592 /www/nrich/html/content/98/08/six5/ 1 2011-02-01T00:00:01 <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> <p>A trapezium is divided into four triangles by its diagonals. Suppose the two triangles containing the parallel sides have areas <em>a</em> and <em>b</em>, what is the area of the trapezium?</p> </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"><br></br> <p><span class="editorial">A trapezium is divided into four triangles by its diagonals. Suppose the two triangles containing the parallel sides have areas</span> <em class="editorial">a</em> <span class="editorial">and</span> <em class="editorial">b</em> <span class="editorial">, what is the area of the trapezium?</span></p> <p class="editorial">The following solution was done by Ling Xiang Ning, Allan from Tao Nan School, Singapore</p> <mdo:image alt="Same Height diagram" src="trapezium.gif"></mdo:image><br></br> <p>First note that triangles SPR and SQR are equal in area (same base and height) so triangles SPT and RQT are equal in area; suppose this area is <em>c</em> .</p> <p>Now triangles SPT and TPQ have the same height (with their common base on SQ) and the ratio of their areas is:<br></br> $$ \frac{c}{b} = \frac{\text{Area}(SPT)}{\text{Area}(TPQ)} = \frac{ST}{TQ} $$<br></br> $$ \frac{a}{c} = \frac{\text{Area}(SRT)}{\text{Area}(TRQ)} = \frac{ST}{TQ} $$<br></br> hence $ c/b = a/c $.<br></br> <br></br> Then $ c^2 = ab$ and $c = \sqrt{ab}$</p> <p>Therefore, the total area of the trapezium is $ a + b + 2\sqrt{ab}$.</p></mdoxml> 2 4 0 0 0 1 0 A trapezium is divided into four triangles by its diagonals. Suppose the two triangles containing the parallel sides have areas a and b, what is the area of the trapezium? Measures and Mensuration Area 2D Geometry, Shape and Space Trapezia