Question

Q.1 In an Isosceles triangle ABC, AB=AC and AD is the altitude to side BC. Is triangle ABD congruent to ACD? Use two different rules of congruency to prove this.

Q.2 Gina goes straight up North 15 km and then turns ninety degrees left to go 20 km West. Finally, she turns ninety degrees left again and travels 15 km. How far is she from her starting point?

Note: Draw a rough sketch for both the questions

Answer 1

Answer:

Answered both below

Step-by-step explanation:

we use sss and sad

AD=AD (Common)

ADB= ADC(90°) angles

BD=CD (AD is an altitude)

so  triangle ABD is congruent to triangle ACD (by SAS congruence condition)

∆ABC is an isosceles (given)

so AB =AC

OR

∆ABD=~ ∆ACD (Proved above)

hence, AB=AC (by CPCT)

AD=AD (common)

BD=BC (AD is an altitude)

therefore, ∆ABD is congruent to ∆ACD (By SSS congruence condition)

Hence,Proved.

Q2)

Sketch will be roughly like this

Let A be her starting position and b final positon

15+20+(-15)=20

  |----20----|

  |              |  15

15|             |

  b            a

Answered by gauthmath

mark brainliest if satisfied with answer

Answer 2

Answer:

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