Triangles Test

Result

Triangles Test - 24

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Question 1
1 / -0

State the property by which$${\triangle }$$ADB $${\cong }$$ $${\triangle }$$ADC in the following figure.

A
SAS property

B
SSS property

C
RHS property

D
ASA property

Solution

$$AB=AC$$ (Given)
$$\angle BAD=\angle CAD$$ (given)
$$AD=AD$$ (common)
$$\therefore \triangle ADB\cong \triangle ADC$$
By $$SAS$$ property.
Question 2
1 / -0

The side opposite to an obtuse angle of a triangle is

A
Smallest

B
Greatest

C
Half of the perimeter

D
None of these

Solution

Angles between $$ {90}^{o} $$ and $$ {180}^{o} $$ are obtuse angles.
Since, sum of angles in a triangle $$ = {180}^{o} $$
If one angle is obtuse, it will be the greatest angle as the other two angles will be acute angles such that the sum of the interior angles is $$ {180}^{o} .$$
The angles opposite to the sides will also have the same relation.
Hence, the side opposite to the obtuse angle will be the greatest side of the triangle.
Question 3
1 / -0

If a, b and c are the sides of a $$\Delta$$ le then

A
a - b > c

B
c > a + b

C
c = a + b

D
b < c + a

Solution

$$b < c + a$$
$$\because$$ Sum of any two sides is greater than the third side.
Question 4
1 / -0

Sum of the lengths of any two sides of a triangle is always ____ than the length of the third side.

A
less than

B
equal to

C
greater than

D
None of these

Solution

It is a property of a triangle that:
Sum of the lengths of any two sides of a triangle is always greater than the length of the third side.
Question 5
1 / -0

Two sides of a triangle are $$7$$ and $$10$$ units. Which of the following length can be the length of the third side?

A
$$19$$ cm

B
$$17$$ cm

C
$$13$$ cm

D
$$3$$ cm

Solution

In a triangle the sum of any two sides is always greater than the third side.

Consider each option :

A) $$7+10 \ngtr 19$$

B) $$7+10 \ngtr 17$$

C) $$7+10> 13, 13+10>7, 7+13>10$$
It satisfies the condition.

D) $$7+3 \ngtr 10$$

Hence : Option C is the correct answer.
Question 6
1 / -0

$$\Delta$$ABC is congruent of $$\Delta$$DEF, if

A
AB = 4 = DE, AC = 6 = DF, $$\angle$$A = $$\angle$$D

B
AB = 4 = DE, AC = 6 = DF, $$\angle$$B = $$\angle$$E

C
AB = 6 = DE, AC = 4 = DF, $$\angle$$C = $$\angle$$F

D
AB = 4 = DE, AC = 6 = DF, $$\angle$$C = $$\angle$$F

Solution

If $$\triangle ABC$$ and $$\triangle DEF$$ are congruent, then
$$AB=DE$$
$$AC=DF$$
$$\angle B=\angle E$$

Hence, this is the answer.
Question 7
1 / -0

In the figure, AB and CD are two walls of equal height. A ladder BO is shifted to reach the other wall at D. Then, the triangles are congruent by the postulate

A
SSS

B
SAS

C
ASA

D
RHS

Solution
Question 8
1 / -0

Can $$6$$ cm, $$5$$ cm and $$3$$ cm form a triangle?

A
Yes

B
No

C
Sometimes

D
None of these

Solution

Given, $$6$$ cm, $$5$$ cm and $$3$$ cm are the sides of triangle.
Lets check if the triangle is possible or not.
$$(6 + 5=11)$$ $$11 > 3$$ Inequality property
$$(5 + 3=8)$$ $$8 > 6$$
$$(3 + 6=9)$$ $$9 > 5$$
They can form $$\Delta $$
Question 9
1 / -0

Can the three sides of length $$6 cm, 5 cm,$$ and $$3 cm$$ form a triangle?

A
Yes

B
No

C
Sometimes

D
None

Solution

Taking two sides at a time to check the inequality property of a triangle that is the sum of two sides of a triangle is always greater than the third side.
(1) $$(6 + 5) =11 > 3$$ {Satisfying}
(2) $$(5 + 3)= 8 > 6$$ {Satisfying}
(3) $$(3 + 6) =9 > 5$$ {Satisfying}
Hence, they can form a triangle.
Question 10
1 / -0

In the figure, $$\displaystyle AB=CD$$ and $$\displaystyle \angle A={ 90 }^{ o }=\angle D$$. Then

A
$$\displaystyle \Delta ABC\cong \Delta DBC$$ by SAS postulate

B
$$\displaystyle \Delta ABC\cong \Delta DCB$$ by RHS postulate

C
$$\displaystyle \Delta ABC\cong \Delta DBC$$ by AAS postulate

D
$$\displaystyle \Delta ABC\cong \Delta DCB$$ by SSS postulate

Solution

In $$\displaystyle \Delta ABC$$ and $$\displaystyle \Delta DBC$$
$$\displaystyle \angle A={ 90 }^{ o }=\angle D$$ given
$$\displaystyle BC=BC$$ common side
$$\displaystyle AB=CD$$ given
$$\displaystyle \therefore \Delta ABC\cong \Delta DCB$$ (By RHS congruence postulate)