8th Grade > Mathematics
UNDERSTANDING QUADRILATERALS MCQs
Total Questions : 464
| Page 41 of 47 pages
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=106⇒46+2x=106⇒2x=106−46⇒2x=60⇒x=30 m
Hence, the lengths of the other three sides are 23m, 30m and 30m.
:
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=106⇒46+2x=106⇒2x=106−46⇒2x=60⇒x=30 m
Hence, the lengths of the other three sides are 23m, 30m and 30m.
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∘.
Â
In ΔRMI,
∠RMI+∠MRI+∠MIR=180∘ . [Angle sum property]
⇒∠RMI=180∘−55∘−55∘=180∘−110∘=70∘
Also, ∠PMA=∠RMI  [vertically opposite angles]
⇒∠PMA=70∘
:
Given, ∠RAI=35∘.
∴∠PRA=35∘ .  [PR∥AI and AR is transversal]
⇒∠ARI=90∘−∠PRA=90∘−35∘=55∘.
∵RM=IM,∠MRI=∠MIR=55∘.
Â
In ΔRMI,
∠RMI+∠MRI+∠MIR=180∘ . [Angle sum property]
⇒∠RMI=180∘−55∘−55∘=180∘−110∘=70∘
Also, ∠PMA=∠RMI  [vertically opposite angles]
⇒∠PMA=70∘
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∘.
:
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∘.
Answer: Option A. ->
:
Given, ∠RQY=60∘
∠SRQ=∠RQY    [SR∥PY, RQ is the transversal]
⇒∠R=60∘ Â
∴∠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
:
Given, ∠RQY=60∘
∠SRQ=∠RQY    [SR∥PY, RQ is the transversal]
⇒∠R=60∘ Â
∴∠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
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]
:
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]
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∘
:
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∘
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∘.
:
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∘.
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∘
:
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∘
Answer: Option A. ->
:
Yes, ∠MYO=∠RXE
Here, MY and RX are perpendicular to OE.
Since, ∠RXO=90∘⇒∠RXE=90∘ and ∠MYE=90∘⇒∠MYO=90∘.
:
Yes, ∠MYO=∠RXE
Here, MY and RX are perpendicular to OE.
Since, ∠RXO=90∘⇒∠RXE=90∘ and ∠MYE=90∘⇒∠MYO=90∘.
Answer: Option A. ->
:
Yes, RE = OM
Given, MORE is a rectangle. Therefore, opposite sides are equal.
:
Yes, RE = OM
Given, MORE is a rectangle. Therefore, opposite sides are equal.