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.
But Area (ΔACB)=1/2×6×8= 24 cm²[ACB is a right angled triangle]
→ Area(ΔMNC)=24÷4
→Area(ΔMNC)=6 cm²
Answer:
n = 84.1
Step-by-step explanation:
Answer:
3/4−2/4 = 1/4
Step-by-step explanation:
tienes que encontrar un denominador común para restar las dos fracciones. espero que esto te haya ayudado.
f1×f2-f3=(-5x+4)×(-6x-5)-(-9x+2)=30x^2-24x+25x-20+9x-2=30x^2+10x-22=(5x+1)^2-23
Answer:
20 minutes
Step-by-step explanation:
15 minutes past 10 is 10:15
25 minutes to 11 am is 10:35
(60-25=35)
35-15=20