Given : In Right triangle ABC, AC=6 cm, BC=8 cm.Point M and N belong to AB so that AM:MN:NB=1:2.5:1.5.
To find : Area (ΔMNC)
Solution: In Δ ABC, right angled at C,
AC= 6 cm, BC= 8 cm
Using pythagoras theorem
AB² =AC²+ BC²
=6²+8²
= 36 + 64
→AB² =100
→AB² =10²
→AB =10
Also, AM:MN:NB=1:2.5:1.5
Then AM, MN, NB are k, 2.5 k, 1.5 k.
→2.5 k + k+1.5 k= 10
→ 5 k =10
Dividing both sides by 2, we get
→ k =2
MN=2.5×2=5 cm, NB=1.5×2=3 cm, AM=2 cm
As Δ ACB and ΔMNC are similar by SAS.
So when triangles are similar , their sides are proportional and ratio of their areas is equal to square of their corresponding sides.
![\frac{Ar(ACB)}{Ar(MNC)}=[\frac{10}{5}]^{2}](https://tex.z-dn.net/?f=%5Cfrac%7BAr%28ACB%29%7D%7BAr%28MNC%29%7D%3D%5B%5Cfrac%7B10%7D%7B5%7D%5D%5E%7B2%7D)
But Area (ΔACB)=1/2×6×8= 24 cm²[ACB is a right angled triangle]
→ Area(ΔMNC)=24÷4
→Area(ΔMNC)=6 cm²
Answer:
27 miles.
Step-by-step explanation:
Here I attach the draw of the coordinates.
Tony traveled 3 segments. The first was from (12,6) to (12, 15), where, leting 12 constant, he moved from 6 to 15 in the ordinates axis, which implies 9 units. This is the section 1 in the draw.
Then he moved from point B to C. If you notice, this distance is the hypotenuse on the the triangle DBC. We can find this value using Pitagoras' theorem:
DB^2 + CD^2 = CB^2
With DB=15 and CD=8 (12 minus 4 = 8)
15^2 + 8^2 = 289
So CB^2=289
Applying sqr root:
CB = 17
So, the second section has a measure of 17 units.
Finally, the 3rd section is the hypotenuse of the DAC triangle and we can use Pitagoras to solve it:
CD^2 + AD^2 = CA^2
8^2 + 6^2 = CA^2
64 + 36 = 100
So, CA=10
In the 3r section we traveled 10 units.
So, in total he traveled 10 + 17 + 9 = 36 units
As every unit is 0.75 miles he traveled 36*0.75 miles:
36*0.75 = 27 miles
He traveled in total 27 miles!!
Answer:
-7/32
Step-by-step explanation:
Multiply,Simplify
Answer: x > -8
Step-by-step explanation: the line is going to 8 then 9, 10, etc. so greater than ; the circle on #8 isn't filled in so there won't be a line by the sign
