Question
In the given figure, $AB \| CD \| EF , \angle D B G=x, \angle E D H=y, \angle A E B=z, \angle E A B=90^{\circ}$ and $\angle B E F=65^{\circ}$. Find the values of $x , y$ and $z$ .
✓
Answer
$EF \| CD$ and $ED$ is the transversal.
$\therefore \angle F E D+\angle E D H=180^{\circ} \ [$co $-$ interior angles$]$
$\Rightarrow 65^{\circ}+y=180^{\circ}$
$\Rightarrow y=\left(180^{\circ}-65^{\circ}\right)=115^{\circ}$
Now $CH \| AG$ and $DB$ is the transversal
$\therefore x = y =115^{\circ}\ [$corresponding angles$]$
$\therefore \angle A B E+\angle E B G=180^{\circ} \ [$sum of linear pair of angles is $180^{\circ}]$
$\Rightarrow \angle A B E+x=180^{\circ}$
$\Rightarrow \angle A B E+115^{\circ}=180^{\circ}$
$\Rightarrow \angle A B E=\left(180^{\circ}-115^{\circ}\right)=65^{\circ}$
We know that the sum of the angles of a triangle is $180^{\circ}$.
From $\triangle E A B$, we get
$\angle E A B+\angle A B E+\angle B E A=180^{\circ}$
$\Rightarrow 90^{\circ}+65^{\circ}+ z =180^{\circ}$
$\Rightarrow z =\left(180^{\circ}-155^{\circ}\right)=25^{\circ}$
$\therefore x =115^{\circ}, y =115^{\circ} $ and $z =25^{\circ}$
$\therefore \angle F E D+\angle E D H=180^{\circ} \ [$co $-$ interior angles$]$
$\Rightarrow 65^{\circ}+y=180^{\circ}$
$\Rightarrow y=\left(180^{\circ}-65^{\circ}\right)=115^{\circ}$
Now $CH \| AG$ and $DB$ is the transversal
$\therefore x = y =115^{\circ}\ [$corresponding angles$]$
$\therefore \angle A B E+\angle E B G=180^{\circ} \ [$sum of linear pair of angles is $180^{\circ}]$
$\Rightarrow \angle A B E+x=180^{\circ}$
$\Rightarrow \angle A B E+115^{\circ}=180^{\circ}$
$\Rightarrow \angle A B E=\left(180^{\circ}-115^{\circ}\right)=65^{\circ}$
We know that the sum of the angles of a triangle is $180^{\circ}$.
From $\triangle E A B$, we get
$\angle E A B+\angle A B E+\angle B E A=180^{\circ}$
$\Rightarrow 90^{\circ}+65^{\circ}+ z =180^{\circ}$
$\Rightarrow z =\left(180^{\circ}-155^{\circ}\right)=25^{\circ}$
$\therefore x =115^{\circ}, y =115^{\circ} $ and $z =25^{\circ}$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
In Fig, name the following:
i. Five line segments
ii. Five rays
iii. Four collinear points
iv. Two pairs of non-intersecting line segments
i. Five line segments
ii. Five rays
iii. Four collinear points
iv. Two pairs of non-intersecting line segments
The angles of a triangle are in the ratio $2 : 3 : 4.$ Find the angles of the triangle.
→In each of the figures given below, $A B \| C D$. Find the value of $x^{\circ}$ in each other case.
→The following table shows the number of illiterate persons in the age group ($10-58$ years) in a town:
Draw a histogram to represent the above data.
→|
Age group (in years)
|
$10-16$
|
$17-23$
|
$24-30$
|
$31-37$
|
$38-44$
|
$45-51$
|
$50-58$
|
|
Number of illiterate persons
|
$175$
|
$325$
|
$100$
|
$150$
|
$250$
|
$400$
|
$525$
|
The following data on the number of girls $($to the nearest ten$)$ per thousand boys in different sections of the society is given below :
$i.$ Represent the information above by a bar graph.
$ii.$ In the classroom discuss what conclusion can be arrived at from the graph.
→| Section | Number of girls per thousand boys |
|---|---|
| Scheduled caste | $940$ |
| scheduled tribe | $970$ |
| Non $SC/ST$ | $920$ |
| Backward districts | $950$ |
| Non$-$backward districts | $920$ |
| Rural | $930$ |
| Urban | $910$ |
$ii.$ In the classroom discuss what conclusion can be arrived at from the graph.
The image of an object placed at a point $A$ before a plane mirror $LM$ is seen at the point $B$ by an observer at $D$, as shown in the figure. Prove that the image is as far behind the mirror as the object is in front of the mirror.
→The difference between inside and outside surfaces of a cylindrical tube $14\ cm$ long, is $88\ cm^2.$ If the volume of the tube is $176\ cm^3$, find the inner and outer radii of the tube.
→In the given figure, $AB\ ||\ CD$ and $EF\ ||\ GH$. Find the value of $x, y, z$ and $t.$
→In the given figure, $D$ is a point on side $BC$ of a $\triangle\text{ABC}$ and $E$ is a point such that $CD = DE$. Prove that $AB + AC > BE$.
→Draw the graph for each of the equations $x + y = 6$ and $x - y = 2$ on the same graph paper and find the coordinates of the point where the two straight lines intersect.
→