Math Problem Statement

ABCD is a parallelogram. AB = 3√ 2, AD = 8, ∠BAD = 45◦ , M is a point on CD, CM : MD = 1 : 2. N is a point on AD, AN : ND = 3 : 1. Find: a) the length of AM; b) the length of BN;

Solution

Given a parallelogram ABCDABCD, the sides and angles provided give us enough information to compute the required lengths. Let's break down the information and the steps:

1. Given Data:

  • AB=32AB = 3\sqrt{2}, AD=8AD = 8, and BAD=45\angle BAD = 45^\circ.
  • MM is a point on CDCD, and CM:MD=1:2CM : MD = 1 : 2.
  • NN is a point on ADAD, and AN:ND=3:1AN : ND = 3 : 1.

We are tasked with finding the lengths of:

  1. AMAM
  2. BNBN

Step 1: Analyze the Parallelogram

In any parallelogram:

  • Opposite sides are equal: AB=CDAB = CD, AD=BCAD = BC.
  • Opposite angles are equal: BAD=BCD\angle BAD = \angle BCD, and adjacent angles are supplementary.
  • Vectors can be helpful in breaking down the problem geometrically.

Step 2: Work on Length of AMAM

Let’s use vector coordinates. Assign coordinates for the vertices:

  • A=(0,0)A = (0, 0),
  • B=(8,0)B = (8, 0) (since AD=8AD = 8),
  • D=(0,8)D = (0, 8) (perpendicular from AA),
  • C=(8,8)C = (8, 8).

Now, since CM:MD=1:2CM : MD = 1 : 2, the point MM divides CDCD in the ratio 1:21:2. Hence, the coordinates of MM can be found using the section formula: M=(2(8)+1(8)1+2,2(8)+1(0)1+2)=(8,163).M = \left( \frac{2(8) + 1(8)}{1+2}, \frac{2(8) + 1(0)}{1+2} \right) = \left( 8, \frac{16}{3} \right).

The length of AMAM is the distance between A(0,0)A(0, 0) and M(8,163)M(8, \frac{16}{3}). Using the distance formula: AM=(80)2+(1630)2=64+(163)2=64+2569.AM = \sqrt{(8-0)^2 + \left( \frac{16}{3} - 0 \right)^2} = \sqrt{64 + \left( \frac{16}{3} \right)^2} = \sqrt{64 + \frac{256}{9}}. Simplify: AM=5769+2569=8329=8323=4523.AM = \sqrt{\frac{576}{9} + \frac{256}{9}} = \sqrt{\frac{832}{9}} = \frac{\sqrt{832}}{3} = \frac{4\sqrt{52}}{3}. Hence, AM=8133AM = \frac{8\sqrt{13}}{3}.

Step 3: Work on Length of BNBN

Since NN divides ADAD in the ratio AN:ND=3:1AN : ND = 3:1, the coordinates of NN can be found as: N=(3(0)+1(0)3+1,3(8)+1(0)3+1)=(0,6).N = \left( \frac{3(0) + 1(0)}{3+1}, \frac{3(8) + 1(0)}{3+1} \right) = (0, 6).

The length of BNBN is the distance between B(8,0)B(8, 0) and N(0,6)N(0, 6). Using the distance formula: BN=(80)2+(06)2=82+(6)2=64+36=100=10.BN = \sqrt{(8-0)^2 + (0-6)^2} = \sqrt{8^2 + (-6)^2} = \sqrt{64 + 36} = \sqrt{100} = 10.

Final Answers:

  • AM=8133AM = \frac{8\sqrt{13}}{3},
  • BN=10BN = 10.

Would you like further explanation or details on any step?

Here are some related questions:

  1. How do we calculate section points using the section formula?
  2. What is the geometrical significance of ratios in division of a line?
  3. How can vectors be used to find distances in geometry problems?
  4. What are the properties of parallelograms that are useful in solving geometric problems?
  5. How do we find distances between points in the coordinate plane?

Tip: When dividing a line segment in a given ratio, the section formula simplifies computations in geometry.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Geometry
Coordinate Geometry
Parallelograms
Section Formula
Distance Formula

Formulas

Section formula: P(x, y) = [(m * x2 + n * x1) / (m + n), (m * y2 + n * y1) / (m + n)]
Distance formula: d = √[(x2 - x1)^2 + (y2 - y1)^2]

Theorems

Properties of Parallelograms
Pythagorean Theorem
Section Formula

Suitable Grade Level

Grades 9-12