Additional Questions - Chapter 4 Geometry 10th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Additional Questions
Question $1 .$
In figure if $\mathrm{PQ} \| \mathrm{RS}$, Prove that $\triangle \mathrm{POQ} \sim \triangle \mathrm{SOR}$
Solution:

$\mathrm{PQ} \| \mathrm{RS}$
So, $\angle P=\angle S$ (A Hernate angles)
and $\angle Q=\angle R$
Also, $\angle \mathrm{POQ}=\angle \mathrm{SOR}$ (vertically opposite angle)
$\therefore \triangle \mathrm{POQ} \sim \triangle \mathrm{SOR}$ (AAA similarity criterion)
 

Question 2.
In figure $\mathrm{OA}, \mathrm{OB}=\mathrm{OC}$. $\mathrm{OD}$ Show that $\angle \mathrm{A}=\angle \mathrm{C}$ and $\angle \mathrm{B}=\angle \mathrm{D}$
Solution:

$\mathrm{OA}, \mathrm{OB}=\mathrm{OC} . \mathrm{OD}$ (Given)
Also we have $\angle \mathrm{AOD}=\angle \mathrm{COB}$
(vertically opposite angles)(2)

From (1) and (2)
$\therefore \triangle \mathrm{AOD} \sim \triangle \mathrm{COB}$ (SAS similarity criterion)
So, $\angle \mathrm{A}=\angle \mathrm{C}$ and $\angle \mathrm{B}=\angle \mathrm{D}$
(corresponding angles of similar triangles)

Question $3 .$
In figure the line segment $\mathrm{XY}$ is parallel to side $\mathrm{AC}$ of $\triangle \mathrm{ABC}$ and it divides the triangle into two parts of equal areas. Find the ratio $\frac{\mathrm{AX}}{\mathrm{AB}}$
Solution:

Given XY $\| A C$

So, $\quad \underline{B X Y}=\lfloor A$ and
$\underline{B Y X}=\angle C$ (corresponding angles)
$\therefore \triangle \mathrm{ABC} \sim \triangle \mathrm{XB} \quad$ (AAA similarity criterion)
So, $\frac{\operatorname{ar}(\mathrm{ABC})}{\operatorname{ar}(\mathrm{XBY})}=\left(\frac{\mathrm{AB}}{\mathrm{XB}}\right)^{2}$
$a r(\mathrm{ABC})=2 a r(\mathrm{XBY})$
$\frac{\operatorname{ar}(\mathrm{ABC})}{a r(\mathrm{XBY})}=\frac{2}{1}$
From (1) and (2),
$\begin{aligned}
\left(\frac{\mathrm{AB}}{\mathrm{XB}}\right)^{2} &=\frac{2}{1} \text { i.e., } \frac{\mathrm{AB}}{\mathrm{XB}}=\frac{\sqrt{2}}{1} \\
\frac{\mathrm{XB}}{\mathrm{AB}} &=\frac{1}{\sqrt{2}} \\
1-\frac{\mathrm{XB}}{\mathrm{AB}} &=1-\frac{1}{\sqrt{2}} \\
\frac{\mathrm{AB}-\mathrm{XB}}{\mathrm{AB}} &=\frac{\sqrt{2}-1}{\sqrt{2}} \\
\frac{\mathrm{AX}}{\mathrm{AB}} &=\frac{\sqrt{2}-1}{\sqrt{2}}=\frac{2-\sqrt{2}}{2}
\end{aligned}$

Question $4 .$
In $\mathrm{AD} \perp \mathrm{BC}$, prove that $\mathrm{AB}^{2}+\mathrm{CD}^{2}=\mathrm{BD}^{2}+\mathrm{AC}^{2}$.
Solution:
From $\triangle \mathrm{ADC}$, we have
$\mathrm{AC}^{2}=\mathrm{AD}^{2}+\mathrm{CD}^{2} \ldots \ldots \ldots \ldots \ldots .(1)$
(Pythagoras theorem)
From $\triangle \mathrm{ADB}$, we have
$\mathrm{AB}^{2}=\mathrm{AD}^{2}+\mathrm{BD}^{2}$
(Pythagoras theorem)
Subtracting (1) from (2) we have,
$\begin{aligned}
&A B^{2}-A C^{2}=B D^{2}-C D^{2} \\
&A B^{2}+C D^{2}=B D^{2}+A C^{2}
\end{aligned}$

Question $5 .$
$\mathrm{BL}$ and $\mathrm{CM}$ are medians of a triangle $\mathrm{ABC}$ right angled at $\mathrm{A}$.
Prove that $4\left(\mathrm{BL}^{2}+\mathrm{CM}^{2}\right)=5 \mathrm{BC}^{2}$.
Solution:
$\mathrm{BL}$ and $\mathrm{CM}$ are medians at the $\triangle \mathrm{ABC}$ in which
$\mathrm{A}=\angle 90^{\circ}$.
From $\triangle \mathrm{ABC}$
$B C^{2}=A B^{2}+A C^{2} \ldots \ldots \ldots \ldots \ldots . .(1)$

From $\triangle \mathrm{ABL}$,
$\begin{aligned}
&\mathrm{BL}^{2}=\mathrm{AL}^{2}+\mathrm{AB}^{2} \\
&\mathrm{BL}^{2}=\left(\frac{\mathrm{AC}}{2}\right)^{2}+\mathrm{AB}^{2}
\end{aligned}$
(L is the mid-point at $\mathrm{AC}$ )
$\begin{aligned}
\mathrm{BL}^{2} &=\frac{A C^{2}}{4}+A B^{2} \\
4 B^{2} &=A C^{2}+4 A B^{2}
\end{aligned}$
From $\triangle \mathrm{CMA}$,
$\begin{aligned}
\mathrm{CM}^{2} &=\mathrm{AC}^{2}+\mathrm{AM}^{2} \\
\mathrm{CM}^{2} &=\mathrm{AC}
\end{aligned}$
$(\mathrm{M}$ is the mid-point at $\mathrm{AB})$
$\mathrm{CM}^{2}=\mathrm{AC}^{2}+\frac{\mathrm{AB}^{2}}{4}$
Adding (2) and (3), we have
$\begin{aligned}
&4\left(\mathrm{BL}^{2}+\mathrm{CM}^{2}\right)=5\left(\mathrm{AC}^{2}+\mathrm{AB}^{2}\right) \\
&4\left(\mathrm{BL}^{2}+\mathrm{CM}^{2}\right)=5 \mathrm{BC}^{2}[\text { From (1) }]
\end{aligned}$

Question $6 .$
Prove that in a right triangle, the square of the hypotenure is equal to the sum of the squares of the others two sides.
Solution:
Proof:

We are given a right triangle $\mathrm{ABC}$ right angled at $\mathrm{B}$.
We need to prove that $A C^{2}=A B^{2}+B C^{2}$
Let us draw $\mathrm{BD} \perp \mathrm{AC}$
Now, $\triangle \mathrm{ADB} \sim \triangle \mathrm{ABC}$
$\frac{\mathrm{AD}}{\mathrm{AB}}=\frac{\mathrm{AB}}{\mathrm{AC}}$ (sides are proportional)
$\mathrm{AD}, \mathrm{AC}=\mathrm{AB}^{2} \ldots \ldots \ldots(1)$
Also, $\triangle \mathrm{BDC} \sim \triangle \mathrm{ABC}$
$\frac{C D}{B C}=\frac{B C}{A C}$
$\mathrm{CD} \cdot \mathrm{AC}=\mathrm{BC}^{2}$
(2)
Adding (1) and (2)
$\begin{aligned}
&A D \cdot A C+C D \cdot A C=A B^{2}+B C^{2} \\
&A C(A D+C D)=A B^{2}+B C^{2} \\
&A C \cdot A C=A B^{2}+B C^{2} \\
&A C^{2}=A B^{2}+B C^{2}
\end{aligned}$

Question $7 .$
A ladder is placed against a wall such that its foot is at a distance of $2.5 \mathrm{~m}$ from the wall and its top reaches a window $6 \mathrm{~m}$ above the ground. Find the length of the ladder.
Solution:
Let $\mathrm{AB}$ be the ladder and $\mathrm{CA}$ be the wall with the window at $\mathrm{A}$.

Also, $\mathrm{BC}=2.5 \mathrm{~m}$ and $\mathrm{CA}=6 \mathrm{~m}$
From Pythagoras theorem,
$\begin{aligned}
&\mathrm{AB}^{2}=\mathrm{BC}^{2}+\mathrm{CA}^{2} \\
&=(2.5)^{2}+(6)^{2} \\
&=42.25
\end{aligned}$
$\mathrm{AB}=6.5$
Thus, length at the ladder is $6.5 \mathrm{~m}$.
 

Question $8 .$
In figure $\mathrm{O}$ is any point inside a rectangle $\mathrm{ABCD}$. Prove that $\mathrm{OB}^{2}+\mathrm{OD}^{2}=\mathrm{OA}^{2}+\mathrm{OC}^{2}$.
Solution:
Through $\mathrm{O}$, draw $\mathrm{PQ} \| \mathrm{BC}$ so that $\mathrm{P}$ lies on $\mathrm{AB}$ and $\mathrm{Q}$ lies on $\mathrm{DC}$.
Now, $\mathrm{PQ} \| \mathrm{BC}$
$\mathrm{PQ} \perp \mathrm{AB}$ and $\mathrm{PQ} \perp \mathrm{DC}\left(\because \angle \mathrm{B}=90^{\circ}\right.$ and $\left.\angle \mathrm{C}=90^{\circ}\right)$
So, $\angle \mathrm{BPQ}=90^{\circ}$ and $\angle \mathrm{CQP}=90^{\circ}$
Therefore BPQC and APQD are both rectangles.
Now from $\triangle \mathrm{OPB}$,
$\mathrm{OB}^{2}=\mathrm{BP}^{2}+\mathrm{OP}^{2}$
Similarly from $\triangle \mathrm{OQD}$,
$\mathrm{OD}^{2}=\mathrm{OQ}^{2}+\mathrm{DQ}^{2}$
From $\triangle \mathrm{OQC}$, we have
$\mathrm{OC}^{2}=\mathrm{OQ}^{2}+\mathrm{CQ}^{2}$
$\triangle \mathrm{OAP}$, we have
$\mathrm{OA}^{2}=\mathrm{AP}^{2}+\mathrm{OP}^{2}$
Adding (1) and (2)

$\begin{aligned}
&\mathrm{OB}^{2}+\mathrm{OD}^{2}=\mathrm{BP}^{2}+\mathrm{OP}^{2}+\mathrm{OQ}^{2}+\mathrm{DQ}^{2}(\mathrm{As} \mathrm{} \mathrm{BP}=\mathrm{CQ} \text { and } \mathrm{DQ}=\mathrm{AP}) \\
&=\mathrm{CQ}^{2}+\mathrm{OP}^{2}+\mathrm{OQ}^{2}+\mathrm{AP}^{2} \\
&=\mathrm{CQ}^{2}+\mathrm{OQ}^{2}+\mathrm{OP}^{2}+\mathrm{AP}^{2} \\
&=\mathrm{OC}^{2}+\mathrm{OA}^{2}[\text { From (3) and (4) }
\end{aligned}$

Question 9 .
In $\angle \mathrm{ACD}=90^{\circ}$ and $\mathrm{CD} \perp \mathrm{AB}$. Prove that $\frac{\mathrm{BC}^{2}}{\mathrm{AC}^{2}}=\frac{\mathrm{BD}}{\mathbf{A D}}$
Solution:
$\triangle \mathrm{ACD} \sim \triangle \mathrm{ABC}$
So,
$\frac{\mathrm{AC}}{\mathrm{AB}}=\frac{\mathrm{AD}}{\mathrm{AC}}$

$\mathrm{AC}^{2}=\mathrm{AB} \cdot \mathrm{AD}$
Similarly $\triangle \mathrm{BCD} \sim \triangle \mathrm{BAC}$
So,
$\begin{aligned}
&\frac{B C}{B A}=\frac{B D}{B C} \\
&B C^{2}=B A \cdot B D
\end{aligned}$
From (1) and (2)
$\frac{\mathrm{BC}^{2}}{\mathrm{AC}}=\frac{\mathrm{BA} \cdot \mathrm{BD}}{\mathrm{AB} \cdot \mathrm{AD}}=\frac{\mathrm{BD}}{\mathrm{AD}}$

Question $10 .$
The perpendicular from $\mathrm{A}$ on side $\mathrm{BC}$ at a $\triangle \mathrm{ABC}$ intersects $\mathrm{BC}$ at $\mathrm{D}$ such that $\mathrm{DB}=3 \mathrm{CD}$. Prove that $2 \mathrm{AB}^{2}=2 \mathrm{AC}^{2}+\mathrm{BC}^{2}$.
Solution:
We have $\mathrm{DB}=3 \mathrm{CD}$
$\begin{aligned}
&\mathrm{BC}=\mathrm{BD}+\mathrm{DC} \\
&\mathrm{BC}=3 \mathrm{CD}+ \\
&\mathrm{BC}=4 \mathrm{CD} \\
&\mathrm{CD}=\frac{1}{4} \mathrm{BC}
\end{aligned}$
$C D=\frac{1}{4} B C$ and
$\mathrm{BD}=3 \mathrm{CD}=\frac{3}{4} \mathrm{BC}$
Since $\triangle \mathrm{ABD}$ is a right triangle (i) right angled at $\mathrm{D}$

$\mathrm{AB}^{2}-\mathrm{AD}^{2}+\mathrm{BD}^{2}$
By $\triangle \mathrm{ACD}$ is a right triangle right angled at $\mathrm{D}$
$\mathrm{AC}^{2}=\mathrm{AD}^{2}+\mathrm{CD}^{2}$
Subtracting equation (iii) from equation (ii),
we got
$\begin{aligned}
\mathrm{AB}^{2}-\mathrm{AC}^{2} &=\mathrm{BD}^{2}-C \mathrm{D}^{2} \\
\Rightarrow \quad \mathrm{AB}^{2}-A \mathrm{C}^{2} &=\left(\frac{3}{4} \mathrm{BC}\right)^{2}-\left(\frac{1}{4} \mathrm{BC}\right)_{\text {From (i) }}^{2}
\end{aligned}$
$\left.C D=\frac{1}{4} B C, B D=\frac{3}{4} B C\right)$
$\Rightarrow \quad \mathrm{AB}^{2}-\mathrm{AC}^{2}=\frac{9}{16} \mathrm{BC}^{2}-\frac{1}{16} \mathrm{BC}^{2}$
$\Rightarrow \mathrm{AB}^{2}-\mathrm{AC}^{2}=\frac{1}{2} \mathrm{BC}^{2}$
$\Rightarrow 2\left(A B^{2}-A C^{2}\right)=B C^{2}$
$\Rightarrow \quad 2 \mathrm{AB}^{2}=2 \mathrm{AC}^{2}+\mathrm{BC}^{2}$