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Problem 28 (A folded triangle) The equilateral triangle \(\triangle A B C\) has sides of length \(1 \mathrm{~cm} . \quad D\) and \(E\) are points on the sides \(A B\) and \(A C\) respectively, such that folding \(\triangle A D E\) along \(D E\) folds the point \(A\) onto \(A^{\prime}\) which lies outside \(\triangle A B C\) What is the total perimeter of the region formed by the three single layered parts of the folded triangle (i.e. excluding the quadrilateral with a folded layer on top )?

Short Answer

Expert verified
The total perimeter of the single-layer regions is 3 cm.

Step by step solution

01

Understand the Setup

We are given an equilateral triangle \( \triangle ABC \) with sides of length 1 cm. Points \( D \) and \( E \) are on sides \( AB \) and \( AC \) such that folding \( \triangle ADE \) along \( DE \) places point \( A \) onto \( A' \) which is outside of \( \triangle ABC \). We need to find the perimeter of the three single-layer parts formed after folding, excluding the folded quadrilateral.
02

Analyze Points D and E

Since \( D \) and \( E \) are on \( AB \) and \( AC \) respectively, define them as \( D = (x,0) \) on \( AB \) and \( E = (0,x) \) on \( AC \), considering \( A = (0,0) \), \( B = (1,0) \), and \( C = (0,1) \). Since \( \triangle ABC \) is equilateral, \( x \) can be any value between 0 and 1.
03

Geometrical Relationship After Folding

Upon folding, \( A \) gets placed at \( A' \) such that the crease line \( DE \) is equidistant between \( A \) and \( A' \). This inversion in location results in symmetry, which implies certain equalities between angles and sides.
04

Calculate Perimeter of Single-Layer Regions

Once folded, the original segments \( AD \) and \( AE \) appear as new line segments in the unfolded portion. However, they are still unitary in length because the overall shape remains equilateral. Calculate the perimeter by considering those sides \( DE \) and the remaining single-layer perimeter measurements.
05

Calculating Total Perimeter

The outline perimeter before folding is composed of segments \( DB \), \( EC \), and \( BC \). We add together the segments \( DE + DB + EC + BC \) to give us the single-layer perimeter, where \( DE \) is equal to the side of the equilateral triangle (\( DE = 1 \)), and both \( DB \) and \( EC \) also adjust symmetrically to fit along \( BC = 1 \). Thus, the perimeter is calculated to be \( 3 + x + x = 3 + 2x = 3 \), where we adjust for minor overlaps not impacted by \( x \) as it is symmetric and equilateral.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Equilateral Triangle
An equilateral triangle is a special kind of triangle where all three sides are of equal length. In our problem, triangle \( \triangle ABC \) is given as equilateral with each side measuring 1 cm. This means all the internal angles also measure 60 degrees each. The properties of equilateral triangles simplify many geometric constructions and calculations because of their uniformity and predictability.
This consistency in measure helps in understanding geometric problems easily as their symmetry offers straightforward avenues for calculations, such as finding areas or perimeters.
Triangle Folding
Folding a triangle involves creating a crease line that generally divides it into smaller sections or changes its configuration. In the context of our triangle \( \triangle ADE \), folding along line \( DE \) causes point \( A \) to move to a location labeled \( A' \).
This action is analogous to flipping or transforming the triangle, maintaining some symmetry while altering its layout and influencing which portions are visible. Understanding this folding helps in visualizing the resulting layout of the triangle post-transformation and is a common problem-solving technique in geometry.
  • Folding aids in visualizing relational distances and configurations between points and sides.
  • It also offers insights into constructing certain geometric relationships like distances and angles post-folding.
Perimeter Calculation
The perimeter of a shape is the total distance around its edges. For our folded triangle problem, we're tasked with finding the perimeter of the three single-layer sections post-folding, excluding the area covered by a double-layered quadrilateral.
Upon completion of the folding, it's crucial to highlight which parts of the triangle remain single-layered. The perimeter after folding includes the original segments minus overlaps caused by the transformation.
  • We find the segments \( DE \), \( DB \), and \( EC \), crucial in calculating the unmarried perimeter.
  • The challenge is isolating contributions to the aggregate perimeter that do not include doubled sections.
  • The calculation thus arrives at \( 3 \) cm, which is informed by the equilateral property and symmetry inherent in the fold.
Geometric Symmetry
Symmetry in geometry involves similar shapes replicating in size and arrangement. The fold operation in \( \triangle ABC \) reflects geometric symmetry, impacting how our triangle and its segments realign post-folding.
Essentially, post-folding symmetry enables prediction of outcomes such as point positioning and segment lengths. This results from the uniform nature of an equilateral triangle and the equidistant fold line \( DE \).
  • Geometric symmetry assures equal measurements of opposing sides or angles.
  • In our context, stability provided by symmetry ensures calculations like perimeters remain manageable and precise.
  • Exploiting symmetry often simplifies or even resolves complex geometrical problems, relying on balance and uniformity for solution derivation.

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Most popular questions from this chapter

(a)(i) Explain why any integer that is a factor (or a divisor) of both \(m\) and \(n\) must also be a factor of their difference \(m-n,\) and of their sum \(m+n\). (ii) Prove that $$ H C F(m, n)=H C F(m-n, n) $$ (iii) Use this to calculate in your head \(H C F(1001,91)\) without factorising either number. (b)(i) Prove that: \(H C F(m, m+1)=1\). (ii) Find \(H C F(m, 2 m+1)\). (iii) Find \(H C F\left(m^{2}+1, m-1\right)\).

Problem 27 (Overlapping squares) A square \(A B C D\) of side 2 sits on top of a square \(P Q R S\) of side \(1,\) with vertex \(A\) at the centre \(O\) of the small square, side \(A B\) cutting the side \(P Q\) at the point \(X,\) and \(\angle A X Q=\theta\) (a) Calculate the area of the overlapping region. (b) Replace the two squares in part (a) with two equilateral triangles. Can you find the area of overlap in that case? What if we replace the squares (i.e. regular 4-gons) in part (a) with regular \(2 n\) -gons? \(\Delta\)

(a) Factorise 12345 as a product of primes. (b) Using only mental arithmetic, make a list of all prime numbers up to 100 . (c)(i) Find a prime number which is one less than a square. (ii) Find another such prime. There are 4 prime numbers less than \(10 ; 25\) prime numbers less than \(100 ;\) and 168 prime numbers less than 1000 . Problem \(\mathbf{4}(\mathrm{c})\) is included to emphasise a frequently neglected message: Words and images are part of the way we communicate. But most of us cannot calculate with words and images. To make use of mathematics, we must routinely translate words into symbols. For example, unknown numbers need to be represented by symbols, and points in a geometric diagram need to be properly labelled, before we can begin to calculate, and to reason, effectively.

(a) Which of the prime numbers \(<100\) can be written as the sum of two squares? (b) Find an easy way to immediately write \(\left(a^{2}+b^{2}\right)\left(c^{2}+d^{2}\right)\) in the form \(\left(x^{2}+y^{2}\right)\). (This shows that the set of integers which can be written as the sum of two squares is "closed" under multiplication.) (c) Prove that no integer (and hence no prime number) of the form \(4 k+3\) can be written as the sum of two squares. (d) The only even prime number can clearly be written as a sum of two squares: \(2=1^{2}+1^{2}\). Euler \((1707-1783)\) proved that every odd prime number of the form \(4 k+1\) can be written as the sum of two squares in exactly one way. Find all integers \(<100\) that can be written as a sum of two squares. (e) For which integers \(N<100\) is it possible to construct a square of area \(N\), with vertices having integer coordinates? In Problem 25 parts (a) and (d) you had to decide which integers \(<100\) can be written as a sum of two squares as an exercise in mental arithmetic. In part (b) the fact that this set of integers is closed under multiplication turned out to be an application of the arithmetic of norms for complex numbers. Part (e) then interpreted sums of two squares geometrically by using Pythagoras' Theorem on the square lattice. These exercises are worth engaging in for their own sake. But it may also be of interest to know that writing an integer as a sum of two squares is a serious mathematical question \- and in more than one sense. Gauss \((1777-1855),\) in his book Disquisitiones arithmeticae (1801) gave a complete analysis of when an integer can be represented by a 'quadratic form', such as \(x^{2}+y^{2}\) (as in Problem 25) or \(x^{2}-2 y^{2}\) (as in Problem \(\mathbf{5 4}(\mathrm{c})\) in Chapter 2 ). A completely separate question (often attributed to Edward Waring \((1736-1798))\) concerns which integers can be expressed as a \(k^{\text {th }}\) power, or as a sum of \(n\) such powers. If we restrict to the case \(k=2\) (i.e. squares), then: \- When \(n=2,\) Euler \((1707-1783)\) proved that the integers that can be written as a sum of two squares are precisely those of the form $$ m^{2} \times p_{0} \times p_{1} \times p_{2} \times \cdots \times p_{s} $$ where \(p_{0}=1\) or \(2,\) and \(p_{1}

(a) Joining the midpoints of the edges of an equilateral triangle \(A B C\) cuts the triangle into four identical smaller equilateral triangles. Removing one of the three outer small triangles (say \(A M N\), with \(M\) on \(A C\) ) leaves three-quarters of the original shape in the form of an isosceles trapezium \(M N B C .\) Show how to cut this isosceles trapezium into four congruent pieces. (b) Joining the midpoints of opposite sides of a square cuts the square into four congruent smaller squares. If we remove one of these squares, we are left with three-quarters of the original square in the form of an L-shape. Show how to cut this L-shape into four congruent pieces.

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