Triangles Test

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Triangles Test - 39

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Question 1
1 / -0

Identify the largest angle of the triangle.

A
A

B
B

C
C

D
D

Solution

In a triangle inequality theorem, the largest angle is across from the longest side.
So, $$9$$ is the longest side in the triangle, $$< A$$, across from it, is the largest angle.
Therefore, $$A$$ is the largest angle.
Question 2
1 / -0

$$AD$$ is the bisector of BC.
Which property of triangle applies here?

A
$$SAS$$

B
$$SSS$$

C
$$AAS$$

D
none

Solution

In $$\triangle ABD$$ & $$\triangle ADC$$
$$AD = AD$$
$$BD = DC$$
$$\angle ADB = \angle ADC$$
$$\triangle ABD \cong \triangle ADC$$
$$\therefore$$ Both triangles are congruent by $$SAS$$ congruence condition.
Question 3
1 / -0

$$\Delta ABC \cong \Delta DEF$$ by ASA congruence condition. Find the value of $$x$$.

A
$$x = 20$$

B
$$x = 30$$

C
$$x = 40$$

D
$$x = 50$$

Solution

We know $$\Delta ABC \cong \Delta DEF$$ by ASA congruence.
Therefore, $$\angle B = \angle F$$
$$\Rightarrow 60 = x + 30$$
$$\Rightarrow x = 60 - 30$$
$$\Rightarrow x = 30$$
Question 4
1 / -0

What is the largest angle of the triangle?

A
X

B
Y

C
Z

D
P

Solution

In a triangle inequality theorem, the largest angle is across from the longest side.
So, $$20$$ is the longest side in the triangle, $$\angle Z$$, across from it is the largest angle.
Therefore, $$Z$$ is the largest angle.
Question 5
1 / -0

If $$\Delta ABC$$ and $$\Delta DEF$$ are congruent, then find the value of $$x$$ and $$y$$.

A
$$x = 40, y = 60$$

B
$$x = 30, y = 60$$

C
$$x = 60, y = 60$$

D
$$x = 60, y = 40$$

Solution

We know $$\Delta ABC \cong \Delta DEF$$ by ASA congruence.
Therefore, $$\angle A = \angle D=40$$
$$\therefore x = 40$$
$$\angle B = \angle F=60$$
$$\therefore y=60$$
So, $$x = 40, y = 60$$
Question 6
1 / -0

In triangle $$PQR$$, an exterior angle at $$A$$ measures $$160^o$$, and $$\angle$$ $$Q = 70 ^o$$. Which is the longest side of the triangle?

A
$$\overline{PR}$$

B
$$\overline{PA}$$

C
$$\overline{PQ}$$

D
$$\overline{QR}$$

Solution

By using exterior angle property,
$$\angle P+\angle Q = \angle A$$
$$\Rightarrow \angle P +70^o=160^o$$
$$\Rightarrow \angle P = 160^o-70^o$$
$$\Rightarrow \angle P = 90^o$$

According to the triangle inequality theorem, the side that is opposite to the largest angle is largest.
So, in a right angle triangle, $$90^o$$ is the longest angle in any, $$\overline{QR}$$ is the side opposite to it.

Therefore, $$\overline{QR}$$ is the largest side.
Question 7
1 / -0

Identify the largest side of the triangle. If the angles are given as:
$$\angle A = 50 ^o$$, $$\angle B = 10^o$$ and $$\angle C = 22^o$$

A
$$\overline{AB}$$

B
$$\overline{BC}$$

C
$$\overline{CA}$$

D
$$\overline{AC}$$

Solution

In a triangle inequality theorem, the largest side is opposite to the longest angle.
So, $$50$$ degree is the longest angle in the triangle, $$\overline{BC}$$, opposite to it, is the largest side.
Therefore, $$\overline{BC}$$ is the largest side.
Question 8
1 / -0

In $$\triangle ABC$$, if $$\angle A = 45^{\circ}$$ and $$\angle B = 70^{\circ}$$, find the shortest and the largest sides of the triangle respectively

A
$$AB, BC$$

B
$$BC, AC$$

C
$$AB, AC$$

D
Either (A) or (C)

Solution

Given, $$ \angle A = 45^{0}$$ and $$ \angle B = 70^{0}$$
By angle sum property, $$ A+B+C = 180^{0}$$
$$\implies 45+70+C=180$$
$$\implies C=65^{0}$$
Angle opposite to longest side is larger and the side opposite to $$\angle B$$ is AC .
Also, shortest side is opposite to the shorter angle, $$ \angle A$$ is BC.
Question 9
1 / -0

The distance from town A to town B is five miles. C is six miles from B. Which of the following could be the distance from A to C?
I. $$11$$
II. $$1$$
III. $$7$$

A
I only

B
II only

C
I and II only

D
II and III only

E
I, II or III

Solution

The distance from two A nad B is five miles. C is six miles from B To find distance between A to C there are three possibilities.
Case: $$1$$ In triangle $$ABC,AB=5$$ and $$BC=6$$
Here , the length of AC must be less than the sum of other two sides and greater than the difference of the other sides
$$6.5<6+5$$
$$\Rightarrow 1<Ac<11$$
Thus $$AC=7$$ is a possible
$$\therefore $$ distance from $$A$$ to $$C$$ could be $$11,1 $$ or $$7$$ miles.
Question 10
1 / -0

Two sides of a triangle have lengths $$7$$ and $$9$$. Which of the following could not be the length of the third side?

A
$$4$$

B
$$5$$

C
$$7$$

D
$$11$$

E
$$16$$

Solution

An important rule to remember about triangles is called the third side rule:
The length of the third side of a triangle is less than the sum of the lengths of the other two sides and greater than the (positive) difference of the lengths of the other two sides.

For this triangle, the length of the third side must be greater than $$9-7=2$$ and less than $$9+7=16$$. All the answers are possible except for answer E, which is equal to $$16$$ but not less than $$16$$.