You visited us 4 times! Enjoying our articles? Unlock Full Access!

Lines and Angles

In the adjoin...

Question

In the adjoining figure AB || CD and PQ, QR intersects AB ad CD both at E, F and G, H respectively. Given that $m ∠ P E B = 80^{\circ}$ , $m ∠ Q H D = 120^{\circ}$ and $m ∠ P Q R = x^{\circ}$ , find the value of x:

40^{\circ}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

20^{\circ}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

100^{\circ}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

30^{\circ}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $20^{\circ}$
$∠ P G H = 80^{\circ} \Rightarrow ∠ Q G H = 100^{\circ}$
$∠ Q H D = 120^{\circ} \Rightarrow ∠ C H Q = 60^{\circ}$
$∴ ∠ x + 100 + 60 = 80^{\circ}$
$\Rightarrow x = 20^{\circ}$

Suggest Corrections

Similar questions

Q. In the adjoining figure,

A B | | C D

and

P Q, Q R

intersects

A B

and

C D

both at

E, F

and

G, H

respectively. Given that

∠

PEB

= 80^{o}

∠

QHD

= 120^{o}

and

∠

PQR

= x^{o}

, find the value of

x

Q. In the adjoining figure

A B ∥ C D

and PQ,QR intersects AB and CD both at EF, and G,H respectively. Given that

P E R = 80^{\circ}, Q H D = 120^{\circ}

and

P Q R = x^{\circ}

, then the value of 'x' is

In the adjoining figure, $A B C D$ is a ||gm in which $E$ and $F$ are the midpoints of $A B$ and $C D$ respectively. If $G H$ is a line segment that cuts $A D, E F$ and $B C$ at $G, P$ and $H$ respectively, prove that $G P = P H$ .

Q. In the adjoining figure, ABCD is a || gm in which E and F are the midpoints of AB and CD respectively. If GH is a line segment that cuts AD, EF and BC at G, P and H respectively, prove that GP = PH.

In the above figure, DM, HN and EG are parallel lines and AB || CD. Find the value of angle x using the given angles.

Join BYJU'S Learning Program

In the adjoining figure AB || CD and PQ, QR intersects AB ad CD both at E, F and G, H respectively. Given that m∠PEB=80∘, m∠QHD=120∘ and m∠PQR=x∘, find the value of x:

The correct option is B 20∘∠PGH=80∘⇒∠QGH=100∘ ∠QHD=120∘⇒∠CHQ=60∘ ∴∠x+100+60=80∘ ⇒x=20∘

In the adjoining figure AB || CD and PQ, QR intersects AB ad CD both at E, F and G, H respectively. Given that $m ∠ P E B = 80^{\circ}$ , $m ∠ Q H D = 120^{\circ}$ and $m ∠ P Q R = x^{\circ}$ , find the value of x:

The correct option is B $20^{\circ}$
$∠ P G H = 80^{\circ} \Rightarrow ∠ Q G H = 100^{\circ}$
$∠ Q H D = 120^{\circ} \Rightarrow ∠ C H Q = 60^{\circ}$
$∴ ∠ x + 100 + 60 = 80^{\circ}$
$\Rightarrow x = 20^{\circ}$