Coordinated Polygons II

Draw $QR$ parallel to $OT$ so that $R$ is on $OP$ .

Then $\triangle PQR \sim \triangle PTO$ by AA similarity, so $\frac{TP}{QP} = \frac{TO}{QR}$ , or $\frac{1+4}{4} = \frac{5}{QR}$ , or $QR = 4$ .

Therefore, the area of $\triangle OQP$ is $A = \frac{1}{2} \cdot OP \cdot QR = \frac{1}{2} \cdot 4 \cdot 10 = \boxed{20}$ .

From the coordinates we note that $OP=10$ and $OT=5$ . By Pythagorean theorem we have $TP^2 = OT^2+OP^2 = 5^2+ 10^2$ $\implies TP = 5\sqrt 5$ . Since $TQ:QP = 1:4$ , $QP = \frac 45 TP = \frac 45 \times 5\sqrt 5 = 4\sqrt 5$ . Then, the area of $\triangle OQP$ is given by:

$\begin{aligned} A & = \frac 12 \times QP \times OP \times \sin \angle TPO \\ & = \frac 12 \times 4 \sqrt 5 \times 10 \times \frac 5{5\sqrt 5} \\ & = \boxed{20} \end{aligned}$

The diagram looks like this:

If we rotate the blue triangle and drop a perpendicular from point O we get:

The light blue triangle $\bigtriangleup OQP$ and the big triangle $\bigtriangleup OTP$ both have the same height $h$ , but the light blue triangle has $\frac{4}{5}$ the base of the large triangle.

Therefore, $\bigtriangleup OQP$ has $\frac{4}{5}$ the area of $\bigtriangleup OTP$ . The area of $\bigtriangleup OTP = \frac{5 \times 10}{2} = 25$

Thus the area of $\bigtriangleup OQP = 25 \times \frac{4}{5} = \fbox{20}$

This problem can be solved without finding out the value of any additional lengths.

Given the coordinates of T and P, the area of △OTP = $\frac{1}{2}$ ⋅ 5 ⋅10 = 25

Construct height OH, with TP as the base of the triangle, OH is the height for both △OQT and △OQP Since TQ:QP=1:4, the ratio between area of △OQT and area of △OQP is also 1:4 (they have the same height, OH)

Therefore, area of △OQP = $\frac{4}{1+5}$ ⋅ 25 = 20

(The area of two triangles are the same if the corresponding base and height of the two triangles have the same length. If a ratio exists, the areas have the same ratio as their bases or heights)

Consider TP the base of the TOP triangle, which has an area of 25. OQP triangle is 80% of TOP (that is, 4/5, as is has 4/5 of the base of TOP). Therefore the area of OQP is 20.

Draw a line OH perpendicular to the TP. Thus,we can know that S {△OTP}=0.5*TP*OH=25 and S {△OQP}=0.5 OH QP. Since we have TQ:QP=1:4,we can get S {△OTP}: S {△OQP}=TP:QP=5:4 So S_{△OQP}=20