Mathematics Questions and Answers

From the diagram, XYZQ is a parallelogram. Thus, |YZ| = |XQ| = |PX|; \(\bigtriangleup\)PXY, Let the area of XYZQ = A₁, the area of \(\bigtriangleup\)PXY

= Area of \(\bigtriangleup\)YZR = A₂

Area of \(\bigtriangleup\)PQR = A = A₁ + 2A₂

But from similarity of triangles

\(\frac{\text{Area of PQR}}{\text{Area of PXY}} = (\frac{PQ}{PX})^2 = (\frac{QR}{XY})^2\)

\(\frac{A}{A_2} = (\frac{2}{1})^2 = \frac{2}{1}\)

A = 4A₂ But, A = A₁ + 2A

A₁ = 4A₂ - 2A₂

A₁ = 2A₂

\(\frac{A_1}{A_2}\) = 2

A₁:A₂ = 2:1


Area of XYZQ:Area of \(\bigtriangleup\)YZR = 2:1

From the diagram, < QPS = x^o (Corresponding angles)

Also, < QPS = < PSR(Alternate angles)

x = 42^o

< QPR + < PRQ = < RQS

(Sum of two interior angles of a triangle = Opposite exterior angles)

70^o + < PRQ = 148

< PRQ = 148^o - 70^o

= 78^o

For a rhombus, all the sides are equal and diagonals bisect each other at 90o. Hence, the triangles formed are congruent: (under RHS).

Thus: \(\bar{PS}^2\) = 52 + 122

\(\bar{PS} = \sqrt{25 + 144}\)

= \(\sqrt{169}\)

\(\bar{PS}\) = 13cm

perimeter = 4 x length of a side

= 4 x 13cm

= 52cm

Since |PR| = |RS| = |SP|

\(\bigtriangleup\) PRS is equilateral and so < RPS = < PRS = < PSR = 70^o

But < PQR + < PSR = 180^o (Opposite interior angles of a cyclic quadrilateral)

< PQR + 60 = 180^o

< PR = 180 - 60 = 120^o

But in \(\bigtriangleup\) PQR, PQ = PR, hence < QPR = < PRQ(Base angles of an Isosceles triangle)

< QPR + < PRQ + < PQR = 180^o (Angles in a triangle)

2 < QPR + 120 = 18-

2 < QPR = 180 - 120

QPR = \(\frac{60}{2}\) = 30^o

From the diagram, < QRS = < PRQ + < PRS

30 + 60 = 90^o