# Geometry(similarity) ?

How can I solve for PA on this problem? Relevance

OR=sqr(12^2+5^2)=13

Tri. AOP ~ Tri. AOR

=>

AP/AR=AO/OR

=>

AP/12=5/13

=>

AP=5(12)/13~4.62.

• Anonymous
4 weeks ago

By areas.  Area of ROA = 30 sq units (5 x 12/2)

base  RA = square root 169 = 13,

height is PA, half base = 6.5

30/6.5 = 4.62 units

• PA/OA = 12/OR = 12/(12^2+5^2)^(1/2) = 12/(144+25)^(1/2) = 12/13;

PA =(12/13)OA=(12/13)5=(60/13)=4.615384615=4.62 rounded to nearest hundredth

• Whatever the triangle, the sum of the 3 angles is always 180 °. So in the triangle SQU, you can write:

In the right triangle (ORA):

yellow + red + 90 = 180

yellow = 90 - red

In the right triangle (OPA):

blue + yellow + 90 = 180

blue = 90 - yellow

blue = 90 - (90 - red)

blue = red

AP = AO.cos(blue) → we've just seen that: blue = red

AP = AO.cos(red) → where: cos(red) = AR/OR

AP = AO.(AR/OR)

AP = 5 * (12/13)

AP ≈ 4.615 • OR=sqrt(12^2+5^2)=13

12/13=.9230

.9230*5=  4.615

• For rectangle BOAR:

OA = BR = 5  and  BO = RA = 12

In ΔROA:

OR² = OA² + RA²  (Pythagorean theorem)

OR = √(5² + 12²)

OR = √13

Area of ΔROA:

(1/2) × PA × OR = (1/2) × OA × RA

(1/2) × PA × 13 = (1/2) × 5 × 12

PA = 5 × 12 / 13

PA = 4.62

• RO = 13

The area of ∆ RAO = 30,

RO/2 * PA = 30

PA = 60/13 = 4.62

∆RPA is similar to  ∆PAO

OA/RA = PA/RP

5/12 = PA/RP

• I'm guessing a bit at this one, but OA is 5 (same as BR)

By pythagorean theorem, PO^2 + PA^2 = OA^2,

Thus PO^2 + PA^2 = 25

Ratio of PO to PA is same as the larger triangle (ie same as BR to BO) so 5 to 12

Thus (PO) * 12 = (PA) * 5.

I've gotten you 2 equations, you figure it out from there.

• Since RO cuts the rectangle in half through the diagonal, ROA and BRO are identical.

So we need the height of the triangle with lengths 5 and 12 when the hypotenuse is the base.

Let's do this.  If we put the triangle upright, we have the base of 5 and height of 12.  We can find the area:

A = bh/2

A = 5(12)/2

A = 30 unit²

We can use Pythagarean theorem to find the length of the hypotenuse:

a² + b² = c²

5² + 12² = c²

25 + 144 = c²

169 = c²

13 = c

With the same triangle on its hypotenuse for a base, the area is unchanged from when we solved for it above.  We can use that information to solve for the unknown height:

A = bh/2

30 = 13h/2

60 = 13h

h = 60/13

h = 4.62 units (rounded to 2DP)