Triangles Test

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Triangles Test - 30

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Weekly Quiz Competition

Question 1
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In the given figure, $$\angle$$B = $$90^{\circ}$$, $$XY|| BC$$, $$AB = 12\ cm$$, $$AY = 8\ cm$$ and $$AX : XB = 1 : 2$$. Find the length of $$AC$$ and $$BC$$.

A
$$AC = 12\ cm$$ and $$BC = 10\sqrt3 \ cm $$.

B

$$AC = 24\ cm$$ and $$BC = 12\sqrt3\ cm $$.

C
$$AC = 19 \ cm$$ and $$BC = 10\sqrt3\ cm $$.

D
$$AC = 34 \ cm$$ and $$BC = 14\sqrt3\ cm$$.

Solution

Given: $$XY \parallel BC$$, $$\angle B = 90^{\circ}$$, $$AX : XB = 1: 2$$
And, $$AB=12\ cm$$, $$AY=8\ cm$$

Now,
$$\dfrac{AX}{XB} = \dfrac{1}{2}$$
$$\Rightarrow \dfrac{AX}{AB - AX} = \dfrac{1}{2}$$
$$\Rightarrow2 AX = AB - AX$$
$$\Rightarrow 3 AX = AB$$
$$\Rightarrow 3 AX = 12\ cm$$
$$\Rightarrow AX = 4\ cm$$

Now, In $$\triangle AXY$$ and $$\triangle ABC$$
$$\angle XAY = \angle BAC$$ (Common)
$$\angle AXY = \angle ABC$$ (Corresponding angles)
$$\angle AYX = \angle ACB$$ (Corresponding angles)
Thus, $$\triangle AXY \sim \triangle ABC$$ ($$AAA$$ rule)

Hence, $$\dfrac{AX}{AB} = \dfrac{AY}{AC}$$
$$\Rightarrow \dfrac{4}{12} = \dfrac{8}{AC}$$
$$\Rightarrow AC = \dfrac{8\times 12}{4}$$
$$\Rightarrow AC = 24$$

Hence, $$AC= 24$$ cm

Now, applying Pythagoras theorem in $$\triangle ABC$$,
$$AC^2 = AB^2 + BC^2$$
$$\Rightarrow 24^2 = 12^2 + BC^2$$
$$\Rightarrow BC^2 = 24^2 - 12^2$$
$$\Rightarrow BC^2 = 432$$
$$\Rightarrow BC = \sqrt{432}$$
$$\Rightarrow BC = 12 \sqrt{3}$$

Hence, $$BC=12\sqrt{3}\ cm$$
Question 2
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Triangle $$ABC$$ is right -angled at $$C$$. Find $$BC$$, If $$AB = 9 \ cm$$ and $$AC = 1\ cm$$.
In each case, answer correct to two place of decimal.

A
$$6\sqrt {7} \ cm$$

B
$$6\sqrt {5} \ cm$$

C
$$4\sqrt {5} \ cm$$

D
$$4\sqrt {3} \ cm$$

Solution

$$\Delta ABC$$ is right angled at $$C$$.
$$AB=9\ cm$$
$$AC=1\ cm$$

Applying pythagoras theorem,

$$AB^2 = BC^2 + AC^2$$

$$\Rightarrow 9^2 = BC^2 + 1^2$$

$$\Rightarrow BC^2 = 80$$

$$\Rightarrow BC =\sqrt{80} = 4\sqrt{5} cm$$
Question 3
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Find AE, if BD = 4.1 cm

A
8.2 cm

B
0.82 cm

C
82 cm

D
8.1 cm

Solution

In $$\triangle ACE$$ and $$\triangle BCD$$,
$$\angle ACE = \angle BCD $$ (Common angle)
$$\angle CBD =\angle CAE$$ (Corresponding angles, since BD || AE)
$$\angle CDB =\angle CEA$$ (Corresponding angles, since BD || AE)
Hence, $$\triangle ACE \sim \triangle BCD$$, BY AAA
Thus, $$\dfrac{AE}{BD} = \dfrac{CE}{CD}$$
$$\dfrac{AE} {4.1} = \dfrac{2 CD}{CD} $$ (Given, CD = DE)
$$AE = 4.1 \times 2$$
$$AE = 8.2 cm $$
Question 4
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Find $$DF,$$ if $$CG = 11\ cm$$

A
$$5.6\ cm$$

B
$$5.5\ cm$$

C
$$0.55\ cm$$

D
$$6\ cm$$

Solution

In $$\triangle CEG$$ and $$\triangle DEF$$,
$$\angle CEG = \angle DEF $$ (Common angle)
$$\angle ECG =\angle EDF$$ (Corresponding angles, since $$BD\ |\ |\ AE$$)
$$\angle EGC =\angle EFD$$ (Corresponding angles, since $$BD\ |\ |\ AE$$)

So, by $$AAA$$ criteria of similarity,
$$\triangle CEG \sim \triangle DEF$$

$$\Rightarrow \dfrac{DF}{CG} = \dfrac{DE}{CE}$$

$$\Rightarrow \dfrac{DF}{11} = \dfrac{DE}{DE + CD} $$ (Given, $$CD = DE$$)

$$\Rightarrow \dfrac{DE}{11} = \dfrac{1}{2} $$

$$\Rightarrow DE = 5.5\ cm$$
Question 5
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Triangle ABC is right -angled at C. Find $$AC, $$ If $$AB = 2.2 \ cm\,\, $$and$$\,\, BC = 1.8 \ cm$$.
In each case, answer correct to two place of decimal.

A
$$\sqrt {1. 6} \ cm= 1.17 \ cm$$

B
$$\sqrt {1.6} \ cm= 1.26 \ cm$$

C
$$\sqrt {1.6} \ cm= 1.33 \ cm$$

D
$$\sqrt {1.6} \ cm= 1.50 \ cm$$

Solution

ABC is right angled at C.

$$\therefore AB^2 = BC^2 + AC^2$$

Let's put the given values,

$$(2.2)^2 = (1.8)^2 + AC^2$$

$$4.84 = 3.24 + AC^2$$

$$AC = \sqrt{1.6} = 1.26 \quad cm$$
Question 6
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Find GE, if FE = 4 cm

A
80 cm

B
8 cm

C
0.8 cm

D
8 m

Solution

In $$\triangle CEG$$ and $$\triangle DEF$$,
$$\angle CEG = \angle DEF $$ (Common angle)
$$\angle ECG =\angle EDF$$ (Corresponding angles, since BD || AE)
$$\angle EGC =\angle EFD$$ (Corresponding angles, since BD || AE)
Hence, $$\triangle CEG \sim \triangle DEF$$, BY AAA
Thus, $$\dfrac{GE}{FE} = \dfrac{CE}{DE}$$
$$\dfrac{GE}{FE} = \dfrac{CD + DE}{DE} $$ (Given, CD = DE)
$$\dfrac{GE}{4} = 2 $$
$$GE = 8 cm$$
Question 7
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If the measures of sides of a triangle are $$(x^2-1) \ cm, \ (x^2 +1) \ cm$$, and $$2x \ cm$$, then the triangle will be:

A
right angled

B
obtuse angled

C
equilateral

D
isosceles

Solution

Given that,
A triangle has three sides as $$(x^2-1) \ cm, \ (x^2 +1) \ cm$$, and $$2x \ cm$$
To find out,
The type of triangle.

We can see that,
$$(x^2 -1)^2+(2x)^2 = x^4+2x+1\quad \quad [\because \ (a-b)^2=a^2-2ab+b^2]$$

$$ (x^2+1)^2 =x^2+2x+1\quad \quad [\because \ (a+b)^2=a^2+2ab+b^2]$$

Here, the square of the largest side of the triangle is equal to the sum of the squares of the other two sides.

Hence, by the converse of Pythagoras theorem, we can say that the given triangle is a right angled triangle.
Question 8
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Two isosceles triangles have equal vertical angles and their areas are in the ratio $$16:25$$. Find the ratio of their corresponding heights.

A
$$4:5$$

B
$$25:16$$

C
$$5:4$$

D
$$16:25$$

Solution

$$\triangle ABC$$ and $$\triangle DEF$$ be the given triangles in which $$AB=AC, DE=DF$$, $$\angle A=\angle D$$
and $$\cfrac{Area\quad (\triangle ABC)}{Area\quad (\triangle DEF)}=\cfrac{16}{25}$$
Draw $$AL\bot BC$$ and $$DM\bot EF$$
Now, $$\cfrac{AB}{BC}=1$$ and $$\cfrac{DE}{DF}=1$$ ($$\because \quad AB=AC;\quad DE=DF$$)
$$\Rightarrow \cfrac{AB}{AC}=\cfrac{DE}{DF}$$,
$$\therefore$$ $$\ln \triangle ABC$$ and $$\triangle DEF$$, we have
$$\cfrac{AB}{DE}=\cfrac{AC}{DF}$$ and $$\angle A=\angle D$$
$$\Rightarrow$$ $$\triangle ABC\sim \triangle DEF$$ [By SAS similarity axiom)
But, the ratio of the areas of two similar $$\triangle s$$ is the same as the ratio of the squares of their corresponding heights.
$$\cfrac{Area\quad (\triangle ABC)}{Area\quad (\triangle DEF)}=\cfrac {{AL}^{2}}{{DM}^{2}}$$
$$\Rightarrow$$ $$\cfrac{16}{25}={ \left( \cfrac {AL}{DM} \right) }^{ 2 }$$
$$\Rightarrow$$ $$\cfrac{4}{5}$$
$$\therefore$$ $$AL:DM=4:5$$, i.e., the ratio of their corresponding heights$$=4:5$$
Question 9
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In figure, $$\angle BAC={90}^{o}$$ and $$AD\bot BC$$. Then

A
$$BD.CD={BC}^{2}$$

B
$$AB.AC={BC}^{2}$$

C
$$BD.CD={AD}^{2}$$

D
$$AB.AC={AD}^{2}$$

Solution

Given, In $$\triangle ABC $$, $$\angle A = 90 $$ and $$ AD \perp BC $$
In $$\triangle ABC$$,
$$\angle BAC + \angle ABC + \angle ACB = 180 $$
$$90 + \angle ABC + \angle ACB = 180 $$
$$\angle ABC + \angle ACB = 90 $$ --------(I)

In $$\triangle CAD$$,
$$\angle CAD + \angle ACD + \angle ADC = 180 $$
$$\angle CAD + \angle ACD + 90 = 180 $$
$$\angle CAD + \angle ACD = 90 $$ ---------(II)
Equating (I) and (II),
$$\angle ABC + \angle ACB = \angle CAD + \angle ACD $$
$$\angle ABC = \angle CAD $$ --------(III)
Similarly, $$\angle ACB = \angle BAD $$ --------(IV)
Now, In $$\triangle$$s, ABD and CAD
$$\angle ABC = \angle CAD $$ -------(From III)
$$\angle BAD = \angle ACB $$ --------(From IV)
$$\angle ADB = \angle ADC$$ (Each $$90^{\circ})$$
Thus, $$\triangle ABD \sim \triangle CAD$$ (AAA rule)
Thus, $$\dfrac{AD}{CD} = \dfrac{BD}{AD}$$ (Sides of similar triangles are in proportion)

$$AD^2 = BD \times CD $$
Question 10
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$$E$$ and $$F$$ are respectively, the mid points of the sides $$AB$$ and $$AC$$ of $$\Delta ABC$$ and the area of the quadrilateral $$BEFC$$ is $$k$$ times the area of $$\Delta ABC$$. The value of $$k$$ is :

A
$$\displaystyle\frac{1}{2}$$

B
$$3$$

C
$$\displaystyle\frac{3}{4}$$

D
$$4$$

Solution

$$\Delta ABC \sim \Delta AEF$$
$$\displaystyle\frac{A(\Delta AEF)}{A(\Delta ABC)}=\left[\frac{AE}{AB}\right]^2=\frac{1}{4}$$
$$\Rightarrow A( \Box BEFC)=A(\Delta ABC)- A(\Delta AEF)$$
$$A( \Box BEFC)$$ $$\displaystyle=A(\Delta ABC)-\frac{1}{4} A(\Delta ABC)$$
$$A(\Box BEFC)$$$$=\displaystyle\frac{3}{4} A(\Delta ABC)$$.