5/30/2023 0 Comments Rectangle abcd![]() ![]() Instead of a square and a rectangle, let’s consider a triangle and an equilateral triangle. If a square IS a rectangle, finding out the perimeter of the square is equivalent to finding out the perimeter of the rectangle, isn’t it? That essentially means a square IS a rectangle, right? All the squares are rectangle but it's not the other way round.ĭon’t you think you have answered your own question here? You have yourself agreed that “All squares are rectangles”. If that's the case then the rectangle will become square. ![]() In reference to statement I, Is it possible that AB=BD in case of a rectangle. The combination of statements is sufficient to find the perimeter of the rectangle. Therefore, perimeter of rectangle = 2(AB + BC) = 2(5√3 + 5) units. If BD = 10, BC = 5 units and CD = 5√3 units. We also know that BC:CD:BD = 1: √3 : 2.įrom statement I alone, we know that BD = 10 units. Possible answer options are C or E.Ĭombining statements I and II, we have the following:įrom statement II, we know that the rectangle has been divided into two “30-60-90” right angled triangles. Statement II alone is insufficient to find the exact lengths of the sides and hence the perimeter of the rectangle ABCD.Īnswer option B can be eliminated. However, knowing the ratio of the sides is not the same as knowing the sides themselves. BC, CD and BD are in the ratio of 1: √3 :2. This means the sides opposite to these angles i.e. This tells us that the rectangle is divided into 2 “30-60-90” right angled triangles. ![]() This means that angle CBD = 60 degrees since angle BCD = 90 degrees. The possible answer options are B, C or E.įrom statement II alone, angle BDC = 30 degrees. ![]() Answer options A and D can be eliminated. Statement I alone is insufficient to find a unique value for the perimeter of the rectangle. In this case, the perimeter of rectangle = 2(5√2+ 5√2 ) = 20√2. IN this case, the perimeter of rectangle = 2(6+8) = 2(14) = 28. Since every angle in a rectangle is a right angle, we have a right angled triangle BDA with BD, the hypotenuse, being 10 units in length. The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.Perimeter of rectangle ABCD = 2(AB + BC).įrom statement I alone, diagonal BD = 10 units. Let's assume, for simplicity, that the sides of the rectangle are and The area of the 3 triangles would then beĪdding these up, we get, and subtracting that from, we get, so the answer is Labelling and as the right and lower midpoints respectively, and redoing all the work above, we get: However this is an alternative if you don't get the above answer. All in all, it's a very good answer though. The above answer is fast, but satisfying, and assumes that the area of is independent of the dimensions of the rectangle. Labelling the right midpoint as, and the bottom midpoint as, we know that, and. Give the rectangle dimensions of and, which is the easiest way to avoid fractions. To quickly solve this multiple choice problem, make the (not necessarily valid, but very convenient) assumption that can have any dimension. If point and the midpoints of and are joined to form a triangle, the area of that triangle is ![]()
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