On this webpage you can find the solved tests for geometry of the 8th and 9th grades:
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=
EK || AC are parallel and EK=
Then MN || EK are parallel and MN=EK, therefore
MNKE is the parallelogram (by the first theorem of a parallelogram).
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
H
Solve:
The area of the triangle = Area ∆ABC = ?
Solution:
Area ∆ABC =
By the Pythagorean theorem
BH =
By the Pythagorean theorem
CH =
BC = BH + CH = 3 +7.5 = 10.5 cm
Area ∆ABC =
Answer: Area ∆ABC = 21 cm2
***
Given:
ABCD is an isosceles trapezoid
Prove: NE
Proof:
Draw perpendiculars BH and CH1, i.e. BH
But BH and CH1 cross NE
The sides BH = CH1 and
Therefore BH = KM = CH1
Therefore the angles are equal
Then
***
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
+
Adding equalities (1) and (2), we get
***
Given:
a, b, c are vectors
Three vectors
The sums and differences of vectors.
Construction:
By the rule of a polygon
a)
b)
=
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
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
By the property of the midline of a triangle ∆ it follows that
EF || KR.
Then EF
***
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
BH2 is a median
Prove: O
Proof:
Draw perpendiculars OH1 ; OH2 ; OH3 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
***
Prove that the center of the circle circumscribed around an isosceles triangle lies on the median drawn to the base or to its continuation.
The circumcircle around an isosceles triangle
∆ ABC is an inscribed isosceles triangle
BH3 is a median
Prove: O
Proof:
Draw perpendiculars from the center of the circle
OH1 ; OH2 ; OH3 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
***
The lemma is a theorem that is subsidiary for the proof of the following theorem.
The lemma about collinear vectors:
If the vectors
Given: vector a, vector b
The vectors
Prove: there is a such number k that the following equality is true
Proof:
Therefore
***
Case 2.
Let a, b vectors be opposite vectors, i.e.
Let us take
Consequently,
***
Given:
vector m, vector n
1)
2)
Solve: k – ?
Solution: 1) Since
Answer: k = – 4.
Solution: 2) Since
Answer: k = 20.
***
ABCD is a parallelogram
BD
M is the midpoint of the segment AO
1)
2)
Solve: k – ?
Solution:
1) Since
By the property of a parallelogram
Answer: k=
2) Since
We see that AM=MO=ON=NC
Since k<0, we see that
Answer: k=
***
Given:
1)
2)
Solve: k – ?
Solution: 1) Since
Answer: k = –1.
Solution: 2) Since
Answer: k = 5.
***
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=
***
Definition: If
Solution:
a) By the parallelogram rule
b)
c)
d) Since
***
Given: ABCD is a parallelogram
M
Find:
Solution:
By the parallelogram rule
But
Answer:
***
Given: vectors
a)
b)
Solve: the resolution coefficients of the x, y – ?
Solution:
a)
3 – y = 0, x+1=0
b)
4 – x = 0, 5+y=0
Answer: a) x= –1, y= 3 б) x = 4, y= –5
***
Given: a figure ABCD is a trapezoid
EF is a middle line of the trapezoid
Prove: EF
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
Therefore EF || AD and
Theorem: Any vector
Given:
vector a, vector b
Prove:
Proof:
Through the point A and the point B we draw lines parallel to the straight lines containing the vectors
Then by the rule of the triangle
Let us remark that the vectors
By the lemma about collinear vectors
Then
Uniqueness of resolution
Proof:
We know that
Let there be
As a result of the difference between expressions (1) and (2) we get
This equality is possible
I.e.
***
Definition: An unit vector is a vector of length is equal to 1.
i and j are coordinate vectors
Since
I.e.
If
then
***
Find the coordinates of the vectors.
Solution:
***
Find the coordinates of the vectors.
Solution:
***
Find the sum of the vector from its coordinates.
Solution:
***
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
Given:
Prove:
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:
***
Example 1 is about the addition of vectors, how to find the coordinates of vectors:
If the coordinates of the vectors are given
2. The difference of the vectors
3. The product of a vector
Given:
k is an arbitrary number
Prove:
is the product of a vector by a number
Proof:
Therefore, the vector
Example 2 is about how to find the vector coordinates:
Find the coordinates of the vector
Solution:
{0;6}
{0;6}
Answer:
***
Find the coordinates of a vector
Solution:
= {–21;–14}
Answer:
***
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
***
Given: the coordinates of the vectors
1)
2)
Solve: the difference of vectors
Solution:
1)
2)
***
Given: the coordinates of the vectors
Solve: the coordinates of the vectors are opposite to the given coordinates.
Solution:
***