Straight Lines Test 16

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Straight Lines Test 16

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Weekly Quiz Competition

Question 1
1 / -0

In figure, if the midpoints of segments $$\overline{GH}, \overline{JK}$$, and $$\overline{LM}$$ are connected, calculate the area of the resulting triangle.

A
$$20$$

B
$$23$$

C
$$26$$

D
$$33$$

Solution

Midpoint of $$GH$$ is $$\left (0, \dfrac {15}{4}\right)$$
Midpoint of $$JK$$ is $$(-5,-2)$$
Midpoint of $$ML$$ is $$(3,-2)$$
If the coordinates of triangle is $$(a,b) , (c,d) , (e,f)$$ then the area formed by the coordinates of triangle is $$\dfrac{1}{2} \times |a(d-f) + c(f-b) + e(b-d)|$$
Therefore, the area enclosed by those midpoints will be
$$ \dfrac12 \times \left |0(-2+2) -5\left (-2-\dfrac {15}{4}\right) +3\left (\dfrac {15}{4}+2\right)\right| $$
$$= \dfrac {1}{2} \times |0+46| $$
$$= 23$$
Question 2
1 / -0

Calculate the area of a triangle with vertices $$(1, 1), (3, 1)$$ and $$(5, 7)$$.

A
$$6$$

B
$$7$$

C
$$9$$

D
$$10$$

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 $$\cfrac{1}{2}\left | x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1}) +x_{3}(y_{1}-y_{2})\right |$$
Area of triangle $$=\dfrac 12 \left|1(21-5)-1(7-5)+1(1-3) \right|$$
$$=\dfrac 12 \left| 16-2-2\right| = 6$$
Hence, area of the triangle with the given coordinates is $$6$$.
Question 3
1 / -0

If P is a point on the line $$y = 2x$$ in the first quadrant, and the distance from the origin to point P is $$5$$, find the approximate coordinates of point P.

A
$$(2.24, 4.47)$$

B
$$(3.00, 6.00)$$

C
$$(4.00, 8.00)$$

D
$$(4.47, 2.24)$$

E
$$(4.72, 2.36)$$

Solution

Any point on the line $$y=2x$$ can be represented as $$(x,2x)$$. Now the distance of the point from the origin is 5 units.
Hence $$\sqrt{(x^{2} -0)+((2x)^{2}-0)}=5$$
$$\Rightarrow$$ $$\sqrt{5}x=5$$ or $$x=\sqrt{5}$$.
Hence the required point is $$(x,2x)=(\sqrt{5},2\sqrt{5})$$ $$\approx (2.24, 4.47)$$.
Question 4
1 / -0

In the figure, calculate the distance from the midpoint to $$\overline{EF}$$ to the midpoint of $$\overline{GH}$$.

A
$$5.408$$

B
$$5.454$$

C
$$5.568$$

D
$$5.590$$

E
$$5.612$$

Solution

Mid point of point $$E (-3,3)$$ and $$F (-2,-4)$$ is $$\left ( \dfrac{-3-2}{2} \right ),\left ( \dfrac{3-4}{2} \right )=\dfrac{-5}{2},\dfrac{-1}{2}=-2.5,-0.5$$ ....(hint: using mid-point formula)
And mid point of $$G (1,-2)$$ and $$H ( 5,3) $$ is $$\left ( \dfrac{1+5}{2} \right ),\left ( \dfrac{3-2}{2} \right )=\dfrac{6}{2},\dfrac{1}{2}= 3,0.5$$
Then distance between mid point $$EF$$ to mid point $$GH =$$ $$\sqrt{(-2.5-3)^{2}+(-0.5-0.5)^{2}}=\sqrt{(5.5)^{2}+(-1)^{2}}=\sqrt{30.25+1}=\sqrt{31.25}=5.590$$
Question 5
1 / -0

Find the area of a triangle whose vertices are $$(0, 6\sqrt {3}), (\sqrt {35}, 7)$$, and $$(0, 3)$$.

A
$$15.37$$

B
$$17.75$$

C
$$21.87$$

D
$$25.61$$

E
$$39.61$$

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 whose vertices are $$(0,6\sqrt{3}),(\sqrt{35},7) ,(0,3)$$
$$=\dfrac{1}{2}\left [ 0(7-3)+\sqrt{35}(3-6\sqrt{3})+0(6\sqrt{3}-7)) \right ]$$
$$=\dfrac{1}{2}\left [ \sqrt{35}(3-6\sqrt{3}) \right ]$$
$$=\dfrac{1}{2}\left [ 5.91(3-6\times 1.73) \right ]$$
$$=\dfrac{1}{2}\times 5.91\times 7.40$$
$$=\dfrac{1}{2}\times 43.74$$
$$=21.87$$
Question 6
1 / -0

The area of the triangle with coordinates $$(1, 2), (5, 5)$$ and $$(k, 2)$$ is $$15$$ square units. Calculate a possible value for $$k$$.

A
$$-10$$

B
$$-9$$

C
$$-5$$

D
$$5$$

E
$$6$$

Solution

Area of triangle having vertices $$(x_1,y_1), (x_2,y_2)$$ and $$(x_3,y_3)$$ is given by
$$ = \dfrac{1}{2} \times [ x_1 (y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) ] $$
Therefore,
Area of triangle is $$15 = \dfrac {1}{2} \times |1(5-2) + 5(2-2) + k(2-5)| $$
$$ \Rightarrow \dfrac {1}{2} \times |3+0-3k| = \dfrac {1}{2} \times |3-3k| = 15$$
$$\Rightarrow |3-3k|=30$$ => $$|1-k| = 10$$
$$\Rightarrow 1-k = 10$$ and $$1-k=-10$$
$$\Rightarrow k=-9$$ and $$k=11$$
Question 7
1 / -0

Find the perimeter of the triangle with vertices $$(-2, 3), (4, 3),$$ and $$(6, -3)$$.

A
$$4\sqrt {11}$$

B
$$18\sqrt {10}$$

C
$$10 + 4\sqrt {5}$$

D
$$16 + 2\sqrt {10}$$

E
$$16 + 3\sqrt {7}$$

Solution

For perimeter, we have to find length of each side of triangle.
The distance between $$(-2,3)$$ and $$(4,3)$$ is $$\sqrt{36} = 6$$
The distance between $$(-2,3)$$ and $$(6,-3)$$ is $$\sqrt{64+36} = 10$$
The distance between $$(4,3)$$ and $$(6,-3)$$ is $$\sqrt{4+36} = 2 \sqrt{10}$$
$$\therefore$$ perimeter is $$6+10+2 \sqrt{10} = 16+2 \sqrt{10}$$.
Question 8
1 / -0

In fig., the area of triangle ABC (in sq. units) is:

A
$$15$$

B
$$10$$

C
$$7.5$$

D
$$2.5$$

Solution

Given: Coordinates of Point $$A (1,3) ,B (-1,0)$$ and $$C (4,0)$$
Construction: Drop a perpendicular from $$A$$ on $$x-$$ axis, which meets x-axis at $$D\equiv(1,0)$$
Now in $$\Delta ADC, AD = 3, DC = 3$$
Area of $$\Delta ADC = \dfrac12\times DC\times AD$$
$$= \dfrac12\times3\times3 = \dfrac92 \ cm^2$$

Now in $$\Delta ADB, AD = 3, DB = 2$$
Area of $$\Delta ADB = \dfrac12\times DB\times AD$$
$$= \dfrac12\times2\times3 = 3 \ cm^2$$

Area of $$\Delta ABC =$$ Ara of $$\Delta ADC + $$ Area of $$\Delta ABD$$
$$ = \dfrac92 + 3 = \dfrac{15}2 = 7.5\ cm^2$$
Question 9
1 / -0

In the XY-coordinate plane, how many points are at the distance of 4 units from the origin?

A
One

B
Two

C
Three

D
Four

E
More than four

Solution

There are an infinite number of points that are a distance of 4 from the origin.
That’s a circle centered on the origin with a radius of 4.
Then answer is option (E) more then four
Question 10
1 / -0

Find the value of $$x$$, so that the three points, $$(2, 7), (6, 1), (x, 0)$$ are collinear.

A
$$7$$

B
$$4 \dfrac{1}{2}$$

C
$$10$$

D
$$6 \dfrac{2}{3}$$

Solution

The given points are $$(2,7) , (6.1)$$ and $$(x,0)$$
of the points are collinear , they will lie on the same line ,i.e, they will not form triangle .
Area of $$\triangle$$ABC = 0
$$\Longrightarrow \dfrac { 1 }{ 2 } \left[ { x }_{ 1 }({ y }_{ 2 }-{ y }_{ 3 })+{ x }_{ 2 }({ y }_{ 3 }-{ y }_{ 1 })+{ x }_{ 3 }({ y }_{ 1 }-{ y }_{ 2 }) \right] =0\\ =\dfrac { 1 }{ 2 } \left[ 2(1-0)+6(0-7)+x(7-1) \right] =0\\ \dfrac { 1 }{ 2 } \left[ 2-42+6x \right] =0\\ -40+6x=0\Longrightarrow x=\dfrac { 40 }{ 6 } =6\dfrac { 2 }{ 3 } =6.666$$

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Straight Lines Test 16

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Straight Lines Test 16

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SHARING IS CARING

Question 1

$$20$$

$$23$$

$$26$$

$$33$$

Solution

Question 2

$$6$$

$$7$$

$$9$$

$$10$$

Solution

Question 3

$$(2.24, 4.47)$$

$$(3.00, 6.00)$$

$$(4.00, 8.00)$$

$$(4.47, 2.24)$$

$$(4.72, 2.36)$$

Solution

Question 4

$$5.408$$

$$5.454$$

$$5.568$$

$$5.590$$

$$5.612$$

Solution

Question 5

$$15.37$$

$$17.75$$

$$21.87$$

$$25.61$$

$$39.61$$

Solution

Question 6

$$-10$$

$$-9$$

$$-5$$

$$5$$

$$6$$

Solution

Question 7

$$4\sqrt {11}$$

$$18\sqrt {10}$$

$$10 + 4\sqrt {5}$$

$$16 + 2\sqrt {10}$$

$$16 + 3\sqrt {7}$$

Solution

Question 8

$$15$$

$$10$$

$$7.5$$

$$2.5$$

Solution

Question 9

One

Two

Three

Four

More than four

Solution

Question 10

$$7$$

$$4 \dfrac{1}{2}$$

$$10$$

$$6 \dfrac{2}{3}$$

Solution

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