Straight Lines Test 16

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Straight Lines Test 16

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Question 1
1 / -0

The points $$A (2, 9), B (a, 5)$$ and $$C (5, 5) $$ are the vertices of a triangle $$ABC$$ right angled at $$B$$. Find the values of $$a$$ and hence the area of $$\Delta $$ $$ABC$$.

A
$$a = -1$$ , Area $$=$$ $$15$$ sq. unit

B
$$a = 2$$ , Area $$=$$ $$6$$ sq. unit

C
$$a = 0 $$ , Area $$=$$ $$8$$ sq. unit

D
$$a = 1$$ , Area $$=$$ $$12$$ sq. unit

Solution

Given that the vertices of the $$\Delta ABC$$ are $$A\left( { x }_{ 1 },{ y }_{ 1 } \right) =\left( 2,9 \right) , B\left( { x }_{ 2 },{ y }_{ 2 } \right) =\left( a,5 \right) $$ and $$C\left( { x }_{ 3 },{ y }_{ 3 } \right) =\left( 5,5 \right) and \angle B={ 90 }^{ o }$$.
So, $$AC$$ is the hypotenuse.
$$ \therefore$$ by Pythagoras theorem, we have
$$ { AB }^{ 2 }+{ BC }^{ 2 }={ AC }^{ 2 }$$ .........(i)
Now, by distance formula
$$d=\sqrt { { \left( { x }_{ 1 }-{ x }_{ 2 } \right) }^{ 2 }+{ \left( { y }_{ 1 }-y_{ 2 } \right) }^{ 2 } } $$
So, $$AB=\sqrt { { \left( { x }_{ 1 }-{ x }_{ 2 } \right) }^{ 2 }+{ \left( { y }_{ 1 }-y_{ 2 } \right) }^{ 2 } } =\sqrt { { \left( 2-a \right) }^{ 2 }+{ \left( 9-5 \right) }^{ 2 } } =\sqrt { { a }^{ 2 }-4a+20 } $$,
$$BC=\sqrt { { \left( { x }_{ 3 }-{ x }_{ 2 } \right) }^{ 2 }+{ \left( { y }_{ 3 }-y_{ 2 } \right) }^{ 2 } } =\sqrt { { \left( 5-a \right) }^{ 2 }+{ \left( 5-5 \right) }^{ 2 } } =\sqrt { { a }^{ 2 }-10a+25 } $$ and
$$AC=\sqrt { { \left( { x }_{ 3 }-{ x }_{ 1 } \right) }^{ 2 }+{ \left( { y }_{ 3 }-y_{ 1 } \right) }^{ 2 } } =\sqrt { { \left( 5-2 \right) }^{ 2 }+{ \left( 5-9 \right) }^{ 2 } } $$ units $$=\sqrt { 25 } $$ units.
$$ \therefore$$ by (i), we get
$${ a }^{ 2 }-4a+20{ +a }^{ 2 }-10a+25=25$$
$$ \Rightarrow { a }^{ 2 }-7a+10=0$$
$$ \Rightarrow a=5,2$$
We reject $$a=5$$, since $$B$$ and $$C$$ will coincide in that case and $$\Delta ABC$$ will collapse.
So, $$a=2. i.e. AB=\sqrt { { a }^{ 2 }-4a+20 } =\sqrt { 4-8+20 } 3$$ units $$=43$$ units
$$ BC=\sqrt { { a }^{ 2 }-10a+25 } =\sqrt { 4-20+25 } 3$$ units $$=3$$ units.
Now,
We know that the area of the triangle whose vertices are $$\displaystyle (x_{1},y_{1}),(x_{2},y_{2}),$$ and $$(x_{3},y_{3})$$ is $$\cfrac{1}{2}\left | x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1}) +x_{3}(y_{1}-y_{2})\right |$$
Therefore, required area is $$6$$ square units.
Question 2
1 / -0

Area of the triangle formed by the points P(-1.5, 3), Q(6, -2) and R(-3, 4) is 0.

A
True

B
False

C
Ambiguous

D
Data insufficient

Solution

We know that the area of the triangle whose vertices are $$\displaystyle (x_{1},y_{1}),(x_{2},y_{2}),$$ and $$(x_{3},y_{3})$$ is $$\cfrac{1}{2}\left | x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1}) +x_{3}(y_{1}-y_{2})\right |$$
Given $$P(-1.5, 3), Q(6, -2)$$ and $$R(-3, 4)$$
Therefore, area is given by
$$= \frac{1}{2}\times[(-1.5)(-2-4) + 6(4-3)+(-3)(3+2)]$$
$$= \frac{1}{2}\times(9 +6-15)$$
$$= 0$$
Question 3
1 / -0

Directions For Questions

Directions (27-31) : Out of nine cells of a square one cell is left blank and in the rest of the cells numbers are written follow some rule. Get the rule and find out the proper option for the blank cell (?)

...view full instructions

91
64
73
84
76
61
25
60
?

A
66

B
68

C
69

D
71

Solution

In column first $$91+84+25$$$$=$$ $$200$$
In column second $$64+76+60$$ $$=$$$$200$$
In column third $$73+61+66$$ $$=$$200
Question 4
1 / -0

Given: $$\dfrac {20!}{18!}=380$$

A
True

B
False

C
Either

D
Neither

Solution

$$\cfrac { 20! }{ 18! } =?\\ 20!=20\times 19\times 18\times ........3\times 2\times 1\\ =20\times 19\times (18!)\\ \cfrac { 20! }{ 18! } =\cfrac { 20\times 19\times (18!) }{ 18! } =20\times 19=380$$
Hence the value matches so the equation is true.
Question 5
1 / -0

In a class of 80 students it is found that 40 students like Tajmahal and 50 students like Charminar and 18 like both. Then the number of students who do not like neither are.

A
18

B
72

C
50

D
8

Solution

Given that
Total students$$=80$$
Students who like Tajmahal $$=40 \quad n(T)$$
Students who like charminar$$=50 \quad n(C)$$
Students who like both $$=18 \quad n(T\cap C)$$
Say $$'x'$$ like neither then,
$$80=n(T)+n(C)-n(T\cap C)+x\\ \Rightarrow 80=40+50-18+x\\ \Rightarrow x=8$$
Question 6
1 / -0

In the following questions, the numbers are arranged in a particular order or pattern. Choose the missing number from the given alternatives.3, 9, 17, 27, _ ,

A
33

B
37

C
39

D
40

Solution

$$3 + 6 = 9$$
$$9 + 8 = 17$$
$$17 + 10 = 27$$
$$27 + 12 = 39$$
Question 7
1 / -0

In a combination, the ordering of the selected objects is immaterial whereas in a permutation, the ordering is essential.

A
True

B
False

C
Either

D
Neither

Solution

Combination of few things .
The order of selected things is not specified .
Combination is basically selection hence ordering in immaterial .
Permtation of few things the order of selected things is specified .
Permutation basically means selecting and arranging .
Hence ordering is essential .
Question 8
1 / -0

Out of nine cells of a square, one cell is left blank and in the rest of the cells, numbers are written follow some rule. Get the rule and find out the proper option for the blank cell
2
72
56
?
0
42
12
20
30

A
4

B
6

C
8

D
10

Solution

The pattern of numbers are as followed
$$1^2-1=0$$
$$2^2-1=3$$
$$3^2-3=6$$
$$4^2-4=12$$
$$5^2-5=20$$
$$6^2-6=30$$
$$7^2-7=42$$
$$8^2-8=56$$
$$9^2-9=72$$
Question 9
1 / -0

There are $$6$$ boxes numbered $$1, 2 ....... 6$$. Each box is to be filled up either with a red or a green ball in such a way that at least $$1$$ box contains a green ball and the boxes containing green balls are consecutively numbered. The total number of ways in which this can be done is:

A
$$5$$

B
$$21$$

C
$$33$$

D
$$60$$

Solution

(B) The number of ways in which 1 green ball can be put $$=6$$ . The number of ways in which two green balls can be put such that the boxes are consecutive
$$=5$$ $$(i.e., (1, 2),(2, 3),(3, 4),(4, 5),(5, 6))$$

Similarly, the number of ways in which three green balls can be put
$$=4( i.e. (1, 2, 3),(2, 3, 4),(3, 4, 5),(4, 5, 6))$$
$$\cdots \cdots \cdots \cdots \cdots $$ and so on.
$$\therefore $$ Total number of ways of doing this
$$=6+5+4+3+2+1=21$$
Question 10
1 / -0

Identify the progression:
$$A : 4, 7, 10, 13, 16, 19, 22, 25, .....$$
$$B : 4, 7, 9, 10, 13, 14, .........$$

A
Only $$A$$

B
Only $$B$$

C
Both $$A$$ and $$B$$

D
None of these

Solution

Only sequence $$A$$ is progression as its terms follow same pattern

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Straight Lines Test 16

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Straight Lines Test 16

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Question 1

$$a = -1$$ , Area $$=$$ $$15$$ sq. unit

$$a = 2$$ , Area $$=$$ $$6$$ sq. unit

$$a = 0 $$ , Area $$=$$ $$8$$ sq. unit

$$a = 1$$ , Area $$=$$ $$12$$ sq. unit

Solution

Question 2

True

False

Ambiguous

Data insufficient

Solution

Question 3

66

68

69

71

Solution

Question 4

True

False

Either

Neither

Solution

Question 5

18

72

50

8

Solution

Question 6

33

37

39

40

Solution

Question 7

True

False

Either

Neither

Solution

Question 8

4

6

8

10

Solution

Question 9

$$5$$

$$21$$

$$33$$

$$60$$

Solution

Question 10

Only $$A$$

Only $$B$$

Both $$A$$ and $$B$$

None of these

Solution

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