Question

Seth is using the figure shown below to prove Pythagorean Theorem using triangle similarity.

In the given triangle ABC, angle A is 90° and segment AD is perpendicular to segment BC.

The figure shows triangle ABC with right angle at A and segment AD. Point D is on side BC.

Part A: Identify a pair of similar triangles. (2 points)

Part B: Explain how you know the triangles from Part A are similar. (4 points)

Part C: If DB = 9 and DC = 4, find the length of segment DA. Show your work. (4 points)

Answer 1

please correct it ia think there no segment AD perpendicular to segment BC

this AD wan being AB

yes ?

Answer 2

There is a segment AD.

Answer 3

where ?

Answer 4

and segment AD is perpendicular to segment BC.

how is possibly this ?

Answer 5

They both have two lines that meet at the same angle

Answer 6

may be AD perpendicular on DC and not on BC

Answer 7

check this please

Answer 8

sorry wrong image

Answer 9

so than what will be correct ?

Answer 10

the second posted image ?

Answer 11

the second image is the image for this question the first one was for another question

Answer 12

ok so

Part A: Identify a pair of similar triangles. (2 points)

Part B: Explain how you know the triangles from Part A are similar. (4 points)

Part C: If DB = 9 and DC = 4, find the length of segment DA. Show your work. (4 points)

Answer 13

step by step please - i m here to help you and explain

Answer 14

A . triangle ADB similar triangle ADC ?

Answer 15

i would say so they look perpendicular

Answer 16

do you wan saying angle D_1 congruent angle D_2 bc. AD perpendicular BC ?

Answer 17

and again what can you see there ? side AD what is for these triangles ?

Answer 18

AD not is common side ?

Answer 19

Answer: Part A; Triangle ABD is similar to Triangle ACD

Step-by-step explanation: Part B; The triangle is given such that angle A is 90°

If point D is on line BC, then it touches angle A. However line AD is perpendicular to line BC,which means it splits the angle at A into two equal halves.

Hence, you now have two right angled triangles, on the one hand, triangle ABD and on the other hand, triangle ACD. The first one has angle D measuring 90° and line AB as the hypotenuse. The second one also has angle D measuring 90° with it's hypotenuse at line AC.

Part C; From the dimensions given in the question, triangle ABD has side BD measuring 9 while the other two sides are not given. But angle D is given as 90° (perpendicular line from angle A in the original triangle). Also, angle A has been split in two which makes angle A in triangle ABD to measure 45°

Therefore triangle ABD is a right angled triangle with the other two sides measuring 45° each. [180° = (90°+45°+45°)]

We can solve for the unknown side by applying trigonometrical ratios. We have line DB given as 9 units and the angle facing it (angle A) is 45°. Line AD is unknown and is also facing angle B which is 45°

We shall apply the ratio of tangent of angle since we have the opposite to angle B (unknown) and the adjacent to angle B (9 units)

Tan 45° = opposite/adjacent

Tan 45° = DA/9

Multiply both sides by 9

9 × Tan 45° = DA

(Looking up the table of values for trigonometrical ratios, Tan 45°=1)

9 × 1 = DA

Therefore line DA equals 9 units

Answer 20

got it ^

Answer 21

understandably ?

Answer 22

yes sir

Answer 23

ok was my pleasure

have a nice day

Answer 24

thank you.

Answer 25

np

anytime