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CBSE Questions for Class 9 Maths Triangles Quiz 12 - MCQExams.com
CBSE
Class 9 Maths
Triangles
Quiz 12
If two sides and one angle of a triangle are equal to the two sides and angle of another triangle, then the two triangles are congruent.
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0%
True
0%
False
Explanation
False
If two sides including one angle of a triangle are equal to the two sides including the one angle of another triangle, both triangles are congruent.
If two triangles are congruent, then the corresponding angles are equal.
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0%
True
0%
False
Explanation
True
By CPCT
If two angles and a side of a triangle are equal to two angles and a side of another triangle, then the triangles are congruent.
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0%
True
0%
False
Explanation
False
If two angles and included side of a triangle are equal to two corresponding angles and the included side of another triangle, both triangles are congruent.
If the hypotenuse of one right triangle is equal to the hypotenuse of another right triangle, then the triangles are congruent.
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0%
True
0%
False
Explanation
False
If one side and hypotenuse of a triangle are equal to one side and hypotenuse of another triangle, both triangles are congruent.
In Figure 6.28, two triangles are congruent by RHS.
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0%
True
0%
False
Explanation
False The above given triangle is congruent by SAS.
In $$\Delta ABC, \angle A=100^0, \angle B=30^0$$ and $$\angle C=50^0$$ then
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$$AB > AC$$
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$$AB < AC$$
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$$BC < AC$$
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none of these
Explanation
We know that the side opposite to largest angle is longest side of triangle .
$$\angle A$$ is the largest angle and side opposite to it is $$BC$$
So $$BC$$ is the longest side.
Then comes $$\angle C$$ and side opposite to it is $$AB$$.
So $$AB$$ is second longest side.
Then comes $$\angle B$$ and side opposite to it is $$AC$$.
So it is the smallest side.
So the decreasing order of side is
$$BC>AB>AC$$
So option $$A$$ is correct.
If in two triangles $$\Delta ABC$$ and $$\Delta PQR$$, $$AB = QR, BC = PR$$ and $$CA = PQ,$$ then :
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$$\Delta ABC\cong \Delta PQR$$
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$$\Delta CBA\cong \Delta PRQ$$
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$$\Delta BAC\cong \Delta RPQ$$
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$$\Delta PQR\cong \Delta BCA$$
Explanation
In $$\triangle ABC$$ and $$\triangle PQR$$, we are given
$$AB=QR$$,
$$BC=PR$$
and $$CA=PQ$$
$$\therefore \Delta CBA\cong \Delta PRQ$$ .... SSS test
In $$\Delta PQR$$, $$\angle P =$$$$ 70^{\circ}$$ and $$\angle R = 30^{\circ}.$$ Which side of this triangle is the longest?
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$$PR$$
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$$QR$$
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$$PQ$$
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All sides are equal.
Explanation
Given, $$\triangle PQR$$, $$\angle P = 70^{\circ}$$, $$\angle R = 30^{\circ}$$
Sum of angles of triangle $$= 180^{\circ}$$
$$\implies \angle P + \angle Q + \angle R = 180^{\circ}$$
$$\implies 70^{\circ} + 30^{\circ} + \angle Q = 180^{\circ}$$
$$\implies \angle Q = 80^{\circ}$$
Since $$\angle Q$$ is the largest $$\implies PR$$ will be the longest side t
his is because
when the two sides of a triangle are unequal, the angle opposite to the longer side is larger.
If $$D$$ and $$E$$ are the mid-points of the sides $$AB$$ and $$AC$$ respectively of $$\triangle ABC$$. $$DE$$ is produced to $$F$$. To prove that $$CF$$ is equal and parallel to $$DA$$, we need an additional information which is:
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$$\triangle DAE$$ = $$\triangle EFC$$
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$$AE = EF$$
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$$DE = EF$$
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$$\triangle ADE = \triangle ECF$$
Explanation
Given $$CF$$ is equal and parallel to $$DA$$
$$D$$ is the midpoint of $$ AB$$
So, $$ AD $$ =$$ DB$$= $$ CF$$= $$\cfrac { c }{ 2 } $$
Similarly $$ AE$$ =$$ EC $$=$$\cfrac { b }{ 2 } $$
Let $$\angle AED=\alpha $$ $$ADE=\beta \quad $$and$$\quad DAE=\gamma$$
Since $$AD $$ is parallel to $$ FC$$
$$\angle FEC = \alpha \angle EFC=\beta $$and$$ \angle ECF=\gamma$$
Applying $$S.A.S$$ Congruency for triangle$$ ADE$$ and $$ CFE$$, $$AD= CF$$, $$AE= AC$$
$$\angle DAE=\angle ECF$$
So $${ \Delta }^{ k } DAE$$ and
$${ \Delta }^{ k } FCE$$ are congruent
So $$ DE = EF$$
It is given that $$ DE= EF$$,then by $$S.S.S. $$ congruency all angles will be equal and $$ CF$$ and parallel to $$DA$$
$$\therefore $$ Additionsl information needed is $$ DE= EF$$.
$$AB$$ and $$CD$$ are the smallest and largest sides of a quadrilateral $$ABCD$$.
Out of $$\angle B$$ and $$\angle D$$, decide which is greater.
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$$\angle B$$ will be greater.
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$$\angle D$$ will be greater.
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Both the angles are equal.
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Cannot be determined.
Explanation
In the fig. $${ABCD}$$ is a quadrilateral with $${AB}$$ as the smallest side, and $${CD}$$ as the largest side.
We join the diagonal $${BD}$$.
We label the angles as shown in the figure as $$\theta 1$$ , $$\theta 2$$, $$\theta 3$$, $$\theta 4$$ as shown in figure.
Now in $$\Delta {ABD}$$,
$${AB}<{AD}$$
$$\implies \theta 2 < \theta 1 \rightarrow {(1)}$$
Similarly in $$\Delta {BCD}$$,
$${BC}<{CD}$$
$$\implies \theta 4 < \theta 3 \rightarrow {(2)}$$
Adding $${(1)} and {(2)}$$,
$$ \theta 2 + \theta 4 < \theta 1 + \theta 3$$
which is nothing but,
$$\angle{D} < \angle{B}$$
Thus $$\angle(B)$$ will be greater.
In the given figure, $$B < A$$ and $$D > C$$, then:
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$$AD > BC$$
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$$AD = BC$$
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$$AD < BC$$
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$$AD = 2BC$$
Explanation
In $$\triangle AOB$$, $$\angle {A}>\angle {B}$$
Hence, $$OB>OA$$..... (i)
In $$\triangle ODC$$, $$\angle {D}>\angle {C}$$
Hence, $$OC>OD$$.... (ii)
Adding inequalities (i) and (ii), we get
$$OB+OC>OA+OD$$
$$BC>AD$$
In $$\triangle ABC$$ and $$\triangle DEF$$, $$AB = FD$$ 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
$$SAS=$$ If two pairs of sides of two triangles are equal in length, and the included angles are equal in measurement, then the triangles are congruent.
Here, $$AB = FD$$,
$$\angle A = \angle D$$
So, $$AC = DE$$
Hence, option B is correct.
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