Question

Prove that:

4BE²=4BC²+AC²​

Answer 1

Required Answer:-

Given:

• ΔABC is a right angled triangle, right angled at C
• D and E are mid-points of side BC and AC.

To Prove:

• 4BE² = 4BC² + AC²

Proof:

In ΔBCE, we have,

➡ BE² = BC² + CE² (By Pythagoras Theorem)

Multiplying both sides by 4, we get,

➡ 4BE² = 4BC² + 4CE²

➡ 4BE² = 4BC² + (2CE)²

As E is the midpoint of AC, then,

➡ AC = 2CE

➡ 4BE² = 4BC² + AC² (Hence Proved)

Learn More:

• Pythagoras Theorem: In a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Answer 2

Correct Question -

The circumference of two circle are in the ratio 2 : 3. Find the ratio of their areas.

Given -

Ratio of their circumference = 2:3

To find -

Ratio of their areas.

Formula used -

Circumference of circle

Area of circle.

Solution -

In the question, we are provided, with the ratios of the circumference of 2 circles, and we need to find the ratio of area of those circle, for that first we will use the formula of circumference of a circle, then we will use the formula of area of circles. We will be writing 1 equation in it too.

So -

Let the circumference of 2 circles be c1 and c2

According to question -

c1 : c2

Circumference of circle = 2πr

where -

π = $\tt\dfrac{22}{7}$

r = radius

On substituting the values -

c1 : c2 = 2 : 3

2πr1 : 2πr2 = 2 : 3

$\tt\dfrac{2\pi\:r\:1}{2\pi\:r\:2}$ = $\tt\dfrac{2}{3}$

$\tt\dfrac{r1}{r2}$ = $\tt\dfrac{2}{3}$$\longrightarrow$ [Equation 1]

Now -

Let the areas of both the circles be A1 and A2

Area of circle = πr²

So -

Area of both circles = πr1² : πr2²

On substituting the values -

A1 : A2 = πr1² : πr2²

$\tt\dfrac{A1}{A2}$ = $\tt\dfrac{(\pi\:r1)}{(\pi\:r2)}^{2}$

$\tt\dfrac{A1}{A2}$ = $\tt\dfrac{(r1)}{(r2)}^{2}$

$\tt\dfrac{A1}{A2}$ = $\tt\dfrac{(2)}{(3)}^{2}$ [From equation 1]

So -

$\tt\dfrac{A1}{A2}$ = $\tt\dfrac{4}{9}$

$\therefore$ The ratio of their areas is 4 : 9

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