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yan [13]
2 years ago
6

Louis needs to secure his tent.He will use a rope to go down the edge of the tent, straight out to a post and back to the starti

ng think fastener,as shown​

Mathematics
1 answer:
tester [92]2 years ago
4 0

Answer:

i need help

Step-by-step explanation:

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Evaluate.<br> 21 + } + (-)) -<br> Enter your answer as a mixed number in simplest form in the box.
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Evaluate: 2(1/4) + 2/5 ÷ (-1/2) - 1/4

⇛9/4 + 2/5 ÷ (-1/2) - 1/4

Take the LCM of the 4 and 5 is 20.

Again, 2 and 4 is 4.

⇛[(54+8)/20] ÷ [(-2 - 1)/4]

⇛62/20 ÷ (-3/4)

⇛62/20 × 4/-3

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⇛62/-15.

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3 years ago
In the right triangle ?ABC, leg AC=6 cm and leg BC=8 cm. Point M and N belong to AB so that AM:MN:NB=1:2.5:1.5. Find area of ?MN
Licemer1 [7]

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}

\frac{Ar(ACB)}{Ar(MNC)}=4

But Area (ΔACB)=1/2×6×8= 24 cm²[ACB is a right angled triangle]

\frac{24}{Ar(MNC)}=4

→ Area(ΔMNC)=24÷4

→Area(ΔMNC)=6 cm²

4 0
3 years ago
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