4 Mark Question

Question 14 Marks

In the given figure $\text{l}\ ||\ \text{m}$ and t is a transversal. If $\angle5=70^\circ$ find the measure of each of the angles $\angle1,\ \angle3,\ \angle4\ \text{and}\ \angle8.$

Answer

Given: $\text{l}\ ||\ \text{m}$ t is a transversal.

$\angle5=70^\circ$
$\angle5=\angle3=70^\circ$ (alternate interior angles)
$\angle5+\angle8=180^\circ$ (linear pair)
$70^\circ+\angle8=180^\circ$
$\angle8=110^\circ$
$\angle1+\angle3=70^\circ$ (vertically opposite angles)
$\angle3+\angle4=180^\circ$ (linear pair)
$\angle4=180-\angle3=180-70=110^\circ$

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Question 24 Marks

In the given figure, $\text{l || m}$ and $\text{p || q}$ Find the measure of each of the angles $\angle\text{a},\angle\text{b},\angle\text{c}\ \text{and}\ \angle\text{d}.$

Answer

Given: $\text{l || m}$$\text{p || q}$
$\angle1=65^\circ$
$\therefore\ \angle1=\angle\text{a}=65^\circ$ (vertically opposite angles)
$\angle\text{a}+\angle\text{d}=180^\circ$ (consecutive interior angles on the same side of a transversal are supplementary)
$\angle\text{d}=180^\circ-65^\circ=115^\circ$
$\angle\text{c}+\angle\text{d}=180^\circ$ (consecutive interior angles on the same side of a transversal are supplementary)
$\angle\text{c}=180^\circ-115^\circ=65^\circ$
$\angle\text{c}+\angle\text{b}=180^\circ$ (consecutive interior angles on the same side of a transversal are supplementary)
$\angle\text{b}=180^\circ-65^\circ=115^\circ$
$\therefore\ \angle\text{a}=65^\circ$
$\angle\text{b}=115^\circ$
$\angle\text{c}=65^\circ$
$\angle\text{d}=115^\circ$

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Question 34 Marks

Two parallel lines I and $m$ are cut by a transversal $t$. If the interior angles of the same side of $t$ be $(2 x-8)^{\circ}$ and $(3 x-7)^{\circ}$, find the measure of each of these angles.

Answer

Given: $\text{l || m}$ t is a transversal.
Let: $\angle1=(2\text{x}-8)^\circ$
$\angle2=(3\text{x}-7)^\circ$
We know that the consecutive interior angles are supplementary.
$\therefore\ \angle1+\angle2=108^\circ$
$(2\text{x}-8)+(3\text{x}-7)=180$
$5\text{x}-15=180$
$5\text{x}= 195$
$\text{x} = 39$
$\angle1=(2\text{x}-8)=(2\times39-8)=70^\circ$
$\angle2=(3\text{x}-7)=(3\times39-7)=110^\circ$

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Question 44 Marks

In the adjoining figure, it is given that $\text{CE || BA},\ \angle\text{BAC}=80^\circ$ and $\angle\text{ECD}=35^\circ.$ Find

$\angle\text{ACF},$
$\angle\text{ACB},$
$\angle\text{ACF}.$

Answer

Given: $\text{CE || BA}$
$\angle\text{BAC}=80^\circ,\ \angle\text{ECD}=35^\circ$

$\angle\text{BAC}= \angle\text{ACE}=80^\circ$ (alternate angles with AC as a transversal)
$\angle\text{ACB}+ \angle\text{ACD}=180^\circ$ (linear pair)

$\angle\text{ACB}+ \angle\text{ACE}+\angle\text{ECD}=180^\circ$
$\angle\text{ACB}+ 80^\circ+35^\circ=80^\circ$
$\angle\text{ACB}+ 65^\circ$

$\text{In}\ \triangle\text{ABC}:$

$\angle\text{BAC}+ \angle\text{ACB}+\angle\text{ABC}=180^\circ$ (angle sum property)
$80^\circ+65^\circ+\angle\text{ABC}=180^\circ$
$\angle\text{ABC}+ 35^\circ$

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Question 54 Marks

In the given figure $\text{l || m}.$ If s and t be transversals such that s is not parallel to t. find the values of x and y.

Answer

From the given figure:

$\angle1=\angle3=50^\circ$ (corresponding angles)
and $\angle1=\text{x}^\circ=180^\circ$ (linear pair)
$\text{x}^\circ=180^\circ-50^\circ=130^\circ$
$\text{x}=130$
$\angle2=\angle4=65^\circ$ (corresponding angles)
and $\angle2=\text{y}^\circ=180^\circ$ (linear pair)
$\text{y}^\circ=180^\circ-65^\circ=115^\circ$
$\text{y}=115$

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Question 64 Marks

In the given figrue, AB || CD and CA has been produced to E so that $\angle\text{BAE}=125^\circ.$
If $\angle\text{BAC}=\text{x}^\circ, \angle\text{ABD}=\text{x}^\circ,\angle\text{BDC}=\text{y}^\circ,$ and $\angle\text{ACD}=\text{z}^\circ,$ find the values of x, y, z.

Answer

Given: AB || CD $\angle BAE =125^{\circ}$
$\angle CAB +\angle BAE =180^{\circ}$
$125^{\circ}+ x ^{\circ}=180^{\circ}$
$x =55$
$x + z =180^{\circ}$(consecutive interior angles on the same side of transversal are supplementary)
$z=180- x =180-55=125$
$y+x=180^{\circ}$(consecutive interior angles on the same side of transversal are supplementary)
$y=180-x=180-55=125$

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