The Triangle and Its Properties Test

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The Triangle and Its Properties Test - 23

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Question 1
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A ladder $$25m$$ long is leaning against a wall that is perpendicular to the level ground. The bottom of the ladder is $$7m$$ away from the base of the wall. If the top of the ladder slips down $$4m$$, how much will the bottom of the ladder slip?

A
$$7m$$

B
$$8m$$

C
$$10m$$

D
$$15m$$

Solution

R.E.F first image

In $$ \Delta ABC ,$$

$$ AC^{2} = AB^{2}-BC^{2} $$ {Using Pythagorus theorem}

$$ \Rightarrow AC = \sqrt{25^{2}-7^{2}} $$

$$ \Rightarrow \boxed{AC = 24\,m} $$

As after sliping length of ladder remains same $$ \Rightarrow \boxed{DE = 25\,cm} \ and\ AD = 4m$$ {REF second Image}

Let say $$x$$ be the length of how much the ladder slips.

New Height $$= DC = AC-AD = 20\,m; \ EC = (x+7)m $$

In $$ \Delta DEC $$,

$$ DE^{2} = DC^{2}+EC^{2} $$

$$ \Rightarrow 25^{2} = 20^{2}+(x+7)^{2}\Rightarrow (x+7)^{2} = 225 $$

$$ \Rightarrow x+7 = 15 \ \ \ (\because (x+7)> 0 $$ , We can't take negative value $$) \Rightarrow \boxed{x = 8\,m} $$
Question 2
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In the given figure, side $$BC$$ of $$\triangle ABC$$ is produced to $$D$$. Find $$\angle A$$

A
$$80^{\circ}$$

B
$$60^{\circ}$$

C
$$ 30^{\circ}$$

D
None of these

Solution

An exterior angle of a triangle is equal to the sum of the opposite interior angles.
So by this property:
$$\angle A+\angle B=\angle ACD$$
$$\Rightarrow\angle A +40^{\circ}=120^\circ$$
$$\Rightarrow\angle A=120^\circ-40^\circ$$
$$\Rightarrow\angle A=80^\circ$$
Question 3
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Which of the following set of measurements will form a triangle

A
$$11cm,4cm,6cm$$

B
$$13cm,14cm,25cm$$

C
$$8cm,4cm,3cm$$

D
$$5cm.16cm.5cm$$

Solution

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. $$a+b> c,$$ $$b+c> a$$ and $$c+a> b$$ 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
Question 4
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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.

A
$$40$$ m

B
$$30$$ m

C
$$50$$ m

D
$$10$$ m

Solution

Let $$AB=$$ length of ladder, $$AC=$$ height of window
In right angled $$ \triangle ABC$$, we have
$$ AB ^{ 2 }={ AC }^{ 2 }+{ BC }^{ 2 }$$
$$\Rightarrow { (34) }^{ 2 }={ (16) }^{ 2 }+{ BC }^{ 2 }\quad $$
$$\Rightarrow { BC }^{ 2 }={ (34) }^{ 2 }-{ (16) }^{ 2 }=900$$
$$\Rightarrow BC=\sqrt { 900 } =30 m$$
Question 5
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Find the values of x and y in the following figures

A
$$ x=80^0 ; y=120^0$$

B
$$ x=50^0 ; y=30^0$$

C
$$ x=50^0 ; y=130^0$$

D
$$ x=20^0 ; y=10^0$$

Solution

$$\angle CAB = 80^o$$ [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

$$x = 50$$

$$y = 130$$
Question 6
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Find the measure of the angle $$x$$ in the given figure.

A
$${ 50 }^\circ$$

B
$${ 70 }^\circ$$

C
$${ 60 }^\circ$$

D
$${ 30 }^\circ$$

Solution

$$\textbf{Step 1: Find the relation between interior and exterior angles of triangle.}$$
$$ x=\angle{EFD}+\angle{FED}$$
$$\Rightarrow \angle{x}={28^\circ}+{42^\circ}$$ $$\quad \textbf{[Exterior angle is equal to sum of interior angles of a triangle.]}$$
$$\Rightarrow \angle{x}={70^\circ}$$

$$\textbf{Hence, the measure of the angle x is} $$ $$\bf{70^\circ.}$$
Question 7
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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

A
Pairs Q and R show right angle triangle

B
Pairs P and Q show right angle triangle

C
Pairs P and S show right angle triangle

D
Pairs P, Q and S show right angle triangle

Solution

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.
Question 8
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From the given figure, find the values of $$x$$ and $$y$$ respectively.

A
$${ 47 }^{ \circ },{ 66 }^{ \circ }$$

B
$${ 66 }^{ \circ },{ 48 }^{ \circ}$$

C
$${ 68 }^{ \circ },{ 47 }^{ \circ }$$

D
$${ 47 }^{ \circ },{ 68 }^{ \circ }$$

Solution

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)
$$y+{ 66 }^{ \circ }+{ 66 }^{ \circ }={ 180 }^{ \circ }\quad $$
$$y={ 180 }^{ \circ }-\left( { 66 }^{ \circ }+{ 66 }^{ \circ } \right)$$
$$ \Rightarrow y={ 48 }^{ \circ }$$
Question 9
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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.

A
$$10m$$

B
$$15m$$

C
$$13m$$

D
$$18m$$

Solution

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 $$=(13+5)m$$
$$ = 18m$$
Question 10
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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.

A
$$25m$$

B
$$5m$$

C
$$23m$$

D
$$16m$$

Solution

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 $$=(17+8)m = 25m$$