On this webpage you can find the solved tests for geometry of the 8th and 9th grades:

- – the problems 1 - 8 show the practice examples of solutions and answers for mathematics and geometry of the 8th grade. The lesson topics are as follows: midline of a triangle, parallelogram, area of a triangle, isosceles trapezoid, inscribed and circumscribed circumferences;
- – the practice online tests 9 – 12 can help to solve the geometry for the topic "Collinear vectors";
- – problems solutions for topic "Resolution of a vector into two non-collinear vectors" are in the tests 13 - 15;
- – the topic "Vector components" is explained in the workbook exercises 16 - 22. These solutions show answers to questions about how to solve problems if you want to find the components of the sum, the difference of vectors and the product of the vector by a number.

**Problem 1.**

Given:

ABCD is a quadrilateral

M, N, K, E are the midpoints of the sides AB, BC, DC, AD

Prove:

The quadrilateral MNKE is a parallelogram

Proof:

Join a point A to a point C. We obtain the triangle ∆ ABC, where MN is the midline of the triangle ∆ ABC, and the triangle ∆ ADC, where EK is the middle line of the triangle ∆ ADC.

By the property of the midline of a triangle ∆ it follows that

MN || AC are parallel and MN= AC,

EK || AC are parallel and EK= AC.

Then MN || EK are parallel and MN=EK, therefore

MNKE is the parallelogram (by the first theorem of a parallelogram).

***

**Problem 2.**

Given:

∆ ABC is a triangle

The length of the triangle side AB = 8.5 cm

The length of the triangle side AC = 5 cm

The height AH = 4 cm, i.e. the segment AH is perpendicular to the side BC

HBC, i.e. the point H lies on the side BC

Solve:

The area of the triangle = Area _{∆}_{ABC} = ?

Solution:

Area _{∆}_{ABC} = BC ∙ AH

By the Pythagorean theorem

BH = = = = 7.5 cm

By the Pythagorean theorem

CH = = = 3 cm

BC = BH + CH = 3 +7.5 = 10.5 cm

Area _{∆ABC} = ∙ 10.5 ∙ 4 = 21 cm^{2}

Answer: Area _{∆}_{ABC} = 21 cm^{2}

***

**Problem 3.**

Prove that the segments connecting the midpoints of the opposite sides of an isosceles trapezoid are perpendicular.

Given:

ABCD is an isosceles trapezoid

Prove: NE KM =

Proof:

Draw perpendiculars BH and CH_{1}, i.e. BHAD are perpendicular; also CH_{1}AD are perpendicular.

But BH and CH_{1 }cross NE then BRNE and CR_{1}NE are perpendicular.

The sides BH = CH_{1 } and BH || CH_{1} are parallel.

Therefore BH = KM = CH_{1 } BH KM CH_{1} are parallel as segments between parallel lines.

Therefore the angles are equal KON = NR_{1}C = 90º as corresponding angles.

Then KON = EOM = 90º, as vertical angles.

***

**Problem 4.**

Given:

AB is a segment

AC = CB

O is an arbitrary point

Prove:

The OC vector is half of the sum of the other two vectors OA and OB going out of the same point O

Proof: By the rule of a triangle

(1)

+

(2)

Adding equalities (1) and (2), we get

***

**Problem 5.**

Given:

a, b, c are vectors

Three vectors and are non-collinear vectors.

Build:

The sums and differences of vectors.

Construction:

By the rule of a polygon

a)

b)

=

***

**Problem 6.**

Prove that the segments connecting the midpoints of the opposite sides of an isosceles trapezoid are perpendicular.

Given:

The quadrilateral ABCD is an isosceles trapezoid

Prove: EF NM = , i.e. the angle of intersection of two segments in an isosceles trapezoid is 90º.

Proof:

Draw parallel lines

MK || AB

MR || CD

We get an isosceles triangle ∆MKR

AB=MK, since the trapezoid is isosceles,

CD=MR, since the trapezoid is isosceles.

Therefore, EF is the midline of the triangle ∆MKR, hence

MH=HR and OK=MO.

BM=MC=AK=RD, since ABMK and MCDR are parallelograms.

Therefore HR=KO.

Then MN is the median, the bisector and the height of the isosceles triangle ∆MKR.

Since MN is the height, we see that the segments MN AD are perpendicular.

By the property of the midline of a triangle ∆ it follows that

EF || KR.

Then EF NM =

***

**Problem 7.**

Prove that the center of a circle inscribed in an isosceles triangle lies on the median drawn to the base.

Given:

an inscribed circle in an isosceles triangle

∆ABC is an isosceles triangle

BH_{2 } is a median

Prove: O BH_{2}, i.e. the center of the inscribed circle lies on the median of an isosceles triangle

Proof:

Draw perpendiculars OH_{1} ; OH_{2} ; OH_{3 } to the sides of BC, AC, AB.

Here, from the two points, the same perpendicular to the AC side is drawn, but in the triangle, only one perpendicular can be drawn to the side and only from one point.

Therefore that O BH_{2}

***

**Problem 8.**

Prove that the center of the circle circumscribed around an isosceles triangle lies on the median drawn to the base or to its continuation.

Given:

The circumcircle around an isosceles triangle

∆ ABC is an inscribed isosceles triangle

BH_{3 } is a median

Prove: O BH_{3}

Proof:

Draw perpendiculars from the center of the circle

OH_{1 }; OH_{2 }; OH_{3 }to the sides of BC, AC, AB.

Here, a perpendicular to the AC side is drawn from two points, but only one perpendicular can be drawn to the side and only from one point in the triangle.

Therefore, O BH_{3}

***

## Collinear vectors

**The lemma** is a theorem that is subsidiary for the proof of the following theorem.

**The lemma about collinear vectors: **

If the vectors and are collinear (where ), then we can find a such number k that the equality is true (the vector is equal to the product of the number k by the vector )

Given: vector a, vector b

The vectors and are collinear, i.e. the vector b is collinear with the vector a.

Prove: there is a such number k that the following equality is true

Proof:

**Case 1.** Let the vectors b and a be co-directional vectors, i.e.

, where k>0, since . Then and are the co-directional vectors.

Therefore

***

**Case 2.**

Let a, b vectors be opposite vectors, i.e.

Let us take , where k<0

Consequently,

***

**Problem 9.**

Given:

vector m, vector n

1) are oppositely directed vectors,

= 0.5 cm, = 2 cm

2) are co-directional vectors,

= 12 cm, = 240 cm

Solve: k – ?

Solution: 1) Since , we see that k<0. Hence

= – = – 4

Answer: k = – 4.

Solution: 2) Since , we see that k>0. Therefore = = 20.

Answer: k = 20.

***

**Problem 10.**

Given:

ABCD is a parallelogram

BD AC = O

M is the midpoint of the segment AO

1)

2)

Solve: k – ?

Solution:

1) Since , we see that k>0.

By the property of a parallelogram

, we see that

Answer: k=

2) Since , we see that k<0. , are collinear, since they lie on one straight line. Find the midpoint of OC and call it as the point N.

We see that AM=MO=ON=NC

Since k<0, we see that

Answer: k=

***

**Problem 11.**

Given:

1) are oppositely directed vectors,

= 400 mm, = 4dm = 400mm

2) are co-directional vectors, = , =

Solve: k – ?

Solution: 1) Since , we see that k<0. Hence

= – = –1

Answer: k = –1.

Solution: 2) Since , we see that k>0. Therefore = = =5.

Answer: k = 5.

***

**Problem 12.**

Solve the equation: find the values of x, y.

Solution: 1)

y=3

Answer: x=0, y=3

***

Solve the equation: find the values of x, y.

Solution: 2)

–3y = –1 , x= –1

y =

Answer: x= – 1, y=

***

## The resolution of a vector into two non-collinear vectors

**Definition:** If , where and are the given vectors, x and y are some numbers, then we say that the vector is decomposed into vectors and , where x and y are the resolution coefficients.

Express the vector:

through vectors and

through and

through and

through and

Solution:

a) By the parallelogram rule (x= 1, y= 1)

b) , (x=y= 2)

c) = + , = 2 – (x= 2, y = –1)

d) Since = 2 – = 2 +

= – 2(x= 1, y = –2)

***

**Problem 13.**

Given: ABCD is a parallelogram

;

M; AM : MC = 4 : 1

Find:

Solution:

By the parallelogram rule

or

But , then

Answer:

***

**Problem 14.**

Given: vectors and are non-collinear.

a)

b)

Solve: the resolution coefficients of the x, y – ?

Solution:

a)

3 – y = 0, x+1=0 y= 3, x= – 1

b)

4 – x = 0, 5+y=0 x = 4, y= –5

Answer: a) x= –1, y= 3 б) x = 4, y= –5

***

**Problem 15.**

Given: a figure ABCD is a trapezoid

EF is a middle line of the trapezoid

Prove: EFAD, i.e. the midline of the trapezoid is parallel to its base,

i.e. the length of the midline of the trapezoid is a half of the sum of the lengths of the trapezoid bases.

Proof:

By the rule of a polygon

+

Adding both expressions, we get

Since E and F are the midpoints of the sides AB and CD, we see that

Since , we see that , and

Therefore EF || AD and

***

**Theorem:** Any vector can be decomposed into two non-collinear vectors, and the coefficients of the resolution are determined uniquely.

Given:

vector a, vector b

and are non-collinear vectors.

Prove:

Proof:

Through the point A and the point B we draw lines parallel to the straight lines containing the vectors and . Find the point C.

Then by the rule of the triangle

Let us remark that the vectors and are collinear, and the vectors are are collinear too.

By the lemma about collinear vectors

,

Then

**Uniqueness of resolution**

Proof:

We know that (1)

Let there be (2)

As a result of the difference between expressions (1) and (2) we get

This equality is possible

;

I.e. ;

***

## Coordinates of a vector

**Definition:** An unit vector is a vector of length is equal to 1.

i and j are coordinate vectors

Since and are non-collinear vectors, we see that any vector can be decomposed into the vectors and .

I.e. , where x and y are the vector coordinates.

{1:2}

{2:–3}

{0;0}

If and ,

then if and

***

**Problem 16. **

Find the coordinates of the vectors.

Solution:

{2;3}

{–2;3}

{2;0}

{–3;–4}

{2;–2}

{–4;–5}

***

**Problem 17. **

Find the coordinates of the vectors.

Solution:

{2;3}

{–;–2}

{8;0}

{1;–1}

{0;–2}

{–1;0}

***

**Problem 18. **

Find the sum of the vector from its coordinates.

Solution:

{–3;}

{–2;–3}

{–1;0}

{0;3}

{0;1}

***

The rules that allow you to find the coordinates of the sum, the difference of vectors and the product of a vector by a number from the coordinates of the vectors.

1. A vector with coordinates (a_{1}+ b_{1};a_{2} +b_{2}) is called the **sum of vectors** and with coordinates (a_{1};a_{2}) and (b_{1};b_{2}).

Given:

{*a _{1};*

*a*}; {

_{2}*b*

_{1};*b*};

_{2}Prove: { *a _{1}+ *

*b*

_{1};*a*

_{2}+*b*}

_{2}is the sum of the coordinates of the vector, i.e. this is the formula how to find the coordinates of a vector through the addition.

Proof:

{ *a _{1}+ *

*b*

_{1};*a*

_{2}+*b*}

_{2}***

**Example 1** is about the addition of vectors, how to find the coordinates of vectors:

If the coordinates of the vectors are given {3;2}; {2;5}, then

2. The **difference of the vectors** and with coordinates {a_{1}; a_{2}} and {b_{1}; b_{2}} is a vector with coordinates {a_{1} – b_{1}; a_{2} – b_{2}}.

3. The **product of a vector** with coordinates {a_{1}; a_{2}} by an arbitrary number k is a vector with coordinates {ka_{1}; ka_{2}}.

Given:

{a_{1};a_{2}}

k is an arbitrary number

Prove: {ka_{1}; ka_{2}}

is the product of a vector by a number

Proof:

Therefore, the vector {ka_{1}; ka_{2}}

**Example 2** is about how to find the vector coordinates:

Find the coordinates of the vector if

{1;2}; {0;3}; {–2;3}

Solution:

{0;6}

{0;6}

Answer: {0;6}

***

**Problem 19. **

Find the coordinates of a vector if given are the vectors with coordinates

{–7;–1}; {–1;7}; {4;–6}

Solution:

= {–21;–14}

Answer: {–21;–14}

***

**Problem 20. **

Given:

1)

2)

Solve: resolution coefficients of x, y – ?

Solution:

1)

By the theorem about the resolution of a vector into two non-collinear vectors:

x=–3, y=7

2)

By the theorem about the resolution of a vector into two non-collinear vectors:

x= –4, y=0

***

**Problem 21.**

Given: the coordinates of the vectors

1) {3;6}; {4;–3}

2) {–5;–6}; {2;–4}

Solve: the difference of vectors –

Solution:

1) –= = {–1;9}

–{–1;9}

2) –= ={–7;–2}

–{–7;–2}

***

**Problem 22.**

Given: the coordinates of the vectors

{–2;–3}; {2;–3}; {0;5}

Solve: the coordinates of the vectors are opposite to the given coordinates.

Solution:

{–2;–3} {2;3}

{2;–3} {–2;3}

{0;5} {0;–5}

***