8th Grade > Mathematics
UNDERSTANDING QUADRILATERALS MCQs
Total Questions : 464
| Page 46 of 47 pages
Answer: Option A. ->
:
We know that, diagonals of a rectangle and opposite sides are equal.
All the angles of the rectangle are the right angles.
So, AC = 5 cm
AB = 3 cm
Steps of Construction
Step I Draw AB = 3 cm.
Step II Draw ray BX such that ∠ABX=90∘.
Step III Draw an arc such that AC = 5 cm.
Step IV With B as centre, draw an arc of radius 5 cm. With C as centre, draw another arc of radius 3 cm which intersects the first arc at a point, suppose D.
Step V Join CD and AD.
Thus, ABCD is required rectangle.
:
We know that, diagonals of a rectangle and opposite sides are equal.
All the angles of the rectangle are the right angles.
So, AC = 5 cm
AB = 3 cm
Steps of Construction
Step I Draw AB = 3 cm.
Step II Draw ray BX such that ∠ABX=90∘.
Step III Draw an arc such that AC = 5 cm.
Step IV With B as centre, draw an arc of radius 5 cm. With C as centre, draw another arc of radius 3 cm which intersects the first arc at a point, suppose D.
Step V Join CD and AD.
Thus, ABCD is required rectangle.
Answer: Option A. ->
:
Steps of Construction
Step I Draw a line segment BE = 6 cm
Step II With B as center, draw an arc BR = 5 cm and with E as a centre, draw an arc EA = 7 cm.
Step III Now, draw another arc BA = 9 cm with B as a centre, which cut-off arc AE.
Step IV Draw another arc ER = 5 cm with E as a centre, which cut-off arc BR.
Step V Now join BR, EA and AR.
Thus, we have required quadrilateral BEAR.
Also, AR = 5 cm.
:
Steps of Construction
Step I Draw a line segment BE = 6 cm
Step II With B as center, draw an arc BR = 5 cm and with E as a centre, draw an arc EA = 7 cm.
Step III Now, draw another arc BA = 9 cm with B as a centre, which cut-off arc AE.
Step IV Draw another arc ER = 5 cm with E as a centre, which cut-off arc BR.
Step V Now join BR, EA and AR.
Thus, we have required quadrilateral BEAR.
Also, AR = 5 cm.
Answer: Option A. ->
:
We know that, all sides of a square are equal and each side is perpendicular to adjacent side.
So, AB = BC = CD = DA = 4 cm
Steps of Construction
Step I Draw ¯¯¯¯¯¯¯¯AB=4cm
Step II At B, draw ¯¯¯¯¯¯¯¯¯BX such that ∠ABX=90∘.
Step III From ¯¯¯¯¯¯¯¯¯BX, cut-off BC = 4 cm
Step IV With centre C and radius = 4 cm, draw an arc.
Step V With centre A and radius = 4 cm, draw another arc to intersect the previous arc at D.
Step VI Join DA and CD.
Thus, ABCD is the required square.
:
We know that, all sides of a square are equal and each side is perpendicular to adjacent side.
So, AB = BC = CD = DA = 4 cm
Steps of Construction
Step I Draw ¯¯¯¯¯¯¯¯AB=4cm
Step II At B, draw ¯¯¯¯¯¯¯¯¯BX such that ∠ABX=90∘.
Step III From ¯¯¯¯¯¯¯¯¯BX, cut-off BC = 4 cm
Step IV With centre C and radius = 4 cm, draw an arc.
Step V With centre A and radius = 4 cm, draw another arc to intersect the previous arc at D.
Step VI Join DA and CD.
Thus, ABCD is the required square.
Answer: Option A. ->
:
Since,
We have ,
PO = RU = 5.5 cm, OU = RP = 7.2 cm
(Since the opposite sides of a parallelogram are equal)
Steps of Construction
Step I Draw PO = 5.5 cm.
Step II Construct ∠POX=70∘.
Step III With O as centre and radius OU = 7.2 cm, draw an arc.
Step IV With U as centre and radius UR = 5.5 cm, draw an arc.
Step V With P as centre and radius PR = 7.2 cm, draw an arc to cut the arc drawn in Step IV.
Step VI Join PR and UR.
Hence, POUR is the required parallelogram.
:
Since,
We have ,
PO = RU = 5.5 cm, OU = RP = 7.2 cm
(Since the opposite sides of a parallelogram are equal)
Steps of Construction
Step I Draw PO = 5.5 cm.
Step II Construct ∠POX=70∘.
Step III With O as centre and radius OU = 7.2 cm, draw an arc.
Step IV With U as centre and radius UR = 5.5 cm, draw an arc.
Step V With P as centre and radius PR = 7.2 cm, draw an arc to cut the arc drawn in Step IV.
Step VI Join PR and UR.
Hence, POUR is the required parallelogram.
Answer: Option A. ->
:
Steps of Construction
Step I Taking O as centre and radius=OC = 3 cm, draw a circle.
Step II Join A to C and draw a perpendicular bisector of AC that cuts the circumference of the circle at B and D.
Step III Join B and D.
Step IV Thus, ABCD is a cyclic quadrilateral.
Justification:
In quadrilateral ABCD,
∠B=∠D=90∘ [angle in a semi-circle]
∠A=∠C=90∘
⇒∠B+∠D=180∘
and ∠A+∠C=180∘
⇒opposite angles are supplementary
Since, opposite angles are supplementary, thus the quadrilateral ABCD is a cyclic quadrilateral.
:
Steps of Construction
Step I Taking O as centre and radius=OC = 3 cm, draw a circle.
Step II Join A to C and draw a perpendicular bisector of AC that cuts the circumference of the circle at B and D.
Step III Join B and D.
Step IV Thus, ABCD is a cyclic quadrilateral.
Justification:
In quadrilateral ABCD,
∠B=∠D=90∘ [angle in a semi-circle]
∠A=∠C=90∘
⇒∠B+∠D=180∘
and ∠A+∠C=180∘
⇒opposite angles are supplementary
Since, opposite angles are supplementary, thus the quadrilateral ABCD is a cyclic quadrilateral.
Answer: Option A. ->
:
Steps of Construction
Step I Draw HO = 6 cm.
Step II With H as centre and radius HE = 4 cm, draw an arc.
Step III With O as centre and radius OE = 3 cm, draw an arc intersecting the arc drawn in step II at E.
Step IV With E as centre and radius EM = 6 cm, draw an arc opposite to the side HE.
Step V With O as center and radius OM = 4 cm, draw an arc, intersecting the arc drawn in step IV at M.
Step VI Join HE, OE, EM and OM.
Hence, HOME is the required parallelogram.
:
Steps of Construction
Step I Draw HO = 6 cm.
Step II With H as centre and radius HE = 4 cm, draw an arc.
Step III With O as centre and radius OE = 3 cm, draw an arc intersecting the arc drawn in step II at E.
Step IV With E as centre and radius EM = 6 cm, draw an arc opposite to the side HE.
Step V With O as center and radius OM = 4 cm, draw an arc, intersecting the arc drawn in step IV at M.
Step VI Join HE, OE, EM and OM.
Hence, HOME is the required parallelogram.
Answer: Option A. ->
:
Steps of Construction
Step I Draw AC = 5 cm.
Step II With A as center, draw an arc of the length slightly greater than 12 AC above and below the line segment AC.
Step III With C as center, draw an arc of the same length as in step II above and below the line segment AC which intersects the arcs drawn in step II.
Step IV Join both the intersection points obtained in step III by a line segment which intersects AC at O (say).
Step V With O as centre cut-off OB = OD = 2.5 cm along the bisector line.
Step VI Join AD, CD, AB and CB.
This is the required square ABCD.
:
Steps of Construction
Step I Draw AC = 5 cm.
Step II With A as center, draw an arc of the length slightly greater than 12 AC above and below the line segment AC.
Step III With C as center, draw an arc of the same length as in step II above and below the line segment AC which intersects the arcs drawn in step II.
Step IV Join both the intersection points obtained in step III by a line segment which intersects AC at O (say).
Step V With O as centre cut-off OB = OD = 2.5 cm along the bisector line.
Step VI Join AD, CD, AB and CB.
This is the required square ABCD.
Answer: Option A. ->
:
No,
Given measures are
OA = 5 cm, ∠O=120∘,∠R=105∘ and ∠A=135∘
Here, we see that, ∠O+∠R+∠A=120∘+105∘+135∘=360∘
i.e. the sum of three angles of a quadrilateal is 360∘.
This is impossible, as the total sum of four angles is 360∘ in a quadrilateral.
Hence, this quadrilateral cannot be constructed.
:
No,
Given measures are
OA = 5 cm, ∠O=120∘,∠R=105∘ and ∠A=135∘
Here, we see that, ∠O+∠R+∠A=120∘+105∘+135∘=360∘
i.e. the sum of three angles of a quadrilateal is 360∘.
This is impossible, as the total sum of four angles is 360∘ in a quadrilateral.
Hence, this quadrilateral cannot be constructed.
Answer: Option A. ->
:
No,
Given measures are AB = 3 cm, BC = 4 cm, CD = 5.4 cm,
DA = 5.9 cm and AC = 8 cm
Here, we observe that AB + BC = 3 + 4 = 7 cm and AC = 8 cm
i.e. the sum of two sides of a triangle is less than the third side, which is absurd.
Hence, we cannot construct such a quadrilateral.
:
No,
Given measures are AB = 3 cm, BC = 4 cm, CD = 5.4 cm,
DA = 5.9 cm and AC = 8 cm
Here, we observe that AB + BC = 3 + 4 = 7 cm and AC = 8 cm
i.e. the sum of two sides of a triangle is less than the third side, which is absurd.
Hence, we cannot construct such a quadrilateral.
Answer: Option A. ->
:
We know that, the sum of interior angles of an n polygon =(n−2)×180∘
Here, n = 20, then
Sum =(20−2)×180∘=18×180∘=3240∘
∴ The measure of each interior angle =324020=162∘
:
We know that, the sum of interior angles of an n polygon =(n−2)×180∘
Here, n = 20, then
Sum =(20−2)×180∘=18×180∘=3240∘
∴ The measure of each interior angle =324020=162∘
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