Practical Geometry Test

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Practical Geometry Test - 8

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Question 1
1 / -0

Angles to be bisected to obtain an angle of $$90^{\circ}$$ are:

A
$$60^{\circ}$$

B
$$60^{\circ}$$ and $$120^{\circ}$$

C
$$120^{\circ}$$ and $$180^{\circ}$$

D
$$0^{\circ}$$ and $$60^{\circ}$$

Solution

Angles to be bisected to obtain an angle of $$ {90}^\circ$$ are $$ {60}^\circ $$ and $$ {120}^\circ $$ as it exactly lies between these two angles.
$$\dfrac{60^\circ+ 120^\circ}2 = 90^\circ$$
Hence, option $$B$$.
Question 2
1 / -0

Suppose you are given a circle. Give the construction to find its centre.

A
The center lies on the perpendicular bisectors of any two chords of the circle

B
The center lies on the mid point of any chord of the circle

C
The center lies on the bisector of the angle betwee any two chords of the circle

D
Both options B & C.

Solution

$$ Given-\quad \\ A\quad circle.\\ To\quad find\quad out-\\ its\quad cenre\quad by\quad construction.$$

$$ Construction-$$

$$(I)\quad Any\quad three\quad points\quad A,\quad B\quad \& \quad C\quad are\quad taken\quad on\quad the\quad circumference.$$

$$ (II)\quad AB\quad \& \quad BC\quad are\quad joined.$$

$$ (III)\quad The\quad right\quad bisectors\quad OM\quad \& \quad ON\quad of\quad AB\quad \& \quad BC\quad are\quad drawn.$$

$$OM\quad \& \quad ON\quad intersect\quad at\quad O.$$

$$ So\quad O\quad is\quad the\quad centre\quad of\quad the\quad circle.$$

$$ Justification-\\ OA,\quad OB\quad \& \quad OC\quad are\quad joined.$$

$$ OA=OB\quad since\quad \quad OM\quad is\quad the\quad right\quad bisector\quad of\quad AB\quad and$$

$$OB=OC\quad since\quad \quad ON\quad is\quad the\quad right\quad bisector\quad of\quad AB.$$

$$ \therefore \quad OA=OB=OC.$$

$$ i.e\quad A,\quad B\quad \& \quad C\quad are\quad equdistant\quad from\quad O\quad since\quad the\quad centre\quad of\quad a\quad circle\\is\quad eqidistant\quad from\quad all\quad the\quad points\quad on\quad the\quad circumference.$$

$$ \therefore \quad O\quad is\quad the\quad centre\quad of\quad the\quad circle.$$

$$Ans-\quad Option\quad A.\\ \\ $$
Question 3
1 / -0

To draw an angle of $$150^{\circ}$$ using a pair of compass and ruler

A
bisect $$120^{\circ}$$ and $$180^{\circ}$$

B
bisect $$60^{\circ}$$ and $$120^{\circ}$$

C
bisect $$0^{\circ}$$ and $$60^{\circ}$$

D
None

Solution
Question 4
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An angle which can be constructed using a pair of compass and ruler is

A
$$20^{\circ}$$

B
$$80^{\circ}$$

C
$$60^{\circ}$$

D
$$110^{\circ}$$

Solution

An angle which can be constructed using a pair of compass and ruler is $$ {60}^{o} $$ as multiples of $${15}^{o} $$ can be drawn using a compass.
Question 5
1 / -0

Bisecting means dividing into two ......... parts.

A
Unequal

B
Equal

C
Trianguiar

D
None of these

Solution
Question 6
1 / -0

An angle $$\angle$$XYZ = $$75^\circ$$ is bisected by an angular bisector YU, then the measure of $$\angle$$UYZ is .........

A
$$37^\circ$$

B
$$37.5^\circ$$

C
$$47.5^\circ$$

D
$$47^\circ$$

Solution
Question 7
1 / -0

The angles is to be bisected to obtain an angle of 90$$^\circ$$ is

A
$$60^\circ$$ and $$45^\circ$$

B
$$60^\circ$$ and $$120^\circ$$

C
$$120^\circ$$ and $$180^\circ$$

D
$$30^\circ$$ and $$60^\circ$$

Solution

Given: angles to be bisected to make $$90^o$$ by using compass and ruler first we will draw $$60^o$$ angle and then making another $$60^o$$ angle from previous angle made to get $$120^o$$. Now if we bisect $$60^o$$ and $$120^o$$ we will get $$90^o$$.
$$\therefore$$ option B is correct
Question 8
1 / -0

An angle of $$15^\circ$$ is drawn using a pair of compass and ruler by bisecting .........

A
$$60^\circ$$

B
$$60^\circ$$ and $$120^\circ$$

C
$$60^\circ$$ and then bisecting it again

D
$$60^\circ$$ and $$180^\circ$$

Solution

To draw an angle of $$15^o$$ by using compass and ruler, first, we will draw the angle of $$60^o$$, then bisecting it to get the angle of $$30^o$$ and then bisecting it again to get the required angle of $$15^o$$.
$$\therefore$$ option C is correct.
Question 9
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A perpendicular is drawn to a line segment $$\overline{IP}$$ at N using protractor and point M is marked on perpendicular then

A
$$\overline{MP}\, \perp \, \overline{NP}$$

B
$$\overline{MN}\, \perp\, P$$

C
$$M\, \perp\, P$$

D
$$\overline{MN}\, \perp \, \overline{NP}$$

Solution

A perpendicular is drawn to a line segment $$ \overline {IP} $$ at $$ N $$ using protractor and point $$ M $$ is marked on perpendicular then $$ \overline {MN} \bot \overline {NP} $$
Question 10
1 / -0

If $$PQ$$ is the perpendicular bisector of $$AB$$, then $$PQ$$ divides $$AB$$ in the ratio:

A
$$1:2$$

B
$$1:3$$

C
$$2:3$$

D
$$1:1$$

Solution

Perpendicular bisector always divides a segment into two equal parts.
Therefore $$PQ$$ divides $$AB$$ into $$1:1$$.