5 Marks Questions

Question 15 Marks

Find the angles x and y in each figure.

Answer

Let given triangle be $\triangle A B C$.
Then, we have $A B=A C$.
So, $\triangle A B C$ is an isosceles triangle.
$\therefore \quad \angle A C B=\angle A B C=y$
Also, $\angle A C D$ is an exterior angle.
We know that in a triangle an exterior angle and the interior adjacent angle form a linear pair.
$\therefore \angle A C B+\angle A C D=180^{\circ}$$\quad$ [linear pair]
$\Rightarrow \angle A C B+120^{\circ}=180^{\circ} $
$\Rightarrow \angle A C B=180^{\circ}-120^{\circ}=60^{\circ} $
$\therefore \angle A B C=\angle A C B=60^{\circ}$
$\text { In } \triangle A B C, \angle A+\angle B+\angle A C B=180^{\circ}$
[by angle sum property of a triangle]
$\Rightarrow x+y+y =180^{\circ} $
$\Rightarrow x+60^{\circ}+60^{\circ} =180^{\circ} $
$\Rightarrow x =180^{\circ}-120^{\circ}=60^{\circ}$
Hence, the value of $x$ is $60^{\circ}$ and $y$ is $60^{\circ}$.
(ii)

Then, we have $A B=B C$ and $\angle B=90^{\circ}$.
$\therefore \quad \angle A C B=\angle B A C=x$
[since, angle opposite to equal sides are equal]
$\text { In } \triangle A B C, \angle A+\angle B+\angle A C B=180^{\circ}$
[by angle sum property of a triangle]
$\Rightarrow x+90^{\circ}+x =180^{\circ} $
$\Rightarrow 2 x+90^{\circ} =180^{\circ} $
$\Rightarrow 2 x =180^{\circ}-90^{\circ} $
$\Rightarrow x =\frac{90^{\circ}}{2}=45^{\circ}$
Also, $\angle A C D$ is an exterior angles.
By exterior angle property of a triangle,
Exterior angle = Sum of two interior opposite angles
$\Rightarrow y=\angle B A C+\angle B $
$\Rightarrow y=x+90^{\circ} $
$\Rightarrow y=45^{\circ}+90^{\circ}=135^{\circ}$
Hence, the value of $x$ is $45^{\circ}$ and $y$ is $135^{\circ}$.
(iii)

Let given triangle be $\triangle A B C$.
Then, we have $A B=A C$.
So, $\triangle A B C$ is an isosceles triangle.
$\therefore \quad \angle A B C=\angle A C B$
[since, angle opposite to equal sides are equal]
$\Rightarrow \angle A C B=x $
$\text { and } \angle B A C=92^{\circ} \quad \text { [vertically opposite angles] }$
In $\triangle A B C$, by angle sum property of a traingle,
$\angle A B C+\angle B A C+\angle A C B =180^{\circ} $
$\Rightarrow x+92^{\circ}+x =180^{\circ} $
$\Rightarrow 2 x+92^{\circ} =180^{\circ} $
$\Rightarrow 2 x =180^{\circ}-92 \Rightarrow 2 x=88^{\circ} $
$\Rightarrow x =\frac{88^{\circ}}{2} \Rightarrow x=44^{\circ}$
Also, $\angle A C D$ is an exterior angle of $\triangle A B C$.
By exterior angle property of a triangle
Exterior angle = Sum of two interior opposite angles
$\Rightarrow y=\angle B A C+\angle A B C $
$\Rightarrow y=92^{\circ}+x$
$\Rightarrow y=92^{\circ}+44^{\circ}=136^{\circ}$
Hence, the value of $x$ is $44^{\circ}$ and $y$ is $136^{\circ}$.

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Question 25 Marks

Find angle x in each figure.

Answer

(i)

Let the given triangle be $\triangle A B C$, then we have $A B=A C$. We know that angle opposite to equal sides of an isosceles triangle are equal.
$\therefore \quad x=40^{\circ}$
Hence, the value of $x$ is $40^{\circ}$.
(ii)

Let the given triangle be $\triangle A B C$. Then, we have $A B=A C$.
So, $\triangle A B C$ is an isosceles triangle.
$\therefore \angle A=\angle C=45^{\circ}$
[since, the angle opposite to equal sides of an isosceles triangle are equal]
Now, in $\triangle A B C$ by angle sum property of a triangle,
$\angle A+\angle B+\angle C=180^{\circ} \Rightarrow 45^{\circ}+x+45^{\circ}=180^{\circ} $
${[\angle A=\angle B=45 \text { because of equal side}]}$
$\Rightarrow \quad x+90^{\circ}=180^{\circ} \Rightarrow x=180^{\circ}-90^{\circ}=90^{\circ}$
Hence, the value of $x$ is $90^{\circ}$.
(iii) Ans. $x=50^{\circ}$
(iv) Ans. $x=40^{\circ}$
(v) Ans. $45^{\circ}$
(vi) Ans. $x=70^{\circ}$
(vii)

Let the given triangle be $\triangle A B C$.
Then, we have
$A B=A C$
So, $\triangle A B C$ is an isosceles triangle.
$\therefore \quad \angle A C B=\angle B$
[since, the angle opposite to equal sides of an isosceles triangle are equal]
$\angle A C B=x$
We know that linear pair of angles is supplementry.
$\therefore \angle A C B+\angle A C D=180^{\circ} \quad \text { [linear pair] } $
$\Rightarrow x+120^{\circ}=180^{\circ} {[\because \angle A C D=120, \text { given] }} $
$\Rightarrow x=180^{\circ}-120^{\circ}=60^{\circ}$
Hence, the value of $x$ is $60^{\circ}$.
(viii)

Let the given triangle be $\triangle A B C$. Then, we have
$A B=A C$
So, $\triangle A B C$ is an isosceles triangle.
$\therefore \quad \angle B=\angle C$
[since, the angle opposite to equal sides of an isosceles triangle are equal]
Also, $\angle D A C$ is an exterior angle of $\triangle A B C$.
Now, by exterior angle property of a triangle,
Sum of two interior opposite angles = Exterior angle
$\Rightarrow x+x=110^{\circ} \Rightarrow 2 x=110^{\circ}=\frac{110^{\circ}}{2}=55^{\circ}$
Hence, the value of $x$ is $55^{\circ}$.
(ix)

Let given triangle be $\triangle A B C$. Then, we have $A B=B C$ So, $\triangle A B C$ is an isosceles triangle.
$\therefore \quad \angle B A C=\angle B C A=x$
From the figure,
$\angle B C A=\angle D C E=30^{\circ} \text { [vertically opposite angles] }$
From Eq. (i), $\angle B A C=\angle B C A$
$\Rightarrow \quad x=30^{\circ}$
Hence, the value of $x$ is $30^{\circ}$.

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Question 35 Marks

Look at following figures and classify each of the triangles according to its
(a) sides $\quad$ (b) angles

Answer

(i) (a) In $\triangle A B C, A C=B C=8 cm$
i.e. two sides are equal.
Therefore, $\triangle A B C$ is an isosceles triangle.
(b) Also, all the angles of $\triangle A B C$ are less than $90^{\circ}$.
Therefore, $\triangle A B C$ is an acute angled triangle.
(ii) (a) In $\triangle P Q R, P Q \neq Q R \neq R P$ $\quad$ [given]
i.e. all three sides are unequal.
Therefore, $\triangle P R Q$ is a scalene triangle.
(b) Also, $\angle R=90^{\circ}$ $\quad$ [given]
Therefore, $\triangle P Q R$ is a right angled triangle.
(iii) (a) In $\triangle L M N, L N=M N=7 cm$ $\quad$ [given]
i.e. two sides are equal.
Therefore, $\triangle L M N$ is an isosceles triangle.
(b) Also, $\angle N > 90^{\circ}$ $\quad$ [given]
Therefore, $\triangle L M N$ is an obtuse angled triangle.
(iv) (a) $\operatorname{In} \triangle R S T, R S=S T=T R=5.2 cm$ $\quad$ [given]
i.e. all three sides are equal.
Therefore, $\triangle R S T$ is an equilateral triangle.
(b) Also, all the angles of $\triangle R S T$ are acute.
Therefore, $\triangle R S T$ is an acute angled triangle.
(v) (a) Ans. Isosceles triangle
(b) Ans. Obtuse angled triangle
(vi) (a)Ans. Isosceles triangle
(b) Ans. Right angled triangle.

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Question 45 Marks

Find the unknown length x in the following figures.

Answer

(i) Let given triangle be $\triangle A B C$ and $\triangle A B C$ is a right angled at $B$.
Again, let $A C=x$
Given, AB = 3 BC = 4

In $\triangle A B C$, by using Pythagoras property,
$(\text {Hypotenuse})^2=(\text {Base})^2+(\text {Perpendicular})^2$
i.e. $A C^2=A B^2+B C^2
\Rightarrow x^2=(3)^2+(4)^2=9+16$ $\Rightarrow \quad x^2=25 \Rightarrow x=\sqrt{25}=5$
Hence, the unknown length $(x)$ is 5.
(ii) Ans. $x=10$
(iii) Ans. $x=17 cm$
(iv) Ans. $x=25$ units
(v) Let given triangle be $\triangle A B C$ in which $A B=A C$. Then, it is an isosceles triangle and given $A D$ is perpendicular to $B C$, so $D$ is the mid-point of $B C$.

Given, $\quad A B=37=A C, A D=12, B C=x$
Now,$\quad$$B D=D C=\frac{B C}{2}=\frac{x}{2}$
In $\triangle A D B$, using Pythagoras property,
$A B^2 =B D^2+A D^2 $
$\Rightarrow (37)^2 =\left(\frac{x}{2}\right)^2+(12)^2 $
$\Rightarrow 1369 =\frac{x^2}{4}+144 $
$\Rightarrow 1369-144 =\frac{x^2}{4} \Rightarrow 1225=\frac{x^2}{4} $
$\Rightarrow 1225 \times 4 =x^2 $
$\Rightarrow x^2 =1225 \times 4 \Rightarrow x^2=4900 $
$\Rightarrow x =\sqrt{4900}=70$
Hence, the unknown length of $x$ is 70 units.
(vi) Let given triangle be $\triangle P R S$ which is divided into two $\triangle P Q R$ and $\triangle R Q S$.

In $\triangle P Q R$, using Pythagoras property,
$P R^2 =P Q^2+Q R^2 $
$\Rightarrow (12)^2 =x^2+(3)^2 \Rightarrow 144=x^2+9 $
$\Rightarrow 144-9 =x^2 $
$\Rightarrow x^2 =135 \Rightarrow x=\sqrt{135}$
Hence, the unknown length of $(x)$ is $\sqrt{135} cm$.

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Question 55 Marks

Find the values of the unknowns x and y in the following diagrams:

Answer

(i) By exterior angle property of a triangle, Sum of interior opposite angles = Exterior angle
$\Rightarrow x+50^{\circ} =120^{\circ} $
$\Rightarrow x =120^{\circ}-50^{\circ}=70^{\circ}$
Now, by angle sum property of a triangle,
$\Rightarrow x+y+50^{\circ}=180^{\circ} \Rightarrow 70^{\circ}+y+50^{\circ}=180^{\circ} $
$\Rightarrow y+120^{\circ}=180^{\circ} \Rightarrow y=180^{\circ}-120^{\circ}=60^{\circ}$
Hence, the value of the $x$ and $y$ are $70^{\circ}$ and $60^{\circ}$, respectively.
(ii) Ans. $x=50^{\circ}, y=80^{\circ}$
(iii) Ans. $x=110^{\circ}, y=70^{\circ}$
(v) Ans. $x=60^{\circ}, y=90^{\circ}$
(v) Ans, $x=45^{\circ}, y=90^{\circ}$
(vi) From the figure, $y=x \quad$ [vertically opposite angles]
Then, all angles of triangle are equal to $y$.
Now, by angle sum property of a triangle,
$y+y+y=180^{\circ} \Rightarrow 3 y=180^{\circ} \Rightarrow x=\frac{180^{\circ}}{3}=60^{\circ}$
Hence, the value of the $x$ and $y$ are $60^{\circ}$ and $60^{\circ}$ respectively.

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Question 65 Marks

Find the value of the unknown x in the following diagrams.

Answer

(i) Given, in $\triangle A B C, \angle A=x, \angle B=50^{\circ}$ and $\angle C=60^{\circ}$
By angle sum property of a triangle,
$\angle A+\angle B+\angle C=180^{\circ} $
$\Rightarrow x+50^{\circ}+60^{\circ}=180^{\circ} $
$\Rightarrow x+110^{\circ}=180^{\circ} $
$\Rightarrow x=180^{\circ}-110^{\circ}=70^{\circ}$
Hence, the value of the $x$ is $70^{\circ}$.
(ii) Ans. $60^{\circ}$
(iii) Ans. $40^{\circ}$
(iv) Ans. $65^{\circ}$
(v) Ans. $60^{\circ}$
(vi) Ans. $30^{\circ}$

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Question 75 Marks

Find the value of the unknown interior angle x in the following figures.

Answer

(i) By exterior angle property of a triangle, Exterior angle $=$ Sum of interior opposite angles
$\Rightarrow \quad 115^{\circ}=x+50^{\circ} \Rightarrow x=115^{\circ}-50^{\circ}=65^{\circ}$
Hence, the exterior angle $x$ is $65^{\circ}$.
(ii) Ans. $30^{\circ}$
(iii) Ans. $35^{\circ}$.
(iv) Ans. $60^{\circ}$.
(v) Ans. $50^{\circ}$.
(vi) Ans. $40^{\circ}$.

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Question 85 Marks

Find the value of the unknown exterior angle x in the following diagrams.

Answer

(i) By exterior angle property of a triangle, Exterior angle $=$ Sum of interior opposite angles
$
\Rightarrow \quad x=50^{\circ}+70^{\circ}=120^{\circ}
$
Hence, the exterior angle $x$ is $120^{\circ}$.
(ii) Ans. $110^{\circ}$
(iii) Ans. $70^{\circ}$.
(iv) Ans. $120^{\circ}$.
(v) Ans. $100^{\circ}$.
(vi) Ans. $90^{\circ}$.

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