Answer:
c. Definition of altitude.
Step-by-step explanation:
We are given that segment QS is an altitude in ΔPQR and we are asked to find a justification used while proving the similarity of triangles ΔPSQ and ΔQSR.
Since we know that altitude meets opposite side at right angles. When QS will intersect line PR we will get two right triangles QSR and QSP right angled at S.
ΔPQR is similar to ΔPSQ as they both share angle P and right triangle. So their third angle should also be similar.
ΔPQR is similar to ΔQSR as they both share angle R and both have a right triangle at Q and S respectively. So they will have their third angle equal.
ΔPQR is similar to triangles ΔQSR and ΔPSQ. Therefore, ΔQSR is similar to ΔPSQ.
Therefore, by definition of altitude triangles ΔPSQ and ΔQSR are similar as ΔPSQ and ΔQSR are created from ΔPQR by altitude QS.
Answer:−
-20
√
7
Step-by-step explanation:
2 (d(h+w) + h w) is the surface area
1.) 17.1g > 1.71mg
2.) 6.3cm< 63m
3.)1250ml>12.5
4.) 7/12 < 2/3
5.) 7/10 < 11/15
Answer:
126.5
Step-by-step explanation:
To find the solution for KMC use the following geometry theorems:
- Vertical angles are congruent so their measures are equal.
- Supplementary angles form a straight line have when added together equal 180.
HMC and RMK are vertical angles because they are across a vertex from each other. They are congruent and have equal measures. So M can be solved for by:
5m+15=7m+1
5m+15-1=7m+1-1
5m+14=7m
5m-5m+15=7m-5m
15=2m
7.5=m
Now substitute m=7.5 into one of the angles, either HMC or RMK.
7(7.5)+1 = 52.5+1 = 53.5
Notice, RMK and KMC are supplementary because they form a straight line together. This means 53.5 + KMC = 180.
So KMC = 126.5