Question

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Answer 1

Answer:

I am telling no. 2 ok

Step-by-step explanation:

VR= 180-45=135

how I am telling you in next line

straight line is 180° there already given SR is 45° so I do sub that two number

ok

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Answer 2

SOLUTION:

ii) In quadrilateral ABCD,

AQ=1/3 AC

3AQ=AC

3AQ=AQ+QC

3AQ-AQ=QC

2AQ=QC

⇒∴AQ/QC=1/2            ...........(1)

Now in ΔAQD and ΔCQB,

∠AQD = ∠CQB  (Vertically opposite angles)

∠ADQ = ∠QBC   (alternate angles, AD║BC)

By AA similarity,

ΔAQD ~ ΔCQB

∴Corresponding sides are in same proportion,

DQ/BQ=AQ/QC=AD/BC

DQ/BQ=AQ/QC

DQ/BQ=1/2

DQ=1/2 BQ

Hence, proved.

---------------------------------------------------------------------------------

iii) SOLUTION:

In ΔQRS,

SR=6

∠R=45°

Tan45°=1

QS/SR=1

⇒QS=SR=6  ..........(1)

Now, QS=PT=6

In ΔPTV,

∠P=60°

Tan60°=√3

VT/PT=√3⇒VT=√3×PT

VT=6√3

Area(QSR+PQST+PTV)=Area of trapezium PQRV

(1/2×QS×SR)+(PQ×QS)+(1/2×PT×VT)=1/2(Sum of parallel sides)×h

(1/2×6×6)+(9×6)+(1/2×6×6√3)=1/2(9+VR)×6

18+54+18√3=3(9+VR)

72+18√3=3(9+VR)

3(24+6√3)=3(9+VR)

24-9+6√3=VR

⇒3(2√3+5)=VR

        [OR]

15+6(1.732)=VR

⇒25.392=VR

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