Geometry and Mensuration: Test 15

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Geometry and Mensuration: Test 15

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Question 1

In a right-angled triangle, the product of two sides is equal to half of the square of the third side i.e,. hypotenuse. One of the acute angles must be

A	60^o
B	30^o
C	45^o
D	15^o

Question 1 Explanation:

$ \displaystyle \begin{array}{*{35}{l}} Let\text{ }the\text{ }three\text{ }sides\text{ }be\text{ }a,b,c\text{ }units. \\ Where\text{ }c\text{ }is\text{ }the\text{ }hypotenuse. \\ \begin{array}{l}~ab=\frac{1}{2}{{c}^{2}}\\Thus,2ab={{a}^{2}}+{{b}^{2}}Thus\,\,we\,\,can\,\,see\,\,that\,\,a=b.\\i.e.\,\,One\,\,of\,the\,\,angles\,\,is\,\,45.\\The\text{ }correct\text{ }option\text{ }is\text{ }\left( c \right)\end{array} \\ \end{array}$

Question 2

A point D is taken from the side BC of a right-angled triangle ABC, where AB is hypotenuse. Then

A	AB² +CD²= BC² +AD²
B	CD² +BD²= 2 AD²
C	AB² +AC²= 2 AD²
D	AB²= AD² +BD²

Question 2 Explanation:

$ \begin{array}{l}Let\,\,BD=x\,\,and\,\,thus\,CD=BC-x\\A{{B}^{2}}+C{{D}^{2}}=A{{C}^{2}}+B{{C}^{2}}+{{(BC-x)}^{2}}\\=>A{{B}^{2}}+C{{D}^{2}}=A{{D}^{2}}+B{{C}^{2}}\\Correct\,\,option\,\,is\,\,(a)\\\end{array}$

Question 3

In ABC, the internal bisectors of ∠ABC and ∠ACB meet at I and ∠BAC= 50^o. The measure of ∠BIC is

A	105^o
B	115^o
C	125^o
D	130^o

Question 3 Explanation:

$ \displaystyle \begin{array}{l}~\angle BIC\text{ }=\text{ }90\text{ }+\angle BAC/2\\=90\text{ }+50/2\text{ }=\text{ }115.\\Correct\text{ }option\text{ }is\text{ }\left( b \right)\end{array}$

Question 4

Q is a point in the interior of a rectangle ABCD. If QA= 3 cm, QB=4 cm and QC= 5 cm, then the length QD in centimeter is

A	$ \displaystyle 3\sqrt{2}$
B	$\displaystyle 5\sqrt{2}$
C	$ \displaystyle \sqrt{34}$
D	$ \displaystyle \sqrt{41}$

Question 4 Explanation:

$ \displaystyle \begin{array}{l}We\text{ }know\text{ }that\\Q{{D}^{2}}+Q{{B}^{2}}=Q{{C}^{2}}+A{{Q}^{2}}~\\=>~Q{{D}^{2}}=25+9-16=18.\\=>~QD=3\sqrt{2}\end{array}$

Question 5

Each edge of a regular tetrahedron is 3 cm, then its volume is

A	$ \displaystyle \frac{9\sqrt{2}}{4}\,\,c.\,c.$
B	$ \displaystyle 27\sqrt{3}\,\,c.\,\,c.$
C	$ \displaystyle \frac{4\sqrt{2}}{9}\,\,c.\,\,c.$
D	$ \displaystyle 9\sqrt{3}\,\,c.\,\,c.$

Question 5 Explanation:

$ \begin{array}{l}The\,\,formula\,\,for\,this\\\frac{{{a}^{3}}}{6\sqrt{2}}=\frac{27}{6\sqrt{2}}=\frac{3\times 3\times 3}{3\times 2\times \sqrt{2}}\\=\frac{3\times 3\times 3}{3\times 2\times \sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}\\=\frac{9\sqrt{2}}{4}\end{array}$

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