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CBSE Questions for Class 7 Maths The Triangle And Its Properties Quiz 8 - MCQExams.com
CBSE
Class 7 Maths
The Triangle And Its Properties
Quiz 8
Which of the following set of measurements will form a triangle
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$$11cm,4cm,6cm$$
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$$13cm,14cm,25cm$$
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$$8cm,4cm,3cm$$
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$$5cm.16cm.5cm$$
Explanation
Triangle Inequality Theorem states that the sum of two side lengths of a triangle is always greater than the third side.
If this is true for all three combinations of added side lengths.
i.e. $$a+b> c,$$
$$b+c> a$$ a
nd $$c+a> b$$ then the lengths form a triangle
(A) $$11+4=15> 6$$
And $$ 4+6 =10\ngtr 11$$
So, it does not form a triangle
(B) $$13+14=27> 25$$
And $$14+25=39> 13$$
And $$25+13=38> 14$$
So, it forms a triangle
(C) $$8+4=12> 3$$
And $$8+3=11> 4$$
And $$4+3=7\ngtr 8$$
So, it does not form a triangle
(D) $$5+16=21> 5$$
And $$5+5=10\ngtr 16$$
So, it does not form a triangle.
Hence option B is the correct answer
Triangles with sides $$3$$ cm, $$4$$ cm and $$5$$ cm is possible.
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True
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False
Explanation
Yes, as $$3^2+4^2=5^2$$
Find the values of x and y in the following figures
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$$ x=80^0 ; y=120^0$$
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$$ x=50^0 ; y=30^0$$
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$$ x=50^0 ; y=130^0$$
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$$ x=20^0 ; y=10^0$$
Explanation
$$\angle CAB = 80^o$$ [apposite angle]
As $$AB = AC$$
$$\therefore x = \angle ACB$$
Also, by angle sum property, we have
$$ x + x+80 = 180$$
$$2x = 100$$
$$x = 50$$
$$y = 80+x = 130$$ exterior angle property
$$x = 50$$
$$y = 130$$
Find the measure of the angle $$x$$ in the given figure.
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$${ 50 }^\circ$$
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$${ 70 }^\circ$$
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$${ 60 }^\circ$$
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$${ 30 }^\circ$$
Explanation
$$\textbf{Step 1: Find the relation between interior and exterior angles of triangle.}$$
$$ x=\angle{EFD}+\angle{FED}$$
$$\Rightarrow \angle{x}={28^\circ}+{42^\circ}$$ $$\quad \textbf{[Exterior angle is equal to sum of interior angles of a triangle.]}$$
$$\Rightarrow \angle{x}={70^\circ}$$
$$\textbf{Hence, the measure of the angle x is} $$ $$\bf{70^\circ.}$$
Four pair of showing measurements of sides $$\overline{AB}, \overline{BC}$$ and $$\overline{CA}$$ of $$\Delta ABC$$ are given below.
Show which of the following pair/s is/are shows right angle triangle.
Pair P: AB = 25 BC = 7 AC = 24
Pair Q: AB = 8 BC = 6 AC = 10
Pair R: AB = 3 BC = 4 AC = 6
Pair S: AB = 8 BC = 6 AC = 5
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Pairs Q and R show right angle triangle
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Pairs P and Q show right angle triangle
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Pairs P and S show right angle triangle
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Pairs P, Q and S show right angle triangle
Explanation
For pair P:
$$AC^2 + BC^2 = 24^2 + 7^2 = 576 + 49 = 625=25^2$$
and $$AB^2 = 25^2$$
$$\therefore AB^2 = AC^2 + BC^2$$
For pair Q:
$$AB^2 + BC^2 = 8^2 + 6^2 = 64 + 36 = 100$$
$$AC^2 = 10^2 = 100$$
$$\therefore AB^2 + BC^2 = AC^2$$
For pair R:
$$AB^2 + BC^2 = 3^2 + 4^2 = 9 + 16 = 25$$
$$AC^2 = 6^2 = 36$$
$$\therefore AB^2 + BC^2 \neq AC^2$$
For pair S:
$$AC^2 + BC^2 = 5^2 + 6^2 = 25 + 36 = 61$$
$$AB^2 = 8^2 = 64$$
$$\therefore AB^2 \neq AC^2 + BC^2$$
$$\therefore $$ Pairs P and Q show right angled triangle.
A $$34$$ m long ladder reached in the window which is $$16$$ m above from the ground on placing it against a wall. Find the distance of the foot of the ladder from the wall.
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$$40$$ m
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$$30$$ m
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$$50$$ m
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$$10$$ m
Explanation
Let $$AB=$$ length of ladder, $$AC=$$ height of window
In right angled $$ \triangle ABC$$, we have
$$ AB ^{ 2 }={ AC }^{ 2 }+{ BC }^{ 2 }$$
$$\Rightarrow { (34) }^{ 2 }={ (16) }^{ 2 }+{ BC }^{ 2 }\quad $$
$$\Rightarrow { BC }^{ 2 }={ (34) }^{ 2 }-{ (16) }^{ 2 }=900$$
$$\Rightarrow BC=\sqrt { 900 } =30 m$$
From the given figure, find the values of $$x$$ and $$y$$ respectively.
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$${ 47 }^{ \circ },{ 66 }^{ \circ }$$
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$${ 66 }^{ \circ },{ 48 }^{ \circ}$$
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$${ 68 }^{ \circ },{ 47 }^{ \circ }$$
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$${ 47 }^{ \circ },{ 68 }^{ \circ }$$
Explanation
In $$\triangle TCE$$, we have
$$x=\angle TCE+\angle TEC$$ (Exterior angle property)
$$x={ 35 }^{ \circ }+{ 31 }^{ \circ }$$
$$x={ 66 }^{ \circ }\quad \quad $$
In $$\triangle SBD$$ , we have
$$\ \angle AST=\angle SBD+\angle SDB\quad $$
$$\angle AST={ 30 }^{ \circ }+{ 36 }^{ \circ }={ 66 }^{ \circ }$$
In $$\triangle ATS$$, we have
$$ y+x+\angle AST={ 180 }^{ \circ }\quad $$ (Angle sum property)
$$y+{ 66 }^{ \circ }+{ 66 }^{ \circ }={ 180 }^{ \circ }\quad $$
$$y={ 180 }^{ \circ }-\left( { 66 }^{ \circ }+{ 66 }^{ \circ } \right)$$
$$ \Rightarrow y={ 48 }^{ \circ }$$
A tree is broken at a height of $$5$$ m from the ground and its top touches the ground at a distance of $$12$$ m from the base of the tree. Find the original height of the tree.
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$$10m$$
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$$15m$$
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$$13m$$
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$$18m$$
Explanation
Let $$BD$$ be the original height of tree.
$$\therefore AD = AC$$
In $$\triangle ABC$$
$$AC^2=AB^2=BC^2$$
$$=5^2+12^2$$
$$=169$$
$$= 13\times 13$$
$$\Rightarrow AC = 13m$$
$$\Rightarrow AD = 13m$$
$$\therefore$$ Original height of tree $$=(13+5)m$$
$$ = 18m$$
A tree is broken at a height of $$8$$m from the ground and its top touches the ground at a distance of $$15$$m from the base of the tree. Find the original height of the tree.
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$$25m$$
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$$5m$$
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$$23m$$
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$$16m$$
Explanation
Let $$BD$$ be the original height of tree.
$$\therefore AD = AC$$
In $$\triangle ABC$$
$$AC^2=AB^2=BC^2$$
$$=15^2+8^2$$
$$=225+64$$
$$= 289$$
$$\Rightarrow AC = 17m$$
$$\Rightarrow AD = 17m$$
$$\therefore$$ Original height of tree $$=(17+8)m = 25m$$
Find the value of x, y and z in the adjoining figure.
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$$x=160$$, $$y=60$$, $$z=80$$
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$$x=60$$, $$y=80$$, $$z=40$$
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$$x=80$$, $$y=60$$, $$z=140$$
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$$x=140$$, $$y=80$$, $$z=60$$
Explanation
$$In\triangle BCE$$
$$\implies\quad \angle BEC+\angle BCE+\angle CBE={ 180 }^{ \circ }$$
$$\implies\quad { 90 }^{ \circ }+{ 30 }^{ \circ }+\angle CBE={ 180 }^{ \circ }$$
$$\implies\quad \angle CBE={ 60 }^{ \circ }$$
$$ y={ 60 }^{ \circ }$$
$$ In\triangle APC:$$
$$\implies\quad { 50 }^{ \circ }+{ 30 }^{ \circ }+\angle APC={ 180 }^{ \circ }$$
$$\implies\quad \angle APC={ 100 }^{ \circ }$$
$$ \therefore x=180-\angle APC$$
$$ (\because sum\quad of\quad angles\quad on\quad a\quad straight\quad line={ 180 }^{ \circ })$$
$$\implies\quad x={ 180 }^{ \circ }-{ 100 }^{ \circ }={ 80 }^{ \circ }$$
$$ In\triangle BOP:$$
$$ z=x+y\quad (exterior\quad angle\quad of\quad a\quad triangle=sum\quad of\quad two\quad opposite\quad interior\quad angles)$$
$$\implies\quad z=80+60={ 140 }^{ \circ }$$
$$ \therefore x={ 80 }^{ \circ }\quad ,\quad y={ 60 }^{ \circ }\quad ,z={ 140 }^{ \circ }$$
In $$\triangle XYZ$$ ______________is the base.
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$$XY$$
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$$YZ$$
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$$XZ$$
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None of the above
In $$\triangle ABC,\angle A={ x }^{ \circ },\angle B={ \left( 2x-15 \right) }^{ \circ}$$ and $$\angle C ={ \left( 3x+21 \right) }^{ \circ }$$. Find the value of $$x$$ and the measure of each angle of the triangle.
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$$x={ 27 }^{ \circ },\angle A={ 27 }^{ \circ },\angle B={ 41 }^{ \circ },\angle C={ 101 }^{ \circ }$$
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$$x={ 29 }^{ \circ },\angle A={ 29 }^{ \circ },\angle B={ 43 }^{ \circ },\angle C={ 108 }^{ \circ }$$
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$$x={ 27 }^{ \circ },\angle A={ 27 }^{ \circ },\angle B={ 41 }^{ \circ },\angle C={ 108 }^{ \circ }$$
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$$x={ 30 }^{ \circ },\angle A={ 30 }^{ \circ },\angle B={ 41 }^{ \circ },\angle C={ 109 }^{ \circ }$$
Explanation
Given, $$\angle A=x^\circ$$,
$$\angle B=(2x-15)^\circ$$ and
$$\angle C=(3x+21)^\circ$$.
We know, by angle sum property, the sum of angles of a triangle is $$180^\circ$$.
Then, $$\angle A+ \angle B+ \angle C=180^\circ$$
$$\implies$$
$$x^\circ+(2x-15)^\circ+ (3x+21)^\circ=180^\circ$$
$$\implies$$
$$6x^\circ+6^\circ=180^\circ$$
$$\implies$$
$$6x^\circ=180^\circ-6^\circ$$
$$\implies$$
$$6x^\circ=174^\circ$$
$$\implies$$
$$x^\circ=29^\circ$$.
Therefore,
$$\angle A=x^\circ=29^\circ$$,
$$\angle B=(2x-15)^\circ=2\times29^\circ-15^o=58^\circ-15^\circ=43^\circ$$
and
$$\angle C=(3x+21)^\circ=3\times29^\circ+21^\circ=87^\circ+21^\circ=108^\circ$$.
Hence, option $$B$$ is correct.
From the figure, find values of $$x$$ and $$y$$.
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$$25^{o},30^{o}$$
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$$35^{o},31^{o}$$
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$$50^{o},28^{o}$$
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$$45^{o},33^{o}$$
Explanation
In the given triangle,
by angle sum property,
$$40^o+95^o+x^o=180^o.......(i)$$
and $$x^o+y^o+102^o=180^o.......(ii)$$.
From $$(i)$$,
$$40^o+95^o+x^o=180^o$$
$$\implies$$
$$135^o+x^o=180^o$$
$$\implies$$
$$x^o=180^o-135^o$$
$$\implies$$
$$x^o=45^o.......(iii)$$.
Substitute $$(iii)$$ in $$(ii)$$,
$$45^o+y^o+102^o=180^o$$
$$\implies$$
$$y^o+147^o=180^o$$
$$\implies$$
$$y^o=180^o-147^o$$
$$\implies$$
$$y^o=33^o$$.
Therefore,
$$x^o=45^o$$ and
$$y^o=33^o$$.
Hence, option $$D$$ is correct.
Find the length of the hypotenuse in a right angled triangle if the sum of the squares of the sides making right angle is 169.
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$$15$$
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$$13$$
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$$5$$
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$$12$$
Explanation
According to the Pythagoras theorem, the
sum of the squares of the sides making the right angle is equal to the square of the third side (hypotenuse).
$$\therefore\ $$ Square of the hypotenuse $$=169$$
$$\Rightarrow$$ Hypotenuse $$=\sqrt{169}$$
$$=13\space\mathrm{units}$$
Hence, option B is correct.
In a right-angled triangle the lengths of base and perpendicular are 6 cm and 8 cm.What is the length of the hypotenuse?
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9 cm
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10 cm
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11 cm
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12 cm
Explanation
The sides of right angled triangle are $$6,8 cm$$
The hypotenuse is given as
$$a^2+b^2=c^2\\6^2+8^2=c^2\\36+64=c^2\\c^2= 100\\c=\sqrt {100}\\c=10cm$$
If the angles of triangle are in the ratio $$1:4:7$$, then the value of the largest angle is :
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$$135^{\circ}$$
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$$84^{\circ}$$
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$$105^{\circ}$$
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none of these
Mark the correct alternative of the following.
In figure, the value of x is?
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$$84$$
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$$74$$
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$$94$$
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$$57$$
Explanation
Exterior angle = sum of 2 opposite Interior Angles
$$123^0=39^0+x^0$$
$$x^0=123^0-39^0$$
$$\boxed{x^0=84^0}$$
A triangle having sides of different lengths is called
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an isosceles triangles
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an equilateral triangle
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a scalene triangle
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a right triangle
Explanation
A triangle having sides of different lengths is called a scalene triangle
The difference between the length of any two sides of a triangle is smaller than the length of third side.
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True
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False
Explanation
True.
The difference between the length of any two sides of a triangle is smaller than the length of third side.
In an isosceles triangle, one angle is $$70^{o}$$. The other two angles are of
(i) $$55^{o}$$ and $$55^{o}$$
(ii) $$70^{o}$$ and $$40^{o}$$
(iii) any measure
In the given option(s) which of the above statements(s) are true?
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(i) only
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(ii) only
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(iii) only
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(i) and (ii)
Explanation
Case II: Here, triangle $$ABC$$ is an isosceles triangle.
$$AB=AC$$ and vertex angle $$=70^{o}$$
$$\angle 1=\angle 2 \, \, \,...(1)\because AB=AC$$
Now,
$$\angle 1+\angle 2+\angle A=180^{o}$$
$$\Rightarrow 2(\angle 1)=180^{o}-70^{o}$$
$$\Rightarrow \angle 1=\dfrac{110^{o}}{2}=55^{o}$$
Therefore, $$\angle 1= \angle 2=55^{o}$$
In above Figure, Triangle ABC is an isosceles triangle
$$AB=AC$$, Base angle $$=\angle 2=70^{o}$$
$$\angle 1+\angle 2+\angle C=180^{o}$$
$$\Rightarrow \angle 1=180^{o}-70^{o}-70^{o}$$
$$\Rightarrow \angle 1=40^{o}$$
Therefore, $$\angle 1=40^{o}, \angle 2=70^{o}$$
Both statements are true.
It is possible to have a right-angled equilateral triangle.
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0%
True
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False
Explanation
False No, it is not possible to have a right-angled equilateral triangle.
In a triangle, one angle is of $$90^{o}$$. Then
(i) The other two angles are of $$45^{o}$$ each
(ii) In remaining two angles, one angle is $$90^{o}$$ and other is $$45^{o}$$
(iii) Remaining two angles are complementary
In the given option(s) which is always true?
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(i) only
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(ii) only
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(iii) only
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(i) and (ii)
Explanation
If one angle is right angle then the other both angles are a complementary angle in a triangle.
In $$\Delta PQR$$,
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$$PQ - QR > PR$$
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$$PQ + QR < PR$$
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$$PQ - QR < PR$$
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$$PQ + PR > QR$$
Explanation
In a triangle PQR,
$$\Rightarrow \,PQ + QR > PR$$
$$\Rightarrow \,QR + PR > PQ$$
$$\Rightarrow \,PR + PQ > QR$$
Since the third side of the triangle is less then the sum of any two sides of its.
Now,
$$\Rightarrow \,PQ - QR < PR$$
$$\Rightarrow \,QR - PR < PQ$$
$$\Rightarrow \,PR - PQ < QR$$
Since the length of third side of the triangle is always greater than the difference between two sides.
If an isosceles triangle, each of the base angles is $$40^o$$, then the triangle is
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Right-angled triangle
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Acute angled triangle
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Obtuse angled triangle
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Isosceles right-angled triangle
Explanation
Triangle $$ABC$$ is an isosceles triangle in which base angle is $$40^o$$ and $$AB = AC$$
$$\angle A + \angle B + \angle C = 180^o$$
$$\Longrightarrow \angle A = 180^o - 40^o - 40^o$$ ...Angle sum property
$$\Longrightarrow \angle A = 100^o$$
Therefore, triangle $$ABC$$ is an obtuse-angled traingle.
Which of Which of the following statements is not correct?
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The sum of any two sides of a triangle is greater than the third side
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A triangle can have all its angles acute
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A right-angled triangle cannot be equilateral
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Difference of any to sides of a triangle is greater than the third side
Explanation
Since the length of the third side of the triangle is always greater than the difference between two sides.
In Fig. $$BC = CA$$ and $$\angle A = 40$$. Then, $$\angle ACD$$ is equal to
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$$40^o$$
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$$80^o$$
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$$120^o$$
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$$60^o$$
Explanation
$$BC = CA$$
$$\Longrightarrow \angle A = \angle B = 40^o$$
Now,
$$\angle ACD = \angle A + \angle B$$ ...Exterior angle property
$$= 40^o + 40^o = 80^o$$
If the exterior angle of a triangle is $$130^{o}$$ and its interior opposite angles are equal, then measure of each interior opposite angle is
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$$55^{o}$$
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$$65^{o}$$
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$$50^{o}$$
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$$60^{o}$$
Explanation
Let $$y$$ and $$y$$ be the interior opposite angles.
$$\therefore 130^{o}=x+x$$ Exterior angle property
$$\Rightarrow 2x=130^{o}$$
$$\Rightarrow x=65^{o}$$
Therefore, each interior angle if of $$65^{o}$$.
In $$\Delta PQR$$, if $$\angle P=60^{o}$$, and $$\angle Q=40^{o}$$, then the exterior angle formed by producing $$QR$$ is equal to
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$$60^{o}$$
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$$120^{o}$$
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$$100^{o}$$
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$$80^{o}$$
Explanation
Let $$QR$$ is extended to point $$S$$.
(Refer image)
In triangle $$PQR$$
$$\angle PRS=\angle P+\angle Q$$
$$\Rightarrow \angle PRS=60^{o}+40^{o}$$
$$\Rightarrow \angle PRS=100^{o}$$
The measures of $$\angle x$$ and $$\angle y$$ in figure are respectively
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$$30^{o},60^{o}$$
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$$40^{o}, 40^{o}$$
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$$70^{o}, 70^{o}$$
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$$70^{o}, 60^{o}$$
Explanation
$$120^0$$ is exterior of $$\angle R$$
$$\therefore \angle P+\angle Q=120^0$$
$$\Rightarrow 120^{o}=x+50^{o}$$
$$\Rightarrow x=120^{o}-50^{o}$$
$$\Rightarrow x=70^{o}$$
Now in $$\Delta PQR$$
$$x+y+50^{o}=180^{o}$$
$$\Rightarrow 70^{o}+y+50^{o}=180^{o}$$
$$\Rightarrow y=180^{o}-70^{o}-50^{o}$$
$$\Rightarrow y=60^{o}$$
Therefore, $$x=60^{o}$$ and $$y=70^{o}$$.
Hence, option $$D$$ is correct.
If we join a vertex to a point on opposite side, which divides that side in the ratio $$1:1$$, then what is the special name of that line segment?
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Median
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Angle bisector
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Altitude
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Hypotenuse
Explanation
Median from an vertex divides the opposite sides into ratio of $$1:1$$
So, special name of that segment is $$\text{Median}$$
Hence, option $$A$$ is correct.
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