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CBSE Questions for Class 9 Maths Triangles Quiz 8 - MCQExams.com
CBSE
Class 9 Maths
Triangles
Quiz 8
If $$a, b, c$$ are sides of a triangle, then
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$$\sqrt a + \sqrt b > \sqrt c $$
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$$\left| {\sqrt a - \sqrt b } \right| > \sqrt c $$
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$$\sqrt a + \sqrt b < \sqrt c $$
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None of these
Explanation
Given: Sides of the triangle $$a, b$$ and $$c$$.
Triangle inequality theorem: According to this theorem,
[The sum if lengths of any two sides of triangle is greater than the length of its third side.]
i.e. $$a + b > c$$
$$b + c > a$$
$$a + c > b$$
For example, $$a = 3cm, b = 4 cm, c = 5 cm$$ then
$$3 + 4 > 5 ; \, 7 > 5$$
$$4 + 5 > 3 ; \, 9 > 5$$
$$3 + 5 > 4 ; \, 6 > 4$$
$$\sqrt{a} + \sqrt{b} > \sqrt{c}$$
$$\sqrt{3} + \sqrt{4} > \sqrt{5}$$
$$1.73 + 2 > 2.24$$
$$3.73 > 2.24$$
It means the correct option is $$(A)$$. i.e.
$$\sqrt{a} + \sqrt{b} > \sqrt{c}$$
Two sides of a triangle are 7 cm and 25 cm, then the number of all possible integral values of third side of the triangle is
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11
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14
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12
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13
Which of the following pairs of triangle are congruent ?
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$$\triangle ABC$$ and $$\triangle DEF$$ in which : BC = EF, AC = DF and $$\angle C=\angle F$$.
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$$\triangle ABC$$ and $$\triangle PQR$$ in which : AB = PQ, BC = QR and $$\angle C=\angle R$$
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$$\triangle ABC$$ and $$\triangle LMN$$ in which : $$\angle A=\angle L={ 90 }^{ \circ },AB=LM,\angle C={ 40 }^{ \circ }$$ and $$\angle M={ 50 }^{ \circ }$$
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$$\triangle ABC$$ and $$\triangle DEF$$ in which : $$\angle B=\angle E={ 90 }^{ \circ }$$ and AC = DF
Take any pointy $$O$$ in the interior of a triangle $$PQR$$. Is
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$$OP+OQ>PQ?$$
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$$OQ+OR>QR?$$
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$$OR+OP>RP?$$
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All of the above.
If the sides of triangle are 4 cm and 6 cm then the third side cannot be ... cm.
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8
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10
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5
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8.5
If $$\Delta ABC\cong \Delta DEF$$ write the part of $$\Delta ABC$$ that correspond to
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DE
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$$\angle E$$
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DF
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EF
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$$\angle F$$
AM is a median of a triangle ABC. Is $$AB+BC+CA>2 AM$$?
(Consider the sides of triangles $$\triangle ABM$$ and $$\triangle AMC$$.)
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True
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False
In $$\Delta ABC$$ $$AB<AC$$ Then...... holds good.
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$$\angle A<\angle B$$
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$$\angle B<\angle C$$
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$$\angle C<\angle A$$
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$$\angle C<\angle B$$
In a triangle ABC, AB=AC, BA is extended upto D, in such a manner that AC=AD is a circular measure of <BCD:
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$$\dfrac{\pi}{6}$$
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$$\dfrac{\pi}{3}$$
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$$\dfrac{2 \pi}{3}$$
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$$\dfrac{\pi}{2}$$
if ABC and DEF are congruent triangles such that $$\angle A={ 47 }^{ \circ }\quad and\quad \angle E={ 83 }^{ \circ },\quad then\quad \angle C=$$
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$${ 100 }^{ \circ }$$
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$${ 50 }^{ \circ }$$
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$${ 90 }^{ \circ }$$
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NONE OF THESE
In $$\triangle ABC$$, $$P$$ is a point on the side $$BC$$ then which of the following is correct.
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$$ AP<AB+BP$$
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$$ AP<AC+PC$$
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$$ AP<AB+BC$$
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$$ AP> AB+BC$$
The sum of any two sides of a triangle is always
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equal to the third side
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less than
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grater than or equal to the 3rd side
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grater than
Given two right angled triangle ABC and PQR, such that $$<A=20^o, <Q=20^o$$ and $$AC=QP$$. Write the correspondence if triangles are congruent.
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$$\Delta ABC\cong \Delta PQR$$
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$$\Delta ABC\cong \Delta PRO$$
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$$\Delta ABC\cong \Delta ROP$$
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$$\Delta ABC\cong \Delta QRP$$
If one of the two equal sides of a triangle is 5 cm long. Then what can be the measure of the third side?
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8 cm
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14 cm
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12 cm
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16 cm
The sides of a triangle are three consecutive natural numbers and its largest the smallest one then the sides of triangle are
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3, 4, 5
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5, 6, 7
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4, 5, 6
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6, 7, 8
$$ \triangle APQ$$ is such that perpendicular bisector of side $$PQ$$ bisect it at $$M$$ and passes from vertex $$A$$. Then by using which of the following congruence rule it can be proved that $$ \triangle \mathrm{AMP} \cong \triangle \mathrm{AMQ} $$.
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$$RHS$$
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$$SSS$$
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$$SAS$$
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$$AAA$$
Given that $$\Delta ABC = \Delta FDE$$ and $$AB = 5 cm, \angle B = 40^{0}$$ and $$\angle A = 80^{0}$$ then which of the following relation is true?
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$$DF=5cm,\angle F=60^{0}$$
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$$DF=5cm,\angle E=60^{0}$$
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$$DE=5cm,\angle E=60^{0}$$
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$$DE=5cm,\angle D=40^{0}$$
In $$\triangle PQR$$, if $$\angle R\displaystyle>\angle Q$$, then
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$$QR>PR$$
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$$PQ>PR$$
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$$PQ
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$$QR
Explanation
In $$\triangle PQR$$, $$\angle R > \angle Q$$
Hence, $$PQ > PR$$ (Sides opposite larger angles is large)
option B is correct
In a triangle, the difference of any two sides is ____ than the third side.
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Smaller
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Equal
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Greater
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None of these
Explanation
In a triangle, the difference of any two sides is smaller than the third side.
ABC is an isosceles triangle with AB $$= $$AC and D is a point on BC such that $$AD \perp BC$$ (Fig. 7.13). To prove that $$\angle BAD = \angle CAD,$$ a student proceeded as follows:
$$\Delta ABD$$ and $$ \Delta ACD,$$
AB $$=$$ AC (Given)
$$\angle B = \angle C$$ (because AB $$=$$ AC)
and $$\angle ADB = \angle ADC$$
Therefore, $$\Delta ABD \cong \Delta ACD (AAS)$$
So, $$\angle BAD = \angle CAD (CPCT)$$
What is the defect in the above arguments?
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It is defective to use $$\angle ABD = \angle ACD$$ for proving this result.
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It is defective to use $$\angle ADB = \angle ADC$$ for proving this result.
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It is defective to use $$\angle BAD = \angle DCA$$ for proving this result.
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Cannot be determined
Explanation
$$\bigtriangleup ABD$$ and $$\bigtriangleup ACD$$
AB=AC (given )
Then $$\angle ABD=\angle ACD$$ ( because AB=AC )
and $$\angle ADB=\angle ADC=90$$( because AD
⊥BC )
$$\therefore \bigtriangleup ABD=\bigtriangleup ACD$$
$$\angle BAD=\angle CAD$$
It is defective to use $$\angle ABD=\angle ACD$$ for proving this result
Given $$\Delta OAP \cong \Delta OBP$$ in figure, the criteria by which the triangles are congruent is:
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$$SAS$$
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$$SSS$$
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$$RHS$$
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$$ASA$$
Explanation
In $$\triangle OAP$$ and $$\triangle OBP$$
$$OA=OB$$ (given)
$$\angle AOP=\angle BOP$$ (given)
$$OP=OP$$ (common side)
$$\therefore \triangle OAP\cong \triangle OBP$$ by $$SAS$$ congruent rule as two corresponding sides and the included angles of the triangles are equal.
In $$\Delta ABC$$, if $$\angle A = 50^{\circ}$$ and $$\angle B = 60^{\circ}$$, then the greatest side is :
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AB
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BC
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AC
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Cannot say
Explanation
Using angle sum property of triangle
$$\angle A+\angle B+\angle C={ 180 }^{ \circ }\\ \Rightarrow { 50 }^{ \circ }+{ 60 }^{ \circ }+\angle C={ 180 }^{ \circ }\\ \Rightarrow \angle C={ 180 }^{ \circ }-{ 110 }^{ \circ }={ 70 }^{ \circ }$$
Side opposite to the largest angle is the longest side.
Here $$\angle C$$ is largest and side opposite to it is $$AB$$
So $$AB$$ is the longest side.
The construction of a triangle ABC, given that BC = 3 cm is possible when difference of AB and AC is equal to :
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3.2 cm
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3.1 cm
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3 cm
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2.8 cm
Explanation
Let the length of $$AB$$ be $$x$$ and $$AC$$ be $$y$$
A triangle can be formed if the sum of any two sides is greater then the third
$$\Rightarrow BC+AC>AB\\ \Rightarrow 3+AC>AB\\ \Rightarrow 3>AB-AC\\ \Rightarrow AB-AC<3$$
So only option $$D$$is correct.
In $$\Delta PQR,$$ if $$\angle R > \angle Q,$$ then:
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$$QR > PR$$
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$$PQ > PR$$
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$$PQ < PR$$
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$$QR < PR$$
Explanation
We know that the side opposite to the greater angle is greater.
Given: $$\angle R> \angle Q$$
The side opposite to $$\angle R=PQ$$ and the side opposite to $$\angle Q=PR$$.
Thus, $$PQ>PR$$.
Hence, option $$B$$ is correct.
In $$\Delta ABC$$ and $$\Delta DEF$$, AB = DF and $$\angle A = \angle D$$. The two triangles will be congruent by SAS axiom if :
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BC = EF
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AC = DE
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BC = DE
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AC = EF
Explanation
For triangle to be congruent $$SAS$$(Side angle Side) axiom we need to have two equal sides and the included angle same.
In $$\Delta ABC$$ and $$\Delta DFE$$
We have $$AB=DF$$
and $$\angle A=\angle D$$
But $$\angle A$$ is formed by the two sides AB and AC of $$\Delta ABC$$
But $$\angle D$$ is formed by the two sides DE and DF of $$\Delta DFE$$
So we need $$AC=DE$$ for the two triangles to be congruent.
In triangles ABC and DEF, AB $$=$$ FD and $$\angle A = \angle D$$. The two triangles will be congruent by
SAS axiom if :
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BC $$=$$ EF
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AC $$=$$ DE
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AC $$=$$ EF
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BC $$=$$ DE
Explanation
For triangle to be congruent by $$SAS$$ axiom two of the sides and the included angle of the triangles must be equal.
Given $$AB=DF$$ and $$\angle A=\angle D$$
No for $$\triangle ABC\cong \triangle DFE$$ by $$SAS$$ axiom we need $$AC=DE$$
So option $$B$$ is correct.
In $$\Delta ABC, \angle B = 30^{\circ}, \angle C = 80^{\circ}$$ and $$\angle A = 70^{\circ}$$ then,
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$$AB > BC < AC$$
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$$AB < BC > AC$$
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$$AB > BC > AC$$
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$$AB < BC < AC$$
Explanation
In any triangle side opposite to the largest angle is the longest side.
Here $$\angle C$$ is largest and side opposite to it is $$AB$$
$$\therefore AB$$ is the longest side.
Then comes $$\angle A$$ and side opposite to it is $$BC$$.
$$\therefore BC$$ is second longest.
Then comes $$\angle B$$ and side opposite to it is $$AC$$
So it is the smallest side.
So the decreasing order of sides is
$$AB>BC>AC$$
If $$\Delta ABC \cong \Delta DEF$$ by SSS congruence rule then :
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$$AB=EF, BC=FD, CA=DE$$
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$$AB=FD, BC=DE, CA=EF$$
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$$AB=DE, BC=EF, CA=FD$$
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$$AB=DE, BC=EF, \angle C=\angle F$$
Explanation
If triangles are congruent by $$SSS$$ congruence rule then their corresponding sides are equal.
Here $$\triangle ABC\cong \triangle DEF$$
$$\therefore AB=DE,BC=EF,AC=DF$$ or $$(CA=FD)$$
So option $$C$$ is correct.
In the given figure , which of the following statement is true ?
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$$\angle B=\angle C$$
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$$\angle B$$ is the greatest angle in triangle
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$$\angle B$$ is the smallest angle in triangle
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$$\angle A$$ is the smallest angle in triangle
Explanation
In any triangle angle opposite to the longest side is largest and smallest side is smallest.
Here $$BC$$ is the longest side and angle opposite to it is $$\angle A$$ so $$\angle A$$ is the largest angle.
Smallest side is $$AC$$ and angle opposite to it is $$\angle B$$ so $$\angle B$$ is the smallest angle.
So option $$C$$ is correct.
If $$\Delta ABC \cong \Delta DEF$$ by SSS congruence rule then
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$$AB=EF,BC=FD,CA=DE$$
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$$AB=FD,BC=DE,CA=EF$$
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$$AB=DE,BC=EF,CA=FD$$
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$$AB=DE,BC=EF,\angle C = \angle F$$
Explanation
If triangles are congruent by $$SSS$$ then their corresponding sides are equal , therefore
$$AB=DE$$
$$BC=EF$$
$$AC=DF$$ or $$CA=FD$$
So option $$C$$ is correct.
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