Question

can you do this on please? ​

Answer 1

Triangle must be a right angled triangle.

Given:-

$\sf PS = 4cm, SQ = 1cm and PT = 6cm and TR= 1.5$

Now,

$\sf \frac{PS}{SQ} = \frac{4}{1} \: ....(1) \\ \\ </p><p>\sf \frac{PT}{ TR} = \frac{6}{1.5} \\ \\</p><p>\sf = \frac{60}{15} = \frac{4}{1} \: ...(2) \\ \\</p><p>\sf from \ (1) \ and \ (2) \\ \\</p><p></p><p>\sf \therefore \frac{PS}{SQ} = \frac{PT}{TR} \\ \\</p><p></p><p>\textsf { By Converse of Basic proportionality theorem} \\ \\</p><p></p><p> \sf ST || QR$

now,

$</p><p>\sf In \ \triangle PQR, \angle PQR = 90 \\ \\</p><p>\sf By \ pythagoras \ theorem \\ \\</p><p>\sf PQ^2 + QR^2 = PR^2 \\ \\ </p><p>\sf 5^2 + QR ^2 = 7.5^2 \\ \\</p><p>\sf 25 + QR^2 = 56.25 \\ \\</p><p>\sf QR^2 = 56.25 - 25 \\ \\</p><p>\sf QR^2 = 31.25 \\ \\</p><p>\sf QR = \sqrt{31.25} \\ \\</p><p>\sf \boxed { \red { QR = 5.59}} \\ \\ </p><p></p><p>\sf And, \\ \\</p><p></p><p>\sf PS^2 + ST^2 = PT^2 \\ \\</p><p>\sf 4^2 + ST^2 = 6^2 \\ \\</p><p>\sf 16 + ST^2 = 36 \\ \\</p><p>\sf ST^2 = 36 - 16 \\ \\</p><p>\sf ST^2 = 20 \\ \\</p><p>\sf ST = \sqrt{20} \\ \\</p><p>\sf \boxed {\red {ST = 4.47}}$

now,

$\sf Area \ of \triangle PST = \frac{1}{2} \times PS \times ST \\ \\</p><p>\sf = \frac{1}{2} \times 4 \times 4.47 \\ \\</p><p>\sf = 2 \times 4.47 = 8.94sq \ cm \\ \\</p><p></p><p>\sf now, \\ \\</p><p></p><p>\sf Area \ of \ trapezium = \frac{1}{2} \times (ST + QR) \times SQ \\ \\</p><p></p><p>\sf = \frac{1}{2} \times (4.47 + + 5.59) \times 1 \\ \\</p><p></p><p>\sf = \frac{1}{2} \times 10.6 \\ \\</p><p></p><p>\sf = \frac{10.6}{2} = 5.3 \\ \\</p><p></p><p>\sf = 5.3 sq \ cm \\ \\ \\</p><p></p><p>\sf Now, \\ \\</p><p></p><p>\sf \frac{Area ( \triangle \ PST)}{Area ( Quadrilateral \ QRTS)} = \frac{10.6}{5.3} \\ \\</p><p></p><p>\sf = \frac{2}{1} \ is \ Answer.$