Constructions Test

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Constructions Test - 13

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Weekly Quiz Competition

Question 1
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To construct a triangle similar to given $$\displaystyle \Delta ABC$$ with sides equal to $$\dfrac75$$ of the sides of $$\displaystyle \Delta ABC$$, a ray $$BX$$ is drawn such that $$\displaystyle \angle CBX$$ is acute angle and $$\displaystyle { B }_{ 1 },{ B }_{ 2 },{ B }_{ 3 },...$$ are marked at equal distances on $$BX$$. The points to be joined in the next step are:

A
$$\displaystyle { B }_{ 12 },C$$

B
$$\displaystyle { B }_{ 5 },C$$ and $$\displaystyle { B }_{ 7 },Q$$

C
$$\displaystyle { B }_{ 7 },C$$ and $$\displaystyle { B }_{ 5 },Q$$

D
$$\displaystyle { B }_{ 2 },C$$

Solution

$$\displaystyle { B }_{ 7 },Q$$ are joined and then nd $$\displaystyle { B }_{ 5 },C$$
Question 2
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If two tangents are drawn at the end points of two radii that are inclined at an angle of $$\displaystyle { 110 }^{ \circ }$$. Find the angle between the tangents.

A
$$\displaystyle { 110 }^{ \circ }$$

B
$$\displaystyle { 55 }^{ \circ }$$

C
$$\displaystyle { 70 }^{ \circ }$$

D
$$\displaystyle { 20 }^{ \circ }$$

Solution

Clearly from the figure, by joining radii to both the point of contact, we get quadrilateral & the sum of opposite anlges of a quadrilateral is $$180°$$
Since, $$180^\circ - \theta = 110^\circ$$
$$\therefore \theta = 180 - 110 = 70^\circ$$
Question 3
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To construct a triangle similar to given $$\displaystyle \Delta ABC$$ with its sides $$\dfrac45$$ of that of $$\displaystyle \Delta ABC$$, locate points $$\displaystyle { X }_{ 1 },{ X }_{ 2 },{ X }_{ 3 },....$$. on ray $$BX$$ at equal distances such that $$\displaystyle \angle ABX$$ is acute. The points to be joined in the next step are:

A
$$\displaystyle { X }_{ 4 },C$$

B
$$\displaystyle { X }_{ 5 },C$$

C
$$\displaystyle { X }_{ 4 },A$$

D
$$\displaystyle { X }_{ 5 },A$$

Solution

$$\triangle PQB$$ is the required triangle.
Since side $$BQ$$ is $$\dfrac45$$ times side $$BA$$.
$$BQ = \dfrac45 \times (BQ+AQ) \Rightarrow 5BQ= 4BQ+4AQ \Rightarrow BQ = 4AQ \Rightarrow \dfrac{AQ}{BQ} = \dfrac14$$
Therefore, $$Q$$ divides $$BA$$ in ratio $$4:1$$.
So, point $$X_5$$ should be connected to $$A$$, and $$X_4Q$$ should be drawn parallel to $$X_5A$$.
Question 4
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To draw a pair of tangents to a circle which are at right angles to each other, it is required to draw tangents at end points of two radii which are inclined at an angle of?

A
$$\displaystyle { 60 }^{ \circ }$$

B
$$\displaystyle { 90 }^{ \circ }$$

C
$$\displaystyle { 120 }^{ \circ }$$

D
$$\displaystyle { 45 }^{ \circ }$$

Solution

Clearly from the figure, by joining radii to both the point of contact, we get quadrilateral & the sum of opposite angles of a quadrilateral is $$180°$$
Since, $$\theta = 90^\circ$$
Angle between the radii $$=\displaystyle { 180 }^{ \circ }-{ 90 }^{ \circ }={ 90 }^{ \circ }$$
Question 5
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If tangents are drawn from the end points of $$2$$ radii that are inclined at an angle $$\displaystyle { 125 }^{ \circ }$$, what is the angle between the tangents?

A
$$\displaystyle { 55 }^{ \circ }$$

B
$$\displaystyle { 110 }^{ \circ }$$

C
$$\displaystyle { 125 }^{ \circ }$$

D
$$\displaystyle { 90 }^{ \circ }$$

Solution

Clearly from the figure, by joining radii to both the point of contact, we get quadrilateral & the sum of opposite anlges of a quadrilateral is $$180°$$
Since, $$180^\circ - \theta = 125^\circ$$
$$\therefore \theta = (180-125)^\circ =55^\circ$$
Question 6
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What should be the angle between corresponding radii such that the tangents don't intersect?

A
$$\displaystyle { 0 }^{ \circ }$$

B
$$\displaystyle { 90 }^{ \circ }$$

C
$$\displaystyle { 180 }^{ \circ }$$

D
$$\displaystyle { 45 }^{ \circ }$$

Solution

Clearly from the figure, by joining radii to both the point of contact, we get quadrilateral & the sum of opposite anlges of a quadrilateral is $$180°$$
The tangents at $$OA$$ and $$OB$$ won't intersect, i.e, the angle between them is $$0^\circ$$. So, angle between the radii $$=(180-0)^\circ = 0^\circ$$
Question 7
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To construct a triangle similar to given $$\displaystyle \Delta ABC$$ with its sides $$\dfrac45$$ of that of $$\displaystyle \Delta ABC$$. Locate points $$\displaystyle { X }_{ 1 },{ X }_{ 2 },{ X }_{ 3 },....$$ on ray $$BX$$ at equal distances such that $$\displaystyle \angle ABX$$ is acute. The minimum number of points to be located on $$BX$$ is:

A
$$14$$

B
$$5$$

C
$$9$$

D
$$20$$

Solution

$$\triangle PQB$$ is the required triangle.
Since side $$BQ$$ is $$\dfrac45$$ times side $$BA$$.
$$BQ = \dfrac45 \times (BQ+AQ) \Rightarrow 5BQ= 4BQ+4AQ \Rightarrow BQ = 4AQ \Rightarrow \dfrac{AQ}{BQ} = \dfrac14$$
Therefore, $$Q$$ divides $$BA$$ in ratio $$4:1$$.
So, the minimum number of points required on ray $$BX$$ is $$1+4=5$$.
Question 8
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Directions For Questions

$$P$$ is outside circle with center $$O$$.
Here are the steps of construction arranged randomly to construct a pair of tangents from?
$$(a)$$ Taking midpoint of $$OP$$ as center, we draw a circle of radius $$\dfrac{OP}2$$
$$(b)$$ Join $$OP$$
$$(c)$$ Join $$P$$ to the points at which the drawn circle touches the circle with center $$O$$
$$(d)$$ Bisects $$OP$$ and get the midpoint.

...view full instructions

Which is last step?

A
$$(a)$$

B
$$(b)$$

C
$$(c)$$

D
$$(d)$$

Solution

The correct order of the steps of construction will be:
$$(b)$$ Join $$OP$$
$$(d)$$ Bisect $$OP$$ and get the midpoint.
$$(a)$$ Taking midpoint of $$OP$$ as center, we draw a circle of radius $$\dfrac{OP}2$$
$$(c)$$ Join $$P$$ to the points at which the drawn circle touches the circle with center $$O$$.
Question 9
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Directions For Questions

Draw a circle with centre $$O$$ and radius $$6\ cm$$. Take a point $$P$$ outside the circle at a distance of $$10\ cm$$ from $$O$$. Draw tangents to the circle from point $$P$$. Let the tangents intersect the circle in points $$A$$ and $$B$$.

...view full instructions

The length of $$AB$$ is

A
$$7.5\ cm$$

B
$$9.6\ cm$$

C
$$10\ cm$$

D
$$8.4\ cm$$

Solution

Step 1: Place a compass on any point $$O$$ on the paper and draw a circle of radius $$6\ cm$$.
Step 2: Mark a point $$P$$ outside the circle at a distance of $$10\ cm$$ from $$O$$.
Step 3: Place the compass on $$P$$, take radius of more than $$5\ cm$$ and draw two arcs on both sides of line $$OP$$. With the same radius, mark two arcs from point $$O$$ which intersect the arcs drawn from $$P$$.
Step 4: Join the intersection points of the arcs to obtain the perpendicular bisector of $$OP$$. Mark the mid point of $$OP$$ as $$M$$.
Step 5: Place the compass on $$M$$ and draw a circle with radius $$=PM=OM$$
Step 6: Mark the intersection points of the circle obtained in step $$5$$ and the original circle as $$A$$ and $$B$$. Join $$P-A$$ and $$P-B$$.

Draw the segment $$AB$$ and measure the length $$AB$$. We get $$AB = 9.6\ cm$$
Question 10
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$$AP$$ and $$BP$$ are the tangents drawn from an external point $$P$$ to a circle with center $$O$$.
$$\angle AOB$$ and $$\angle APB$$ are always ________.

A
complementary

B
supplementary

C
obtuse angles

D
acute angles

Solution

Tangent is perpendicular to the radius at the point of contact.
Therefore $$\angle{OAP}=\angle{OBP}=90^{\circ}$$

We know that sum of all the four angles of a quadrilateral is $$360^{\circ}$$.
So in quadrilateral AOBP,
$$\angle{OAP}+\angle{AOB}+\angle{OBP}+\angle{APB}=360^{\circ}$$

$$90^{\circ}+\angle{AOB}+90^{\circ}+\angle{APB}=360^{\circ}$$
$$\angle{AOB}+\angle{APB}=180^{\circ}$$
Therefore $$\angle{AOB}$$ and $$\angle{APB}$$ are always supplementary.
So, option B is the answer.