Geometry and Mensuration: Test 19

Let the length of the shortest side be x.
Thus, the hypotenuse be 2x+3.
Let y be the third side.
2x+3+x+y=6x
=>y=3x-3
Thus the length of the sides (in terms of x) are x, 3x-3 and 2x+3.
Now (2x + 3)² = x² + (3x - 3)²
=> 4x² + 9 + 12x = x² + 9x² + 9 - 18x
=>6x² -30x = 0
=> 6x (x - 5) = 0
=> x = 5 (As x cannot be zero)
Hence the sides are 5, 12 and 13.

Since the base is the same therefore to get the same area, the altitude of the triangle must be double of that of the parallelogram. Thus the correct option is (b)

$ \displaystyle \begin{array}{l}MT\bot PQ\\RS\bot PQ\\\vartriangle QMT\approx \vartriangle QRS\\\because T\,is\,\,the\,\,mid\,of\,QS\\TQ+LP=TS+SL\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\\\,\,\,\,\,\,\,\,\,\,\,\,\,\left( \begin{array}{l}\because QT=ST\\and\,\,LP=SL\end{array} \right)\\LT=\frac{1}{2}\times PQ=4\\MT=\frac{1}{2}\times RS=3\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( \vartriangle QMT\approx \vartriangle QRS \right)\\So\,\,if\,\,MT=3,LT=4\\and\vartriangle LMT\,\,is\,\,right\,\,angled\,\,triangle\\therefore\,\,LM=5\end{array}$

$ \displaystyle \begin{array}{l}~The\text{ }length\text{ }of\text{ }corresponding\text{ }sides\text{ }of\text{ }the\text{ }triangle\text{ }\vartriangle ABC\text{ }and\text{ }\vartriangle PQR\text{ }\\are\text{ }in\text{ }the\text{ }ratio\text{ }of\text{ }36:24\text{ }or\text{ }3:2\text{ }respectively.\\If\text{ }PQ\text{ }=\text{ }10,\text{ }\\AB=\frac{3}{2}\times 10=15.\\The\text{ }correct\text{ }option\text{ }is\text{ }\left( d \right)\end{array}$

$ \displaystyle \begin{array}{l}\vartriangle APQ\text{ }is\text{ }also\text{ }equilateral.\text{ }Thus\text{ }all\text{ }sides\text{ }are\text{ }5\text{ }cm.\\The\text{ }area\text{ }=\frac{{{5}^{2}}\sqrt{3}}{4}=25\frac{\sqrt{3}}{4}\end{array}$