Question

length of rectangle is twice its breadth equilateral triangle are drawn on its breadth, length and diagonal. prove that ratio of the areas is 1:4:5​

Answer 1

Answer:

Area of square =a

2

Length of the diagonal =

2a

Area of equilateral triangle with side=√2 a

√3/4×(√2a^2)=√3a^2/2

then,

Required ratio = 3a^2 /2 /a^2= √3/2

Hence, ratio is 3:2

Step-by-step explanation:

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Answer 2

Solution :-

Let us assume that, breadth of rectangle is 1 unit .

Than,

→ Length of rectangle = 2 * 1 = 2 unit .

Now, we know that,

→ Diagonal of a rectangle = √{(length)² + (Breadth)²}

So,

→ Diagonal of rectangle = √(2² + 1²) = √(4 + 1) = √5 unit .

Now, we have given that, Equaliteral ∆'s are drawn taking sides as breadth , length and diagonal .

As we know :-

• Area of Equaliteral ∆ = (√3/4) * (side)²

So,

→ Area of ∆ drawn on Breadth = (√3/4) * (1)² = {(√3/4) * 1} unit² .

→ Area of ∆ drawn on Length = (√3/4) * (2)² = {(√3/4) * 4} unit² .

→ Area of ∆ drawn on Diagonal = (√3/4) * (√5)² = {(√3/4) * 5} unit² .

Therefore,

→ Required Ratio of Area's of all three ∆'s :-

→ {(√3/4) * 1} : {(√3/4) * 4} : {(√3/4) * 5}

{ (√3/4) will be cancel from all ratios. }

→ 1 : 4 : 5. (Hence, Proved).

Learn More :-

in ∆ABC, a=3,b=4,c=5 , then find distance between incentre and circumcentre

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