Straight Lines Test 20

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Straight Lines Test 20

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Weekly Quiz Competition

Question 1
1 / -0

In the following question, the numbers/letters are arranged based on some pattern or principle. Choose the correct answer for the term marked by the symbol (?)

A
$$82$$

B
$$81$$

C
$$48$$

D
$$26$$

Solution

$$\rightarrow 1^2+1=2,\;3^2+1=10,\;5^2+1=26$$
$$\rightarrow 2^2+1=5,\;4^2+1=17,\;6^2+1=37$$
$$\rightarrow 5^2+1=26,\;7^2+1=50,\;9^2+1=82$$
Question 2
1 / -0

Directions For Questions

In this type of questions certain numbers are given, out of which all except one are alike in some manner while one is different, and this number is the answer.

...view full instructions

Choose the one which is different from other

A
119

B
136

C
147

D
153

Solution

Only 119 has different factors 7 and 17 and no factor is repeated
Question 3
1 / -0

Find the missing letters
A, B, D, G, .....

A
L

B
M

C
J

D
K

Solution

Pattern is $$\displaystyle ^{1}A,^{1}A+1,^{2}B+2,^{4}D+3,^{7}G+4,^{11}K$$
$$\displaystyle \therefore $$ Missing letter = K
Question 4
1 / -0

Find the missing letters
M, N, O, L, R, I, V, ........

A
H

B
K

C
E

D
S

Solution

Pattern is combination of two patterns M, O, R, V, A, .... and N, L, I, H, .......
$$\displaystyle ^{13}M,^{14}N,^{15}O,^{11}L,^{18}R,^{9}I,^{22}V,^{8}H,$$ .....
$$\displaystyle \therefore $$ Missing letter = H
Question 5
1 / -0

The area of triangle formed by $$(0, 0), (0, a)$$ and $$(b, 0)$$ is .......... .

A
$$ab$$

B
$$\displaystyle \frac{ab}{2}$$

C
$$\left | \displaystyle \frac{ab}{2} \right |$$

D
$$\left | ab \right |$$

Solution

The area of a triangle formed by joining the points $$(x_1, y_1)$$, $$(x_2, y_2)$$ and $$(x_3, y_3)$$ is

$$A=\dfrac { 1 }{ 2 } \bigg| { y }_{ 1 }({ x }_{ 2 }-{ x }_{ 3 })+{ y }_{ 2 }({ x }_{ 3 }-{ x }_{ 1 })+{ y }_{ 3 }({ x }_{ 1 }-{ x }_{ 2 }) \bigg|$$

Therefore, the area of a triangle formed by joining the points $$(0, 0)$$, $$(0, a)$$ and $$(b, 0)$$ is:

$$A=\dfrac { 1 }{ 2 } \bigg| 0(0-b)+a(b-0)+0(0-0) \bigg| $$

$$=\dfrac { 1 }{ 2 } \bigg| 0+ab \bigg| $$

$$=\bigg| \dfrac { ab }{ 2 } \bigg|$$

Hence, the area of the triangle is $$\left| \dfrac { ab }{ 2 } \right|$$
Question 6
1 / -0

The midpoints of the sides of triangle $$ABC$$ are $$ (-1,-2), (6,1)$$ and $$(3,5) $$. The area of $$\displaystyle \triangle ABC$$ is ____ square units.

A
$$72$$

B
$$73$$

C
$$74$$

D
$$75$$

Solution

Area of triangle having vertices $$(x_1,y_1), (x_2,y_2)$$ and $$(x_3,y_3)$$ is given by
Area $$ = \dfrac{1}{2} \times [ x_1 (y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) ] $$
Area of triangle formed with midpoints
$$\displaystyle =\frac{1}{2}\left [ -1\left ( 1-5 \right )+6\left ( 5+2 \right )+3\left ( -2-1 \right ) \right ]$$
$$\displaystyle =\frac{37}{2}$$
$$\displaystyle \therefore $$ Area of $$\displaystyle \Delta ABC=4\times \frac{37}{2}=74$$ square units.
$$[\because$$ Area of triangle formed by joining mid-points of the sides of given triangle is $$\left(\dfrac{1}{4}\right)^{th}$$ of the area of original triangle $$]$$
Question 7
1 / -0

There is no symmetry in symmetry in legs of Fig

A

B

C

D

E

Solution

There is no symmetry in legs of Fig.(e)
Question 8
1 / -0

$$L, M$$ and $$N$$ are the midpoints of the sides $$BC, CA$$ and $$AB$$ respectively of triangle $$ABC$$. If the vertices are $$A(3,-4), B(5,-2)$$ and $$C(1,3)$$ the area of $$\displaystyle \triangle LMN$$ is ____ square units.

A
2.25

B
2.5

C
2.75

D
3

Solution

Area of triangle having vertices $$(x_1,y_1), (x_2,y_2)$$ and $$(x_3,y_3)$$ is given by
$$ = \dfrac{1}{2} \times [ x_1 (y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) ] $$

Let $$(x_1,y_1)$$ $$=(3,-4)$$, $$(x_2,y_2)$$ $$=(5,-2)$$ and $$(x_3,y_3)$$ $$=(1,3)$$

Area of $$\displaystyle \Delta ABC=\dfrac{1}{2}\left [ 3\left ( -2-3 \right )+5\left ( 3+4 \right )+1\left ( -4+2 \right ) \right ]$$

$$=\dfrac{1}{2}\left ( -15+35-2 \right )$$

$$=9$$
$$\displaystyle \therefore $$ Area of $$\displaystyle \Delta LMN=\frac{1}{4}\times 9=2.25$$ square units.
Question 9
1 / -0

- T - CN - P - NT - C

A
PNTCT

B
TCCTN

C
NPTCP

D
NPCPT

Solution

The series is N T P C / N T P C / N T P C
Question 10
1 / -0

The area of the triangle formed by the points $$(2, 6), (10, 0)$$ and $$(0, k)$$ is zero square units. Find the value of $$k.$$

A
$$\dfrac {15}{2}$$

B
$$\dfrac 32$$

C
$$\dfrac 72$$

D
$$\dfrac {13}{2}$$

Solution

The area of the triangle is:

$$A=\dfrac { 1 }{ 2 } \left| { y }_{ 1 }({ x }_{ 2 }-{ x }_{ 3 })+{ y }_{ 2 }({ x }_{ 3 }-{ x }_{ 1 })+{ y }_{ 3 }({ x }_{ 1 }-{ x }_{ 2 }) \right|$$

We are given that the area of the triangle is $$0$$ and the points are $$(2,6)$$, $$(10,0)$$ and $$(0,k)$$, therefore,

$$\Rightarrow 0=\dfrac { 1 }{ 2 } \left| 6(10-0)+0(0-2)+k(2-10) \right| \\ \Rightarrow 0=\dfrac { 1 }{ 2 } \left| 60-0-8k \right| \\ \Rightarrow 0=\dfrac { 1 }{ 2 } \left| 60-8k \right| $$

$$\Rightarrow 60-8k=0\\ \\ \Rightarrow 8k=60\\ \Rightarrow k=\dfrac { 60 }{ 8 } =\dfrac { 15 }{ 2 }$$

Hence, $$k=\dfrac { 15 }{ 2 }$$.