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## Geometry and Mensuration: Test 19

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Question 1 |

Rama owns a piece of land in the shape of a right triangle. Its hypotenuse is 3 m more than twice the shortest side. If the perimeter of the piece of land is 6 times the shortest side, find the dimensions of the piece of land.

6, 15, 12 | |

5, 12, 13 | |

4, 9, 11 | |

None of these |

Question 1 Explanation:

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)

=> 4x

=>6x

=> 6x (x - 5) = 0

=> x = 5 (As x cannot be zero)

Hence the sides are 5, 12 and 13.

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}= x^{2}+ (3x - 3)^{2}=> 4x

^{2}+ 9 + 12x = x^{2}+ 9x^{2}+ 9 - 18x=>6x

^{2}-30x = 0=> 6x (x - 5) = 0

=> x = 5 (As x cannot be zero)

Hence the sides are 5, 12 and 13.

Question 2 |

A triangle and a parallelogram are constructed on the same base such that their areas are equal. If the altitude of the parallelogram is 100 m, then the altitude of the triangle is

100 m | |

200 m | |

100√2m | |

200√2m |

Question 2 Explanation:

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)

Question 3 |

Let PQR be an acute-angled triangle and RS be the altitude through R. If PQ = 8 and RS= 6, then the distance between the mid-points of PS and QR is

30 | |

25 | |

27 | |

5 |

Question 3 Explanation:

$ \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}$

Question 4 |

The perimeters of two similar triangles ABC and PQR are 36 cm and 24 cm respectively. If PQ = 10 cm, then the length of AB is

16 cm | |

12 cm | |

14 cm | |

15 cm |

Question 4 Explanation:

$ \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}$

Question 5 |

$ \displaystyle \begin{array}{l}ABC\text{ }is\text{ }an\text{ }equilateral\text{ }triangle.\text{ }\\P\text{ }and\text{ }Q\text{ }are\text{ }two\text{ }points\text{ }on\overline{AB}\,\,and\,\,\overline{AC}and~respectively\text{ }such\text{ }that.\text{ }\\\overline{PQ}||\overline{BC},If\overline{PQ}=\text{ }5\text{ }cm\text{ }the\text{ }area\text{ }of\vartriangle APQ\text{ }is:\end{array}$

$ \displaystyle \frac{25}{4}$ | |

$ \displaystyle \frac{25}{\sqrt{3}}$ | |

$ \displaystyle \frac{25\sqrt{3}}{4}$ | |

$ \displaystyle 25\sqrt{3}$ |

Question 5 Explanation:

$ \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}$

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