Triangles Test

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Triangles Test - 22

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Weekly Quiz Competition

Question 1
1 / -0

For any triangle $$ABC$$, the true statement is

A
$${ AC }^{ 2 }={ AB }^{ 2 }+{ BC }^{ 2 }$$

B
$$AC=AB+BC$$

C
$$AC>AB+BC$$

D
$$AC<\,AB+BC$$

Solution

For any $$\triangle ABC$$, the sum of two sides must be greater than the third side.

Hence, $$AB + BC > AC$$.
Question 2
1 / -0

In $$\triangle PQR$$, if $$\angle R\displaystyle>\angle Q$$, then

A
$$QR>PR$$

B
$$PQ>PR$$

C
$$PQ

D
$$QR

Solution

In $$\triangle PQR$$, $$\angle R > \angle Q$$
Hence, $$PQ > PR$$ (Sides opposite larger angles is large)
option B is correct
Question 3
1 / -0

Angles opposite to the equal sides of a triangle are equal.

A
True

B
False

C
Incomplete information

D
None

Solution

Consider $$\triangle ABC$$, $$\angle B = \angle C$$
Construction: Draw a median AD on BC.
Hence, $$BD = DC$$
Now, In $$\triangle ABD$$ and $$\triangle ACD$$
$$\angle ADB = \angle ADC$$ .... Common angle
$$AD = AD$$ ..... (Common)
$$BD = DC$$
Thus, $$\triangle ABD \cong \triangle ACD$$ .... (SAS rule)
Thus, $$\angle B = \angle C$$ .... (By CPCT)
Question 4
1 / -0

Two sides of a triangle are of lengths 5 cm and 1.5 cm, then the length of the third side of the triangle cannot be

A
$$\displaystyle3.6\,cm$$

B
$$\displaystyle4.1\, cm$$

C
$$\displaystyle3.8\, cm$$

D
$$\displaystyle3.4\, cm$$

Solution

In a triangle, the difference between two sides should be less than the third side.
Hence,option D is correct $$3.4\ cm$$
Question 5
1 / -0

In triangles $$ABC$$ and $$PQR$$, $$AB=AC$$, $$\angle C=\angle P$$ and $$\angle B=\angle Q$$. Then the two triangles are :

A
Isosceles but not congruent

B
Isosceles and congruent

C
Congruent but not isosceles

D
Neither congruent nor isosceles

Solution

In $$\triangle ABC$$ and $$\triangle PQR$$
$$\angle C = \angle P$$ (Given)
$$\angle B = \angle Q$$ (Given)
$$\angle A = \angle R$$ (Third angle of the triangle)

Thus, $$\triangle ABC \sim \triangle PQR$$
Also, given, $$AB = AC$$
Thus, $$\angle B = \angle C$$ (Isosceles triangle Property)

But, $$\angle B = \angle Q$$ and $$\angle C = \angle P$$
Hence, $$\angle Q = \angle P$$
or $$PR = QR$$

Thus, both the triangles are Isosceles but not congruent.
Question 6
1 / -0

In a triangle $$ABC$$ and $$DEF$$, $$AB=FD$$ and $$\angle A=\angle D$$. The two triangles will be congruent by SAS axiom if:

A
$$BC=EF$$

B
$$AC=DE$$

C
$$AC=EF$$

D
$$BC=DE$$

Solution

In $$\triangle ABC$$ and $$\triangle DEF$$,
$$\angle A = \angle D$$ ....(Given)
$$AB = FD$$ ....(Given)
and $$\triangle ABC \cong \triangle DFE$$ ....(SAS rule)
It is possible, if corresponding sides are equal, $$AC = ED$$
Question 7
1 / -0

If any two sides of a triangle are unequal, then the larger side has the $$.........$$ angle opposite to it.

A
Greater

B
Lesser

C
Equal

D
May be greater or lesser

Solution

Given: $$\triangle ABC$$, $$AB > AC$$

To Prove: $$\angle ACB > \angle ABC$$

Construction: Make a point D on $$AB$$ such that $$AD = AC$$ and join $$DC$$

In $$\triangle ACD$$,
$$AD = AC$$
$$\angle ACD = \angle ADC$$ (Angles opposite to equal sides)

But $$\angle ADC > \angle ABC$$ (Exterior angle of the triangle is greater than the opposite interior angle)

Again, $$\angle ACB > \angle ACD$$ (D lies in interior of $$\angle ACB$$)
Thus, $$\angle ACB > \angle ABC$$
Question 8
1 / -0

If in two triangles $$ABC$$ and $$PQR$$, $$AB=\, QR$$, $$\angle A=\angle Q$$ and $$\angle B=\angle R$$, then $$\triangle ABC\cong$$ $$\triangle $$

A
$$QRP$$

B
$$PQR$$

C
$$PRQ$$

D
None of the above.

Solution

Given: $$AB=\, QR$$, $$\angle A=\angle Q$$ and $$\angle B=\angle R$$

Now, In $$\triangle ABC$$ and $$\triangle PQR$$
$$AB = QR$$
$$\angle A = \angle Q$$
$$\angle B = \angle R$$
Thus, by $$ASA$$ congruence rule,
$$\triangle ABC \cong \triangle QRP$$

Hence, $$Op-A$$ is correct.
Question 9
1 / -0

If in two triangles $$ABC$$ and $$DEF$$, $$AB=\,DF$$, $$BC=\,DE$$ and $$\angle B=\angle D$$, then $$\triangle ABC\cong $$ $$\triangle $$____.

A
$$FDE$$

B
$$DEF$$

C
$$EDF$$

D
$$EFD$$

Solution

Given:

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

$$AB = DF$$

$$\angle B = \angle D$$

$$BC = DE$$

Thus, $$\triangle ABC \cong \triangle FDE$$ (SAS rule)
Question 10
1 / -0

If the two angles of a triangle are unequal, then the smaller angle has the $$.......$$ side opposite to it.

A
Smaller

B
Larger

C
May be smaller may be larger

D
The side will be equal to one of the opposite sides

Solution

Let, In $$\triangle ABC$$, $$ \angle ACB > \angle ABC$$
To Prove: $$AB > AC$$
Construction: Make a point $$D$$ on $$AB$$, such that $$\angle ACD = \angle ADC$$ and join $$DC$$
Now, in $$\triangle ACD$$,
$$\angle ACD = \angle ADC$$
$$\therefore AD = AC$$ (Sides opposite to equal angles)

But $$AB = AD + DB$$
thus, $$AB > AD$$
Hence, $$AB > AC$$ [as $$AD = AC$$]

Thus, sides opposite smaller angles in a triangle are smaller.