Lakshya Education MCQs

Question:

In a quadrilateral HOPE, PS and ES are bisectors of P and E respectively. Give reason.
Options:

: Data insufficient

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More Questions on This Topic :

Question 1.

___ is a regular quadrilateral.

: Square
Since in square, all the sides are of equal length and all angles are equal.
Question 2.

A quadrilateral can be drawn, if all four sides and one angle is known.

: True
A quadrilateral can be drawn, if all four sides and one angle is known.
Question 3.

If opposite angles of a quadrilateral are equal, it must be a parallelogram.

: True
If opposite angles are equal, it has to be a parallelogram.
Question 4.

The closed curve which is also a non-intersecting polygon, is:

a)



b)



c)



d)


: Figure (a) is a non-intersecting polygon as no two line segments intersect each other.
Question 5.

In parallelogram MODE, the bisectors of M and O meet at Q. Find the measure of MQO

: Let MODE be a parallelogram and Q be the point of intersection of the bisector of M and O

Since, MODE is a parallelogram,
EMO+DOM=180 [ adjacent angles are supplementary]
12EMO+12DOM=90 [dividing both sides by 2]
QMO+QOM=90 ........ (i)
Now, in ΔMOQ,
QOM+QMO+MQO=180 [angle sum property of triangle]
90+MQO=180 [from Eq. (i)]
MQO=18090=90
Question 6.

A regular pentagon ABCDE and a square ABFG are formed on opposite sides of AB. Find BCF?

:
Given, ABCDE is a regular pentagon.
Then, measure of each interior angle of the regular pentagon
=SumofinterioranglesNumberofsides=(x2)×1805=(52)×1805=5405=108CBA=108
Join CF.
Now, FBC=360(90+108)=360198=162
In ΔFBC, by the angle sum property, we have
FBC+BCF+BFC=180BCF+BFC=180162BCF+BFC=18
Since, ΔFBC is an isosceles triangle and BF = BC.
BCF=BFC=9

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