(a) In the diagram, PQST is a parallelogram, PR is a straight line, |TS| = 8cm, |SM| = 6cm and area of triangle PSR = \(36 cm^{2}\). Find the value of |QR|....

Answer Details

(a) Find |QR|.

From the diagram, \(PQST\) is a parallelogram, and \(P,\,Q,\,R\) lie on one straight line. The opposite sides of the parallelogram are equal, so:

\[|PQ|=|TS|=8\text{ cm}\]

Triangle \(PSR\) has its base along the line \(PR\), and \(SM=6\text{ cm}\) is the perpendicular height from \(S\) to that base.

Using the area of a triangle:

\[\text{Area}=\frac{1}{2}\times\text{base}\times\text{height}\]

\[36=\frac{1}{2}\times |PR|\times 6=3\,|PR|\]

\[|PR|=\frac{36}{3}=12\text{ cm}\]

Since \(|PR|=|PQ|+|QR|\):

\[12=8+|QR|\Rightarrow |QR|=4\text{ cm}\]

(b) Height of the flagpole.

Let the horizontal distance from the tree to the flagpole be \(d\). The bird is at the top of the tree, at height \(10.65\text{ m}\).

Bottom of the flagpole (ground level) is seen at an angle of depression of \(60^{\circ}\). The vertical drop is the full \(10.65\text{ m}\):

\[\tan 60^{\circ}=\frac{10.65}{d}\Rightarrow d=\frac{10.65}{\tan 60^{\circ}}=\frac{10.65}{1.7321}=6.149\text{ m}\]

Top of the flagpole is seen at an angle of depression of \(45^{\circ}\). The vertical drop from the bird down to the top of the flagpole is:

\[\text{drop}=d\tan 45^{\circ}=6.149\times 1=6.149\text{ m}\]

The height of the flagpole is the tree height minus this drop:

\[h=10.65-6.149=4.501\text{ m}\]

Correct to 3 significant figures, the height of the flagpole is \(4.50\text{ m}\).