2 Marks Questions

Question 12 Marks

Is it possible to draw a triangle, the lengths of whose sides are given below? $2\ cm, 3\ cm, 4\ cm$

Answer

Clearly, $2 + 3 > 4 3 + 4 > 2 2 + 4 > 3$ Thus, the sum of any two sides is greater than the third side. Hence, it is possible to draw a triangle having sides $2\ cm, 3\ cm$ and $4\ cm.$

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Question 22 Marks

Is it possible to draw a triangle, the lengths of whose sides are given below? $1\ cm, 1\ cm, 1\ cm$

Answer

Consider numbers $1, 1$ and $1.$
Clearly, $1 + 1 > 1 1 + 1 > 1 1 + 1 > 1$
Thus, the sum of any two sides is greater than the third side. Hence, it is possible to draw a triangle having sides $1m, 1cm$ and $1cm.$

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Question 32 Marks

Is it possible to draw a triangle, the lengths of whose sides are given below?$3.4\ cm, 2.1\ cm, 5.3\ cm$

Answer

Consider the numbers $3.4, 2.1$ and $5.3$ Clearly: $3.4 + 2.1 > 5.3 5.3 + 2.1 > 3.4 5.3 + 3.4 > 2.1$ Thus, the sum of any two sides is greater than the third side. Hence, it is possible to draw a triangle having sides $3.4cm, 2.1cm$ and $5.3cm.$

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Question 42 Marks

Is it possible to draw a triangle, the lengths of whose sides are given below$?7\ cm, 8\ cm, 15\ cm$

Answer

Clearly, $7 + 8 = 15$ Thus, the sum of these two numbers is not greater than the third number.
Hence, it is not possible to draw a triangle having sides $7\ cm, 8\ cm$ and $15\ cm.$

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Question 52 Marks

In a $\triangle\text{ABC}$, if $\angle\text{A} = 72^\circ$ and $\angle\text{B} = 63^\circ$, find $\angle\text{C}$.

Answer

In $\triangle\text{ABC}$, $\angle\text{A} = 72^\circ,\angle\text{B} = 63^\circ$ But $\angle\text{A}+\angle\text{B}+\angle\text{C}=360^\circ$ (Sum of angles of a triangle)
$\Rightarrow72^\circ + 63^\circ +\angle\text{C} = 180^\circ$
$\Rightarrow 135^\circ +\angle\text{C} = 180^\circ$
$\Rightarrow\angle\text{C}=180^\circ-135^\circ=45^\circ$

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Question 62 Marks

In a $\triangle\text{XYZ}$, if $\angle\text{X} = 90^\circ$and $\angle\text{Z} = 48^\circ$, find $\angle\text{Y}$.

Answer

In $\triangle\text{XYZ}$, $\angle\text{X} = 90^\circ$, and $\angle\text{Z} = 48^\circ$ But $\angle\text{X}+\angle\text{Y}+\angle\text{Z}=180^\circ$ (sum of angles of a triangle) $\Rightarrow90^\circ+\angle\text{Y}+48^\circ=180^\circ$
$\Rightarrow\angle\text{Y}+138^\circ=180^\circ$
$\Rightarrow\angle\text{Y}=180^\circ-138^\circ$
$\Rightarrow\angle\text{D}=42^\circ$

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Question 72 Marks

In a $\triangle\text{DEF}$, if $\angle\text{E} = 105^\circ$and $\angle\text{F} = 40^\circ$, find $\angle\text{D}$.

Answer

In $\triangle\text{DEF}$, $\angle\text{E} = 105^\circ$, and $\angle\text{F} = 40^\circ$ But $\angle\text{D}+\angle\text{E}+\angle\text{F}=180^\circ$ (sum of angles of a triangle) $\Rightarrow\angle\text{D}+105^\circ+40^\circ=180^\circ$
$\Rightarrow\angle\text{D}+145^\circ=180^\circ$
$\Rightarrow\angle\text{D}=180^\circ-145^\circ$
$\Rightarrow\angle\text{D}=35^\circ$

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Question 82 Marks

In right $\triangle\text{ABC},$ the lengths of its legs are given as $a = 6\ cm$ and $b = 4.5\ cm$. Find the length of its hypotenuse.

Answer

In right $\triangle\text{ABC},$
$\angle\text{C}=90^\circ$
$a=6 \mathrm{~cm}, b=4.5 \mathrm{~cm}$
By Pythagoras Theorem,
$c^2=a^2+b^2$
$=(6)^2+(4.5)^2$
$=36.00+20.25$
$=56.25$
$=(7.5)^2$
$c=7.5 \mathrm{~cm}$

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Question 92 Marks

What is the measure of each angle of an equilateral triangle?

Answer

In an equilateral triangle,
All sides are equal.
All angles are also equal.
Each angle $=\frac{180^\circ}{3}=60^\circ$ (Sum of angles of a triangle $= 180^\circ )$

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Question 102 Marks

Is it possible to draw a triangle, the lengths of whose sides are given below?$6\ cm, 7\ cm, 14\ cm$

Answer

Consider the numbers $6, 7$ and $14$. Clearly, $6 + 7 < 14$ Thus, the sum of these two numbers is not greater than the third number. Hence, it is not possible to draw a triangle having sides $6\ cm, 7\ cm$ and $14\ cm.$

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Question 112 Marks

Find the length of the hypotenuse of a right triangle, the other two sides of which measure $9\ cm$ and $12\ cm.$

Answer

In right triangle $A B C$,
$\angle \mathrm{B}=90^{\circ} \mathrm{AB}=9 \mathrm{~cm}, \mathrm{BC}=12 \mathrm{~cm}$
By Pythagoras Theorem,
$\mathrm{AC}^2=\mathrm{AB}^2+\mathrm{BC}^2=(9)^2+(12)^2$
$\mathrm{AC}^2=81+144$
$\mathrm{AC}^2=225$
$\mathrm{AC}=\sqrt{225}=15 \mathrm{~cm}$

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