Geometry and Mensuration: Test 17

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Geometry and Mensuration: Test 17

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Question 1

In a triangle ABC, the lengths of the sides AB, AC and BC are 3, 5 and 6 cm respectively. If a point D on BC is drawn such that the line AD bisects the ∠A internally, then what is the length of BD?

A	2 cm
B	2.25 cm
C	2.5 cm
D	3 cm

Question 1 Explanation:

geometry-and-mensuration-test-17-question-1-pic-1

Question 2

If P and Q are the mid-points of the sides CA and CD respectively of a triangle ABC right angled at C. Then, the value of 4 (AQ² + BP²) is equal to

A	4 BC²
B	5 AB²
C	2AC²
D	2BC²

Question 2 Explanation:

$ \displaystyle \begin{array}{l}A{{D}^{2}}=\text{ }A{{C}^{2}}+\text{ }C{{D}^{2}}\\In\,\,PCB,\\\begin{array}{*{35}{l}} B{{P}^{2}}=\text{ }P{{C}^{2}}+\text{ }C{{B}^{2}} \\ A{{D}^{2}}+B{{P}^{2}}=\text{ }A{{C}^{2}}+C{{D}^{2}}+P{{C}^{2}}+\text{ }C{{B}^{2}} \\ =\text{ }\left( A{{C}^{2}}+C{{B}^{2}} \right)\text{ }+\text{ }\left( C{{D}^{2}}+\text{ }P{{C}^{2}} \right) \\ =\text{ }A{{B}^{2}}=\text{ }P{{D}^{2}} \\ \end{array}\\=A{{B}^{2}}{{\left( \frac{1}{2}\,AB \right)}^{2}}\\A{{D}^{2}}+\text{ }B{{P}^{2}}=\frac{4A{{B}^{2}}+A{{B}^{2}}}{4}\\Therefore\text{ }4\text{ }\left( A{{D}^{2}}+\text{ }B{{P}^{2}} \right)\text{ }=\text{ }5A{{B}^{2}}\end{array}$

Question 3

If the sides of a right triangle are x,x+1 and x-1,then the hypotenuse is

A	5
B	4
C	1
D	0

Question 3 Explanation:

Since the triangle is a right angled triangle, We have, (x+1)² = x²+(x-1)²
=> x² +2x +1 = x² + x²-2x +1
=> x²- 4x =0
=> x(x-4) =0
=>x = 4 [x cannot be equal to 0 ]
Therefore Hypotenuse = 4+1 = 5

Question 4

Let ABC be an acute-angled triangle and CD be the altitude through C. If AB = 8 and CD= 6, then the distance between the mid-points of AD and BC is

A	30
B	25
C	27
D	5

Question 4 Explanation:

geometry-and-mensuration-test-17-question-4-pic-1

Let P and Q be the mid points of AD and BC respectively.
Draw QR perpendicular on AB. Since Q is the midpoint of BC, so R is the midpoint of BD. Hence by mid point theorem QR = 3.
Also AD = 4 = DB. So DR = PD = 2 Þ PR = 4. Hence by Pythagoras theorem, we have
PQ² = PR² + QR² = 16 + 9 = 25
=>PQ = 5.

Question 5

In ΔABC, ∠B is a right angle, AC =6 cm, and D is the mid-point of AC. The length of BD is: