Target Exam

CUET

Subject

General Test

Chapter

Quantitative Reasoning

Topic

Geometry

Question:

PQRS is a rectangle. T is a point on PQ such that RTQ is an isosceles triangle and PT = 5 QT. If the area of triangle RTQ is $12\sqrt{3}$sq.cm, then the area of the rectangle PQRS is:

Options:

$144 \sqrt{3}$ sq.cm

$134 \sqrt{3}$ sq.cm

$142 \sqrt{3}$ sq.cm

142 sq.cm

Correct Answer:

$144 \sqrt{3}$ sq.cm

Explanation:

PT = 5QT

PT : QT = 5 : 1

Let the ratio of PT : QT be 5x : x

PQ = PT + QT

PQ = 5x + x

PQ = 6x

PQ = RS [PQRS is a rectangle]

RS = 6x

In rectangle PQRS, there are three triangle PST , RTQ and STR

Area of \(\Delta \)RTQ = \(\frac{1}{2}\) x Base x height

= \(\frac{1}{2}\)  × X × RQ = 12√3

RQ = (12√3 x 2)/x

RQ = 24√3/x

Area of PQRS = area of \(\Delta \)RTQ + area of \(\Delta \)PST + area of \(\Delta \)STR

= \(\frac{1}{2}\) x QT x RQ + \(\frac{1}{2}\) x PT x RQ + \(\frac{1}{2}\) x SR x RQ

= \(\frac{1}{2}\) x RQ[QT + PT + SR] = \(\frac{1}{2}\) x(24√3/x)[x + 5x + 6x]

= Area of rectangle PQRS = [12√3/x] x [12x] = 144√3 sq. cm.