[3 marks sum]

Question 13 Marks

If two sides of a triangle are $8 \ cm$ and $13 \ cm,$ then the length of the third side is between $a \ cm$ and $b \ cm.$ Find the values of $a$ and $b$ such that $a$ is less than $b.$

Answer

The sum of any two sides of the triangle is always greater than third side of the triangle.
Third side $< 13 + 8 = 21 \ cm.$
The difference between any two sides of the triangle is always less than the third side of the triangle.
Third side $> 13 - 8 = 5 \ cm.$
Therefore, the length of the third side is between $5 \ cm$ and $21 \ cm,$ respectively.
The value of $a = 5 \ cm$ and $b = 21\ cm.$

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Question 23 Marks

In $\triangle ABC, \text{AB} > \text{AC}$ and $D$ is a point inside $\text{BC}$.Show that$: \text{AB} > \text{AD}.$

Answer

Given that, $\text{AB}>\text{AC}$
$\Rightarrow \angle C>\angle B ...(i$)
Also in $\triangle ADC$
$ \angle ADB =\angle DAC +\angle C \ldots[$ Exterior angle$]$
$ \Rightarrow \angle ADB >\angle C$
$ \Rightarrow \angle ADB >\angle C >\angle B \ldots[$From $(i) ]$
$ \Rightarrow \angle ADB >\angle B$
$ \Rightarrow \text{AB} >\text{AD} .$

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Question 33 Marks

Given: $\text{ED} = \text{EC}$, Prove: $\text{AB}+\text{AD}>\text{BC}$.

Answer

The sum of any two sides of the triangle is always greater than the third side of the triangle.
ln $\triangle CEB,$
$\text{CF} +\text{EB} >\text{ BC}$
$\Rightarrow \text{DE} + \text{EB} > \text{BC} ...[\text{ CE} = \text{DE} ]$
$\Rightarrow \text{DB} > \text{BC} ...(i)$
ln $\triangle ADB,$
$\text{AD}+ \text{AB} >\text{ BD}$
$\Rightarrow \text{AD} +\text{ AB} > \text{BD} >\text{ BC} ...$[ from$(i) ]$
$\Rightarrow \text{AD}+ \text{AB} > \text{BC}$

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Question 43 Marks

In $\triangle ABC,$ side $\text{AC}$ is greater than side $\text{AB}$. If the internal bisector of angle $A$ meets the opposite side at point $D$,prove that$: \angle ADC$ is greater than $\angle ADB.$

Answer

In $\triangle ADC$,
$\angle ADB =\angle 1+\angle C \ldots (i)$
In $ADB _{\text {, }}$
$\angle ADC=\angle 2+\angle B ...(ii)$
But $\text{AC}>\text{AB} ....[$ Given $]$
$\Rightarrow \angle B>\angle C$
Also given, $\angle 2=\angle 1$ .... [ $\text{AD}$ is bisector of $A$ ]
$\Rightarrow \angle 2+\angle B>\angle 1+\angle C ...(iii)$
From $(i), (ii)$ and $(iii)$
$\Rightarrow \angle ADC >\angle ADB .$

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Question 53 Marks

In quadrilateral $\text{ABCD},$ side $\text{AB}$ is the longest and side $\text{DC}$ is the shortest.Prove that$: C > A.$

Answer

In the quadrilateral $\text{ABCD}$,
Since $\text{AB}$ is the longest side and $\text{DC}$ is the shortest side.
$ \angle 1>\angle 2[ \text{AB}> \text{BC} ]$
$ \angle 7>\angle 4[\text{ AD} > \text{DC} ]$
$ \therefore \angle 1+\angle 7>\angle 2+\angle 4$
$ \Rightarrow \angle C>\angle A $

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Question 63 Marks

In the following figure$; \text{AB}$ is the largest side and $\text{BC}$ is the smallest side of $\triangle ABC$.
Write the angles $x^{\circ}, y^{\circ}$ and $z^{\circ}$ in ascending order of their values.

Answer

Given: $\text{AB}$ is the largest side and $\text{BC}$ is the smallest side of the $\triangle ABC$.
To prove: Arrange $x ^{\circ}, y ^{\circ}$ and $z ^{\circ}$ is ascending order
Proof: $\text{AB}>\text{AC}>\text{BC}$
Then, $180^{\circ}- z ^{\circ}>180^{\circ}- y ^{\circ}>180^{\circ}- x ^{\circ}$
$\Rightarrow- z ^{\circ}>- y ^{\circ}>- x ^{\circ}$
$\Rightarrow z ^{\circ}< y ^{\circ}< x ^{\circ} .$

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Question 73 Marks

$P$ is any point inside the $\triangle ABC.$Prove that$: \angle BPC > \angle BAC.$

Answer

Let $\angle PBC = x$ and $\angle PCB = y$
then,
$\angle BPC =180^{\circ}-( x + y )\ldots (i)$
Let $\angle ABP=a$ and $\angle ACP=b$
then,
$\angle BAC =180^{\circ}-( x + a )-( y + b )$
$ \Rightarrow \angle BAC =180^{\circ}-( x + y )-( a + b )$
$ \Rightarrow \angle BAC =\angle BPC -( a + b )$
$ \Rightarrow \angle BPC =\angle BAC +( a + b )$
$ \Rightarrow \angle BPC >\angle BAC$

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MATHEMATICS [3 marks sum]

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