Question

pqrs is a rectangle in which pq=2ps. t and u are the midpoints of ps and pq respectively. qt and us intersect at v. the area of qsrv divided by the area of pqt is?

Answer 1

Given: A rectangle PQRS in which PQ =2 PS. T and U are mid points of PS and PQ respectively.Line segments QT and US intersect at V.

To Find:  Area of quadrilateral (QSRV)÷ Area ofΔ PQT

Solution: Let P Q= 2 x then PS = x

In Δ SPU and Δ QPT

1. ∠P is 90°.

2. $\frac{SP}{PQ}=\frac{TP}{PU}$

Δ SPU ~ Δ QPT   [SAS]

Area(ΔPQT)=$\frac{1}{2} \times PQ \times PT$

                  =$\frac{1}{2} \times 2x \times \frac{x}{2}=\frac{x^{2}}{2}$

Similarly,Area (Δ SPU)=$\frac{x^{2}}{2}$

$\frac{Ar(\triangle PQT)}{Area(\triangle SPU)}=1$

→ AreaΔPQT=AreaΔPSU

→Ar(PUVT)+Ar(ΔTVS)=Ar(PUVT)+Ar(ΔU V Q)

→Ar(ΔT VS)=Ar(ΔU VQ)

Area(SRQU) which is a trapezium =$\frac{1}{2}\times[2x+x]\timesx=\frac{3x^{2}}{2}$

$\frac {Area QSRV}{Area PQT}=$$\frac{{\frac{3x^{2}}{2}-Ar(QVU)}}{\frac{x^{2}}{2}}$

Answer 2

Answer:

ANS IS 8:3

Step-by-step explanation:

U and T are mid-points of PQ and PS respectively.

⇒ SU and QT are medians of ΔPSQ.

⇒ V is the centroid of ΔPSQ.

=AREA OF QUAD QRSV/AREA OF TRIANGLE PQT

=AREA OF TRIANGLE QRS+AREA OF TRIANGLE QSV/AREA OF TRIANGLE PQS/2

=AREA OF TRAINGLE PQS+AREA OF TRAINGLE PQS/3/AREA OF TRIANGLE PQS/2   (QS DIVIDES RECTANGLE INTO 2 EQUAL TRIANGLES PQS AND QRS ALSO AREA OF TRIANGLE QSV=PQS/3)

4/3/1/2=482/1*3=8/3

THIS IS EXTREMELY DIFFICULT