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8th Grade > Mathematics

UNDERSTANDING QUADRILATERALS MCQs

Total Questions : 464 | Page 41 of 47 pages
Question 401.


Question 143
A playground in the town is in the form of a kite. The perimeter is 106 m. If one of its sides is 23 m, what are the lengths of other three sides?


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Answer: Option A. ->
:
In a kite, two pairs of adjacent sides are equal. Since we know that one side measures 23 m in length, we can assume that the side adjacent to it will also be of the same length.
Let the length of the opposite side be x m. Then, the length of the side adjacent to this side will also be x m.
Perimeter of the playground = 106 m
23+23+x+x=10646+2x=1062x=106462x=60x=30 m
Hence, the lengths of the other three sides are 23m, 30m and 30m.
Question 402.


Question 145
In rectangle PAIR, find ∠ARI,∠RMI and ∠PMA.
Question 145In Rectangle PAIR, Find ∠ARI,∠RMI And ∠PMA...


 Discuss Question
Answer: Option A. ->
:
Given, ∠RAI=35∘.
∴∠PRA=35∘  .   [PR∥AI and AR is transversal]
⇒∠ARI=90∘−∠PRA=90∘−35∘=55∘.
∵RM=IM,∠MRI=∠MIR=55∘.
 
In ΔRMI,
∠RMI+∠MRI+∠MIR=180∘ .  [Angle sum property]
⇒∠RMI=180∘−55∘−55∘=180∘−110∘=70∘
Also, ∠PMA=∠RMI  [vertically opposite angles]
⇒∠PMA=70∘
Question 403.


Question 146
In parallelogram ABCD, find ∠B,∠C and ∠D.
Question 146In Parallelogram ABCD, Find ∠B,∠C And ∠D....


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Answer: Option A. ->
:
In a parallelogram, the opposite angles are equal, therefore ∠C=∠A=80∘.
Also, adjacent angles are supplementary.
∴∠A+∠B=180∘80∘+∠B=180∘∠B=180∘−80∘⇒∠B=100∘
Now, ∠B=∠D [Opposite angles are equal]
∴∠D=100∘.
Question 404.


Question 147
In parallelogram PQRS, O is the mid-point of SQ. Find ∠S,∠R, PQ, QR and diagonal PR.
Question 147In Parallelogram PQRS, O Is The Mid-point Of SQ....


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Answer: Option A. ->
:
Given, ∠RQY=60∘
∠SRQ=∠RQY      [SR∥PY, RQ is the transversal]
⇒∠R=60∘   
∴∠S=180∘−∠R=180∘−60∘=120∘     [∵ adjacent angles are supplementary]
Also, SR = 15 cm
⇒ PQ = 15 cm   [∵ opposite sides of a parallelogram are equal]
And PS = 11 cm
⇒ QR = 11 cm   [∵ opposite sides of a parallelogram are equal]
PR is bisected by SQ, so PR=2×PO=2×6=12 cm
Question 405.


Question 148
In rhombus BEAM, find ∠AME and ∠AEM.
Question 148In Rhombus BEAM, Find ∠AME And ∠AEM.


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Answer: Option A. ->
:
Given, ∠BAM=70∘.
We know that, in rhombus, diagonals bisect each other at right angles.
∴∠BOM=∠BOE=∠AOM=∠AOE=90∘.
Now, in ΔAOM.
∠AOM+∠AMO+∠OAM=180∘ .   [angle sum property of triangle]
⇒90∘+∠AMO+70∘=180∘⇒∠AMO=180∘−90∘−70∘⇒∠AMO=20∘
Also, AM = BM = BE = EA.
In ΔAME, we have,
AM = EA
∴∠AME=∠AEM=20∘     [∵ equal sides make equal angles]
Question 406.


Question 149
In parallelogram FIST, find ∠SFT,∠OST and ∠STO.
Question 149In Parallelogram FIST, Find ∠SFT,∠OST And âˆ...


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Answer: Option A. ->
:
Given, ∠FIS=60∘
Now, ∠FTS=∠FIS=60∘    [∵ opposite angles of a parallelogram are equal]
Now, FT∥IS and TI is a transversal, therefore ∠FIO=∠STO=35∘  [alternate angles]
Also, ∠FOT+∠SOT=180∘    [linear pair]
⇒110∘+∠SOT=180∘
⇒∠SOT=70∘
In ΔTOS,∠TSO+∠OTS+∠TOS=180∘   [angle sum property of triangle]
∴∠OST=180∘−(70∘+35∘)=75∘
In ΔFST,∠SFT+∠FTS+∠TSF=180∘  [angle sum property of triangle]
⇒∠SFT=180∘−(60∘+75∘)∴∠SFT=45∘
Question 407.


Question 150
In the given parallelogram YOUR, ∠RUO=120∘ and OY is extended to point S, such that ∠SRY=50∘. Find ∠YSR.
Question 150In The Given Parallelogram YOUR, ∠RUO=120∘ A...


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Answer: Option A. ->
:
Given, ∠RUO=120∘ and ∠SRY=50∘
∠RYO=∠RUO=120∘   [∵ opposite angles of a parallelogram]
Now, ∠SYR=180∘−∠RYO    [linear pair]
=180∘−120∘=60∘
In ΔSRY,
By the angle sum property of a triangle, ∠SRY+∠RYS+∠YSR=180∘
⇒50∘+60∘+∠YSR=180∘⇒∠YSR=180∘−(50∘+60∘)=70∘.
Question 408.


Question 151
In kite WEAR, ∠WEA=70∘ and ∠ARW=80∘. Find the remaining two angles.
Question 151In Kite WEAR, ∠WEA=70∘ And ∠ARW=80∘. Fin...
 


 Discuss Question
Answer: Option A. ->
:
Given, in a kite WEAR, ∠WEA=70∘,∠ARW=80∘
Now, by the interior angle sum property of a quadrilateral,
∠RWE+∠WEA+∠EAR+∠ARW=360∘⇒∠RWE+70∘+∠EAR+80∘=360∘⇒∠RWE+∠EAR=360∘−(70∘+80∘)⇒∠RWE+∠EAR=360∘−150∘⇒∠RWE+∠EAR=210∘……(i)
In ΔRAW, ∠RWA=∠RAW      [∵RW=RA]……(ii)
In ΔEAW, ∠AWE=∠WAE      [∵WE=AE]……(iii)
On adding Eqs. (ii) and (iii), we get ∠RWA+∠AWE=∠RAW+∠WAE
⇒∠RWE=∠RAE
From Eq. (i),
2∠RWE=210∘∠RWE=105∘⇒∠RWE=∠RAE=105∘
Question 409.


Question 152 (ii)
A rectangle MORE is shown below.
Question 152 (ii)A Rectangle MORE Is Shown Below.Answer The ...
Answer the following questions by giving appropriate reason.
Is ∠MYO=∠RXE?


 Discuss Question
Answer: Option A. ->
:
Yes, ∠MYO=∠RXE
Here, MY and RX are perpendicular to OE.
Since, ∠RXO=90∘⇒∠RXE=90∘ and ∠MYE=90∘⇒∠MYO=90∘.
Question 410.


Question 152 (i)
A rectangle MORE is shown below.
Question 152 (i)A Rectangle MORE Is Shown Below.Answer The F...
Answer the following questions by giving an appropriate reason.
Is RE = OM?
 


 Discuss Question
Answer: Option A. ->
:
Yes, RE = OM
Given, MORE is a rectangle. Therefore, opposite sides are equal.

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