Let ABCD be a square with side legnth 7cm. Another square KLMN is inscribed into ABCD such that its vertices lie on the sides of ABCD.
Find the lenghts of parts of ABCD sides which are divided by the vertices of KLMN. If [the area of KLMN]: [area of ABCD] = 25:49
you will have 4 congruent triangles.
each will have legs of x and 7-x and a hypotenuse of 5
thus x^2 + (7-x)^2 = 25
solve as a quadratic, it factors nicely, and you should have sides of 3 and 4
How would you know that the hypotenuse is length 5? Does it have something to do with the ratio?
To find the lengths of the parts of ABCD sides that are divided by the vertices of KLMN, let's label the points as follows:
A: Bottom-left corner of ABCD
B: Bottom-right corner of ABCD
C: Top-right corner of ABCD
D: Top-left corner of ABCD
And let's label the vertices of KLMN as:
K: Bottom-left corner of KLMN (on AB)
L: Bottom-right corner of KLMN (on BC)
M: Top-right corner of KLMN (on CD)
N: Top-left corner of KLMN (on DA)
Since ABCD is a square with side length 7 cm, the length of each side of ABCD is 7 cm.
Let's assume that the length of the segment AK is x cm. Since K is the bottom-left corner of KLMN and lies on AB, the length of BK is also x cm.
Now, we know that the area of KLMN is 25/49 times the area of ABCD. The area of ABCD is 7 cm * 7 cm = 49 cm^2. So, the area of KLMN is (25/49) * 49 cm^2 = 25 cm^2.
The area of a square is equal to the square of its side length. Since the area of KLMN is 25 cm^2, the side length of KLMN is sqrt(25 cm^2) = 5 cm.
Since KLMN is a square, the segments KL, LM, MN, and NK are all equal to 5 cm.
Now, let's consider the lengths of the parts of ABCD sides divided by the vertices of KLMN:
1. The length of the part of AB divided by K and L is equal to AK + BL. Since AK is x cm and BL is 5 cm, this length is x + 5 cm.
2. The length of the part of BC divided by L and M is equal to BL + CM. Since BL is 5 cm and CM is 5 cm, this length is 5 + 5 cm = 10 cm.
3. The length of the part of CD divided by M and N is equal to CM + DN. Since CM is 5 cm and DN is x cm, this length is 5 + x cm.
4. The length of the part of DA divided by N and K is equal to DN + AK. Since DN is x cm and AK is x cm, this length is x + x = 2x cm.
Therefore, the lengths of the parts of ABCD sides that are divided by the vertices of KLMN are:
AB segment: x + 5 cm
BC segment: 10 cm
CD segment: 5 + x cm
DA segment: 2x cm
To find the lengths of the parts of ABCD sides that are divided by the vertices of KLMN, we first need to determine the dimensions of KLMN.
Let's start by calculating the area of ABCD. Since it is a square with a side length of 7 cm, its area is given by (side length)^2 = (7 cm)^2 = 49 cm^2.
The ratio of the area of KLMN to the area of ABCD is given as 25:49. We can write this as (area of KLMN) = (25/49) * (area of ABCD).
Substituting the area of ABCD = 49 cm^2 into the equation, we have (area of KLMN) = (25/49) * 49 cm^2 = 25 cm^2.
Since KLMN is inscribed in ABCD, its area is equal to the area of the square formed by its diagonals.
Let's denote the side length of KLMN as x cm. The area of KLMN is given by (side length)^2 = x^2 = 25 cm^2.
Solving for x, we find x = √(25 cm^2) = 5 cm.
Now that we know the side length of KLMN is 5 cm, we can determine the lengths of the parts of ABCD sides divided by the vertices of KLMN.
Let's consider the side AB. It is divided by the vertices of KLMN into three parts: AK, KL, and LB.
Since KLMN is inscribed in ABCD, AK and BL are diagonals of KLMN, and KL is parallel to one side of ABCD.
The diagonal of a square divides it into two congruent right triangles. Since ABCD is a square, the triangles AKB and BLA are congruent.
Therefore, AK = BL. Let's denote this length as y cm.
Now, we can divide the side length of AB into the three parts:
AB = AK + KL + LB
= y cm + 5 cm + y cm
= 2y + 5 cm.
Similarly, we can determine the lengths of the parts of the other sides of ABCD divided by the vertices of KLMN.
AC = KD = 2y + 5 cm (using the same reasoning as above)
BC = AD = 2y cm (since BC and AD have no parts divided by the vertices of KLMN)
To summarize:
AB = BC = AC = AD = 2y + 5 cm (parts divided by KLMN vertices)
CD = 2y cm (no parts divided by KLMN vertices)
Therefore, the lengths of the parts of ABCD sides divided by the vertices of KLMN are all 2y + 5 cm, except for CD, which is 2y cm.