Triangle : corrected 5th grade math exercises in PDF.

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The answer key to the math exercises in 5th grade on triangles. Know how to draw a triangle using geometry materials. Use the triangular inequality and draw the circumscribed circle of a triangle in fifth grade.

Exercise 1:

1. Let LNI be a triangle such that : $\widehat{I}=76^{\circ}\,\,,\,\,\widehat{L}=45^{\circ}$

Calculate the measure of the angle $\widehat{N}.$

$\widehat{I}+\widehat{L}=76^{\circ}+45^{\circ}=121^{\circ}$

$\widehat{N}=180^{\circ}-121^{\circ}={\color{DarkRed},59^{\circ}}$

2. Let SAC be a triangle such that $\widehat{A}=110^{\circ}\,,\,\widehat{C}=28^{\circ}$

Calculate the measure of the angle $\widehat{S}.$

$\widehat{A}+\widehat{C}=110^{\circ}+28^{\circ}=138^{\circ}$

$\widehat{S}=180^{\circ}-138^{\circ}={\color{DarkRed},42^{\circ}}$

Exercise 2:

Exercise 3:

Exercise 4:

ACD is an equilateral triangle so its three angles measure 60°.

ACE is an isosceles right triangle so $\widehat{EAC}=\widehat{ECA}=45^{\circ}$

Then $\widehat{BCD}$ is a flat angle so $\widehat{BCA}\,=180-45-60=75^{\circ}$

And ABC is isosceles at A so its base angles have the same measure.

$\widehat{CBA}=\widehat{BCA}\,=\,75^{\circ}$

and the last angle measures :

180-150=30°

Exercise 5:

The triangle MNQ is isosceles with main vertex M and base [NQ].

The triangle PMN is isosceles with main vertex P and base [MN].

The angle $\widehat{\,MQN\,}$ measures $35^{\circ}$ .

Determine the measure of the angle $\widehat{PMQ}$ .

MQN is isosceles at M so the angles at the base are equal.

$\widehat{QNM}=\widehat{NQM}=35^{\circ}$

The trangle NPM is isosceles at P so the angles at the base have the same measure.

$\widehat{NMP}=\widehat{QNM}=35^{\circ}$

then

$\widehat{NPM}=180-35-35=110^{\circ}$

and

$\widehat{PMQ}=180-35-110=35^{\circ}$

Exercise 6:

Consider a triangle MNO, isosceles with main vertex N and base [MO].

We know that $\widehat{N}=44^{\circ}$ . Deduce the measure of $\widehat{M}$ and $\widehat{O}$ .

The angles at the base have the same measure for an isosceles triangle.

$\widehat{M}=\widehat{O}=\frac{180-44}{2}=68^{\circ}$

Exercise 7:

Consider an equilateral triangle JKL.

Deduce the measure of its three angles.

$\frac{180}{3}=60^{\circ}$

the three angles of an equilateral triangle are 60°.

Exercise 8:

Consider a triangle GHI, right-angled at H. We know that $\widehat{G}$ = 34°.

Deduce the measure of $\widehat{I}$ .

$\widehat{I}=180-90-34=56^{\circ}$

Exercise 9:

Magalie measured the angles DEF with her protractor.

She found $\widehat{D}$ = 53°, $\widehat{E}$ = 74° and $\widehat{F}$ = 54°.

What do you think of his answer? Justify.

$D+E+F=53+74+54=181^{\circ}$

Magalie’s measurements are wrong because the sum of the three angles of a triangle is 180°.

Exercise 10:

Consider a triangle ABC. We know that $\widehat{A}$ = 28° and $\widehat{B}$ = 73°.

Deduce the measure of $\widehat{C}$ .

$C=180-28-73=79^{\circ}$

Exercise 11:

What is the measure of angle DEF?

$\widehat{ABC}=180^{\circ}-105^{\circ}=75^{\circ}$

$\widehat{ADC}=\widehat{ABC}=75^{\circ}$ (the opposite angles of a parallelogram have the same measure).

$\widehat{EDF}=\widehat{ADC}=75^{\circ}$ (two angles opposite by the vertex have the same measure).

$\widehat{DEF}=180^{\circ}-75^{\circ}-90^{\circ}=180^{\circ}-165^{\circ}=15^{\circ}$ (the sum of the angles of a triangle is $180^{\circ}$ )

Exercise 12:

What is the measure of the angle ADB?

In the triangle ADC :

$\widehat{ADC}=180^{\circ}-45^{\circ}-35^{\circ}$

$\widehat{ADC}=180^{\circ}-80^{\circ}$

$\widehat{ADC}=100^{\circ}$

The angle $\widehat{BDC}$ is a flat angle.

$\widehat{BDA}=180^{\circ}-100^{\circ}=80^{\circ}$

Exercise 13:

Exercise 14:

Exercise 16:

The sum of the angles is 180°.

90°+84,7°=174,7°

180°-174,7°=5,3°

So the angle of inclination to the vertical is 5.3°.

The answers to the exercises on the triangle in 5th grade.

After consulting the answers to these exercises on triangles in 5th grade, you can return to the exercises in 5th grade.

Fifth grade exercises.

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