2 Mark Question

Question 12 Marks

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

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 2cm, 3cm and 4cm.

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Question 22 Marks

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

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?3.4cm, 2.1cm, 5.3cm

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?7cm, 8cm, 15cm

Answer

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

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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 = 6cm and b = 4.5cm. Find the length of its hypotenuse.

Answer

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

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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°)

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Question 102 Marks

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

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 6cm, 7cm and 14cm.

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Question 112 Marks

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

Answer

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

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