM ^2+ RN ^2= MN ^2+ PN ^2+2 PN \cdot NQ + MN ^2+ RM ^2+2 QM \cdot RM$
$ PM ^2+ RN ^2=2 MN ^2+ PN ^2+ RM ^2+2 PN \cdot NQ +2 QM \cdot RM$
$ PM ^2+ RN ^2=2 MN ^2+ NQ ^2+ QM ^2+2\left( QN ^2\right)+2\left( QM ^2\right)$
$ PM ^2+ RN ^2=2 MN ^2+ MN ^2+2 MN ^2$
$ PM ^2+ RN ^2=5 MN ^2$
Hence Proved.
$(ii)$ We consider the $\triangle PQM$,
and applying Pythagoras theorem we get,
$PM ^2= PQ ^2+ MQ ^2$
$4 PM ^2=4 PQ ^2+4 MQ ^2\dots ... [$ Multiply both sides by $4]$
$4 PM ^2=4 PQ ^2+4 \cdot\left(\frac{1}{2} QR \right)^2 \ldots\left[ MQ =\frac{1}{2} QR \right]$
$ 4 PM ^2=4 PQ ^2+4 PQ +4 \cdot \frac{1}{4} QR ^2$
$ 4 PM ^2=4 PQ ^2+ QR ^2$
Hence Proved.
$(iii)$ We consider the $\triangle RQN$,
and applying Pythagoras theorem we get,
$R N^2=N Q^2+R Q^2$
$4 RN ^2=4 NQ ^2+4 QR ^2 \ldots[$ Multiplying both sides by $4]$
$4 RN ^2=4 QR ^2+4 \cdot(1 / 2 PQ )^2 \ldots\left[ NQ =\frac{1}{2} PQ \right]$
$ 4 RN ^2=4 QR ^2+4 \cdot \frac{1}{4} PQ ^2$
$ 4 RN ^2 = PQ ^2+4 QR ^2$
Hence Proved.
$(iv)$ First, we consider the $\triangle PQM$, and applying Pythagoras theorem we get,
$ PM ^2= PQ ^2+ MQ ^2$
$ =( PN + NQ )^2+ MQ ^2$
$ = PN ^2+ NQ ^2+2 PN \cdot NQ + MQ ^2$
$= MN ^2+ PN ^2+2 PN . NQ \ldots[$ From, $\Delta MNQ _1= MN ^2= NQ ^2+ MQ ^2] \ldots \ldots(i)$
Now, we consider the $\triangle RNQ$,
and applying Pythagoras theorem we get,
$RN ^2+ NQ ^2+ RQ ^2$
$=N Q^2+(Q M+R M)^2$
$= NQ ^2+ QM ^2+ RM ^2+2 QM \cdot RM$
$= MN ^2+ RM ^2+2 QM \cdot RM \ldots \ldots .. (ii)$
Adding $(i)$ and $(ii)$ we get,
$PM ^2+ RN ^2= MN ^2+ PN ^2+2 PN \cdot NQ + MN ^2+ RM ^2+2 QM \cdot RM$
$PM ^2+ RN ^2=2 MN ^2+ PN ^2+ RM ^2+2 PN \cdot NQ +2 QM \cdot RM$
$PM ^2+ RN ^2=2 MN ^2+ NQ ^2+ QM ^2+2\left( QN ^2\right)+2\left( QM ^2\right)$
$PM ^2+ RN ^2=2 MN ^2+ MN ^2+2 MN ^2$
$P M^2+ RN ^2=5 MN ^2$
$4\left( PM ^2+ RN ^2\right)=4 \cdot 5 \cdot\left( NQ ^2+ MQ ^2\right)$
$ 4\left( PM ^2+ RN ^2\right)=4.5$
$ {\left[\left(\frac{1}{2} PQ \right)^2+\left(\frac{1}{2} RQ \right)^2\right] \ldots\left[\because NQ =\frac{1}{2} PQ , MQ =\frac{1}{2} QR \right]}$
$4\left(P M^2+R N^2\right)=5 P R^2$
Hence Proved.

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Question 144 Marks

$\text{ABC}$ is a triangle, right$-$angled at $B. M$ is a point on $BC$.Prove that: $AM^2+ BC^2= AC^2+ BM^2$.

Answer

The pictorial form of the given problem is as follows,

Pythagoras theorem states that in a right$-$angled triangle,
the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.
First, we consider the $\triangle A B M$ and applying Pythagoras theorem we get,
$AM ^2= AB + BM ^2$
$A B^2=A M^2-B M^2\dots ......(i)$
Now, we consider the $\triangle A B C$ and applying Pythagoras theorem we get,
$A C^2=A B^2+B C^2$
$A B^2=A C^2-B C^2\dots ......(ii)$
From $(i)$ and $(ii)$ we get,
$A M^2 \cdot B M^2=A C^2-B C^2$
$ A M^2+B C^2=A C^2+B M^2$
Hence Proved

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Question 154 Marks

In $\triangle A B C$, $\angle A=90^{\circ}, C A=A B$ and $D$ is the point on $A B$ produced.Prove that $D C^2-B D^2=2 A B \cdot A D$.

Answer

Pythagoras theorem states that in a right$-$angled triangle,
the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.
We consider the $rt.$ angled $\triangle ACD$ and applying Pythagoras theorem we get,
$C D^2=A C^2+A D^2$
$C D^2=A C^2+(A B+B D)^2 \ldots .[\because A D=A B+B D]$
$C D^2=A C^2+A B^2+B D^2+2 A B \cdot B D\ldots (i)$
Similarly, in $\triangle ABC$,
$B C^2=A C^2+A B^2$
$B C^2=2 A B^2\ldots[ AB = AC ]$
$AB ^2=\frac{1}{2} BC ^2 \dots...(ii)$
Putting, $AB^2$ from $(ii)$ in $(i),$ We get,
$C D^2=A C^2+\frac{1}{2} B C^2+B D^2+2 A B \cdot B D$
$ C D^2-B D^2=A B^2+A B^2+2 A B \cdot(A D-A B)$
$ C D^2-B D^2=A B^2+A B^2+2 A B \cdot A D-2 A B^2$
$ C D^2-B D^2=2 A B \cdot A D$
$ D C^2-B D^2=2 A B \cdot A D$
Hence Proved.

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Question 164 Marks

In an isosceles $\triangle ABC; AB = AC$ and $D$ is the point on $BC$ produced.Prove that: $AD^2= AC^2+ BD.CD$.

Answer

In an isosceles triangle $A B C ; A B=A C$ and $D$ is the point on $B C$ produced.
Construct $AE$ perpendicular $BC$.
Pythagoras theorem states that in a right$-$angled triangle,
the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.
We consider the $rt.$ angled $\triangle AED$ and applying Pythagoras theorem we get,
$ A D^2=A E^2+E D^2$
$ A D^2=A E^2+(E C+C D)^2 \ldots . .(i)\ [\because E D=E C+C D]$
Similarly, in $\triangle A E C$,
$A C^2=A E^2+E C^2$
$AE ^2= AC ^2- EC ^2 \dots....(ii)$
Putting $A E^2=A C^2-E C^2$ in $(i),$ We get,
$A D^2 =A C^2-E C^2+(E C+C D)^2$
$ =A C^2+C D(C D+2 E C)$
$A D^2=A C^2+B D \cdot C D\ldots \ldots[\because 2 E C+C D=B D]$
Hence proved.

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Question 174 Marks

Diagonals of rhombus $\text{ABCD}$ intersect each other at point $O$.Prove that: $O A^2+O C^2=2 A D^2-\frac{B D^2}{2}$

Answer

Diagonals of the rhombus are perpendicular to each other.
In quadrilateral $\text{ABCD} , \angle AOD =\angle COD =90^{\circ}$.
So, $\triangle AOD$ and $\triangle COD$ are right$-$angled triangles.
In $\triangle A O D$ using Pythagoras theorem,
$A D^2=O A^2+O D^2$
$\Rightarrow O A^2=A D^2-O D^2 \dots....(i)$
In $\triangle COD$ using Pythagoras theorem,
$C D^2=O C^2+O D^2$
$\Rightarrow O C^2=C D^2-O D^2 \dots....(ii)$
$ \text { LHS }=O A^2+O C^2$
$ =A D^2-O D^2+C D^2-O D^2 \ldots[$From $(i)$ and $(ii)]$
$ =A D^2+C D^2-2 O D^2$
$ =A D^2+A D^2-2\left(\frac{B D}{2}\right)^2 \ldots[A D=C D$ and $O D=\frac{B D}{2}]$
$ =2 A D^2-\frac{(B D)^2}{2}$
$ =\text { RHS. }$

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Question 184 Marks

In $\triangle ABC, AB = AC = x, BC = 10\ cm$ and the area of the triangle is $60 \ cm^2$.Find $x.$

Answer

Here, the diagram will be,

We have Pythagoras theorem which states that in a right$-$angled triangle,
the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.
Since $\text{ABC}$ is an isosceles triangle,
therefore perpendicular from vertex will cut the base in two equal segments.
First, we consider the $\triangle A B D$, and applying Pythagoras theorem we get,
$ A B^2=A D^2+B D^2$
$ A D^2=x^2-5^2$
$ A D^2=x^2-25$
$AD =\sqrt{x^2-25} \dots.....(i)$
Now,
Area $=60$
$ \frac{1}{2} \times 10 \times AD =60$
$ \frac{1}{2} \times 10 \times \sqrt{x^2-25}=60$
$ x=13$
Therefore, $x$ is $13 \ cm$.

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Question 194 Marks

In $\triangle A B C$, given below, $A B=8 \ cm , B C=6 \ cm$ and $A C=3 \ cm$. Calculate the length of $O C$.

Answer

We have Pythagoras theorem which states that in a right$-$angled triangle,
the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.

In $\triangle AOC _r$
$A C^2=A O^2+C O^2$
$(3)^2=A O^2+x^2$
$9= AO ^2+ x ^2$
$9-x^2=A O^2 \dots...(i)$
In $\triangle AOB _t$
$A B^2=A O^2+B O^2$
$(8)^2=A O^2+(6+x)^2$
$64= AO ^2+(6+x)^2$
$64-(6+x)^2=A O^2\dots ...(ii)$
From equation$ (i)$ and $(ii)$
$9-x^2=64-(6+x)^2$
$9-x^2=64-\left(36+x^2+12 x\right) \ldots\left[(a+b)^2=a^2+2 a b+b^2\right]$
$ 9-x^2=64-36-x^2-12 x$
$ 9=28-12 x$
$ 12 x=28-9$
$ x=\frac{19}{12}$
$ x=1 \frac{7}{12}$

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Question 204 Marks

$AD$ is drawn perpendicular to base $BC$ of an equilateral $\triangle ABC.$ Given $BC = 10 \ cm,$ find the length of $AD,$ correct to $1$ place of decimal.

Answer

Since $\text{ABC}$ is an equilateral triangle therefore, all the sides of the triangle are of the same measure and the perpendicular $AD$ will divide $BC$ into two equal parts.

Pythagoras theorem states that in a right$-$angled triangle,
the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.
Here, we consider the $\triangle A B D$ and applying Pythagoras theorem we get,
$A B^2=A D^2+B D^2$
$AD ^2=10^2-5^2 \ldots \ldots .\left[\right.$ Given, $\left.BC =10 \ cm = AB , BD =\frac{1}{2} BC \right]$
$A D^2=100-25$
$A D^2=75$
$A D=8.7$
Therefore, the length of $A D$ is $8.7 \ cm$

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Question 214 Marks

In $\triangle ABC,$ Find the sides of the triangle, if:

$AB = ( x - 3 ) \ cm, BC = ( x + 4 ) \ cm$ and $AC = ( x + 6 ) \ cm$
$AB = x \ cm, BC = ( 4x + 4 ) \ cm$ and $AC = ( 4x + 5) \ cm$

Answer

$(i)$ In right$-$angled $\triangle ABC$,
$A C^2=A B^2+B C^2$
$ \Rightarrow(x+6)^2=(x-3)^2+(x+4)^2$
$ \Rightarrow\left(x^2+12 x+36\right)=\left(x^2-6 x+9\right)+\left(x^2+8 x+16\right)$
$ \Rightarrow x^2-10 x-11=0$
$ \Rightarrow(x-11)(x+1)=0$
$\Rightarrow x=11 \quad$ or $ x=-1$
But length of the side of a triangle can not be negative.
$\Rightarrow x=11 \ cm$
$\therefore A B=(x-3)=(11-3)=8 \ cm$
$B C=(x+4)=(11+4)=15 \ cm$
$AC =( x +6)=(11+6)=17 \ cm$
$(ii)$ In right$-$angled $\triangle ABC$,
$A C^2=A B^2+B C^2$
$ \Rightarrow(4 x+5)^2=(x)^2+(4 x+4)^2$
$ \Rightarrow\left(16 x^2+40 x+25\right)=\left(x^2\right)+\left(16 x^2+32 x+16\right)$
$\Rightarrow x^2-8 x-9=0$
$\Rightarrow(x-9)(x+1)=0$
$\Rightarrow x=9$ or $x=-1$
But length of the side of a triangle can not be negative.
$\Rightarrow x=9 \ cm$
$\therefore A B=x=9 \ cm$
$B C=(4 x+4)=(36+4)=40 \ cm$
$A C=(4 x+5)=(36+5)=41 \ cm$

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Question 224 Marks

In the given figure, $\angle B=90^{\circ}, X Y\| B C, A B=12 \ cm , A Y=8 \ cm$ and $A X: X B=1: 2=A Y: Y C$.Find the lengths of $A C$ and $B C$.

Answer

Given that $A X: X B=1: 2=A Y: Y C$.
Let $x$ be the common multiple for which this proportion gets satisfied.
So, $A X=1 x$ and $X B=2 x$
$A X+X B=1 x+2 x=3 x$
$\Rightarrow A B=3 x \ldots . .(A-X-B)$
$\Rightarrow 12=3 x$
$\Rightarrow x=4$
$A X=1 x=4$ and $X B=2 x=2 \times 4=8$
Similarly,
$A Y=1 y$ and $Y C=2 y$
$ A Y=8 \ldots ($given$)$
$ \Rightarrow 8=y$
$ \therefore Y C=2 y=2 \times 8=16$
$ \therefore A C=A Y+Y C$
$ A C=8+16$
$AC =24 \ cm$
$\triangle ABC$ is a right angled triangle$\dots. ...($Given$)$
$\therefore$ By Pythagoras Theorem, we get
$ \Rightarrow A B^2+B C^2=A C^2$
$ \Rightarrow B C^2=A C^2-A B^2$
$ \Rightarrow B C^2=(24)^2-(12)^2$
$ \Rightarrow B C^2=576-144$
$ \Rightarrow B C^2=432$
$ \Rightarrow B C=\sqrt{432}$
$ \Rightarrow B C=2 \times 2 \times 3 \sqrt{3}$
$ \Rightarrow B C= 1 2 \sqrt{ 3 } \sim c m $

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Question 234 Marks

In the given figure, $A B \| C D, A B=7 \ cm , B D=25 \ cm$ and $C D=17 \ cm$;find the length of side $BC$.

Answer

Take $M$ to be the point on $C D$ such that $A B=D M$.
So $DM =7 \ cm$ and $MC =10 \ cm$
Join points $B$ and $M$ to form the line segment $BM$.
So $B M \| A D$ also $B M=A D$.

In right$-$angled $\triangle B A D$,
$B D^2=A D^2+B A^2$
$ (25)^2=A D^2+(7)^2$
$ A D^2=(25)^2-(7)^2$
$ A D^2=576$
$ A D=24$
In right$-$angled $\triangle C M B$,
$C B^2=C M^2+M B^2$
$C B^2=(10)^2+(24)^2$
$\ldots[ MB = AD ]$
$ C B^2=100+576$
$ C B^2=676$
$ C B=26 \ cm $

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