Triangles Test

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Triangles Test - 40

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Weekly Quiz Competition

Question 1
1 / -0

Which of the following sets of measurements can be used to construct a triangle?

A
$$4\text{ cm}, 5\text{ cm}, 6\text{ cm}$$

B
$$4\text{ cm}, 3\text{ cm}, 8\text{ cm}$$

C
$$5\text{ cm}, 6\text{ cm}, 12\text{ cm}$$

D
$$6\text{ cm}, 3\text{ cm}, 10\text{ cm}$$

Solution

Because the sum of the length of any $$2$$ sides of the triangle should be greater than the third side, which is only satisfied by option $$A.$$
$$(4+5)>6$$
$$(4+6)>5$$
$$(5+6)>4$$
Question 2
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In the figure above, $$ABCD$$ is a rectangle. If $$AD = 6$$, which of the following could be the length of $$\overline {AC}$$?

A
$$2$$

B
$$4$$

C
$$5$$

D
$$6$$

E
$$7$$

Solution

In triangle $$ACD,$$ angle $$ADC$$ is greater than the other two angles

By sine rule , we get $$\dfrac{AC}{AD}=\dfrac{\angle(ADC)}{\angle(DCA)}$$

We know that $$\angle(ADC) > \angle(DCA)$$

Therefore $$AC>AD=6$$

Therefore option $$E$$ is correct
Question 3
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Mark the triplet that can be the lengths of the sides of a triangle.

A
$$2, 3, 5$$

B
$$1, 4, 2$$

C
$$7, 4, 4$$

D
$$5, 6, 12$$

E
$$9, 20, 8$$

Solution

for triplet to be the length of triangle, it must satisfy triangle property.
sum of any two sides must be greater than third side
So, $$7,4,4$$ is correct answer.
If we consider remaining options, observe that sum of two sides is less than third side, so they cannot form triangle.
Question 4
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In $$\displaystyle \Delta XYZ$$, side $$\displaystyle \overline { XZ } $$ is longer than side $$\displaystyle \overline { YZ } $$, which of the following statements is true?

A
The measure of $$\displaystyle \angle X$$ is always less than the measure of $$\displaystyle \angle Y$$.

B
The measure of $$\displaystyle \angle X$$ is always equal than the measure of $$\displaystyle \angle Y$$.

C
The measure of $$\displaystyle \angle X$$ is always greater than the measure of $$\displaystyle \angle Y$$.

D
The measure of $$\displaystyle \angle X$$ is sometimes less than the measure of $$\displaystyle \angle Y$$ and sometimes equal to the measure of $$\displaystyle \angle Y$$.

E
The measure of $$\displaystyle \angle X$$ is sometimes greater than the measure of $$\displaystyle \angle Y$$ and sometimes equal to the measure of $$\displaystyle \angle Y$$.

Solution

Angles opposite to longer sides are larger.
Here $$XZ>YZ$$
Angle opposite to $$XZ$$ is $$\angle Y$$ and to $$YZ$$ is $$X$$
$$\therefore \angle Y>\angle X$$
So option $$A$$ is correct.
Question 5
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$$\triangle PQR$$ is right angled at $$Q$$, $$PR=5cm$$ and $$QR=4cm$$. If the lengths of sides of another triangle $$ABC$$ are $$3cm$$, $$4cm$$, $$5cm$$, then which one of the following is correct?

A
Area of $$\triangle PQR$$ is double that of $$\triangle ABC$$.

B
Area of $$\triangle ABC$$ is double that of $$\triangle PQR$$.

C
$$\sqrt { B } =\cfrac { \sqrt { Q } }{ 2 } $$

D
Both triangles are congruent.

Solution
Question 6
1 / -0

In a triangle with sides of $$7$$ and $$9$$, the third side must be

A
more than $$16$$

B
between $$7$$ and $$9$$

C
between $$2$$ and $$16$$

D
between $$7$$ and $$16$$

E
between $$9$$ and $$16$$

Solution

Given that two sides of triangle are $$7,9$$
Let the third side be $$x$$
We have $$7+9>x$$ and $$x+9>7$$ and $$x+7>9$$ and $$x>0$$
We get $$x<16$$ and $$x>2$$
Therefore the third side will lie between $$2$$ and $$16$$.
Question 7
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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 8
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In $$\triangle ABC$$, the bisector of $$\angle {A}$$ intersects $$\overline { BC } $$ at a point $$D$$. Then:

A
$$BD\times AC=BC\times AB$$

B
$$BD\times AB=DC\times AC$$

C
$$AC\times AB=DC\times BC$$

D
$$BD\times AC=DC\times AB$$

Solution

In $$\triangle ABC$$, the bisector of $$\angle A$$ intersects $$\overline BC$$ at $$D$$.
$$\therefore$$ $$AD$$ bisects $$BC$$
$$\therefore BD=DC$$ ........ $$(i)$$
In triangles, $$ABD$$ and $$ADC$$
$$BD=CD$$ ......... From $$(i)$$
$$\angle BAD=\angle DAC$$ ....... (Given)
$$AD=AD$$ ........... (Common side)
$$\therefore$$ $$\triangle ABD\cong \triangle ADC$$ ........ [By S.A.S criterion]
$$\implies AB=AC$$
Thus, $$BD\times AC=DC\times AB$$
Question 9
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The two triangles in the figure are congruent by the congruence theorem. Here, it is given $$OQ=OR$$. Which of the following condition, along with the given condition, is sufficient to prove that the two triangles are congruent to each other?

A
$$\angle P=\angle S$$

B
$$\angle Q=\angle R$$

C
$$OP=OS$$

D
$$PQ=SR$$

Solution

Given $$OQ=OR$$ and $$\triangle POQ\cong \triangle ROS$$
We know that, congruent parts of congruent triangles are congruent
$$\angle POQ\cong \angle ROS$$ (vertically opposite angles)
If $$OP=OS$$ then by using the (SAS) congruent, we can conclude the congruency of two triangles.
Hence, option C is sufficient to prove the congruency.
Question 10
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$$\triangle FGC$$ is an isosceles triangle, which of the following method will prove $$\triangle ABC$$ congruent to $$\triangle DEF$$ ?

A
SSS

B
RHS

C
AAS

D
SAS

Solution

In $$\triangle ABC$$ and $$\triangle DEF$$

Given,
$$AB=DE$$
$$\angle ABC=\angle DEF=90^\circ$$

$$\triangle FCG$$ is isoceles, $$GF=GC$$ $$\implies FD=CA$$ ...... (as $$AB||DE$$, $$GD$$ should be equal to $$GA$$)

$$AC=FD$$

Hence, $$\triangle ABC \cong \triangle DEF$$ by $$RHS \ postulate$$