Question

The area of the square with AC as a side is 128 cm? What is the sum of the areas of semicircles drawn on AB and AC as diameters, given ABC is an isosceles right angled triangle and AC is its hypotenuse​

Answer 1

Answer:

Step-by-step explanation: 16 is the answer.

Answer 2

The sum of the areas of semicircles drawn on AB and AC as diameters is 25.36 sq. cm.

Given,

area of square with AC as side = 128 sq. cm

ΔABC is an isosceles right-angled triangle

AC is the hypotenuse of this triangle

To Find,

sum of areas of semicircles drawn on AB and AC as diameters

Solution,

This is a tricky geometric question.

It can be solved using a carefully structured solution.

We have been given that ΔABC is an isosceles right-angled triangle.

Also, we know that AC is the hypotenuse.

Further, we have a square with this AC as one of its sides.

The area of this square is 128 sq. cm.

We know that a square has four equal sides, say 'a'.

The area of this square is a² .

Therefore, the area of the square with AC as one of its side is given by:

⇒ (AC)² = 128 sq. cm

⇒ AC = √(128) cm

⇒ AC = 8√2 cm

Therefore, the length of side AC is 8√2 cm

Now, we know that in an isosceles right-angled triangle, the base and altitude are equal in lengths.

In ΔABC, AC is the hypotenuse.

Let BC be the base and AB be the altitude.

⇒ AB = BC

Now, for a right-angled triangle, we apply the Pythagoras theorem.

Mathematically, Pythagoras theorem could be expressed as:

⇒ (hypotenuse)² = (base)² + (altitude)²

⇒ (AC)² = (AB)² + (BC)²

⇒ (AC)² = (AB)² + (AB)² (since AB = BC)

⇒ (AC)² = 2(AB)²

⇒ AC = AB√2

⇒ 8√2 = AB√2

⇒ AB = 8

Therefore, the length of side AB is 8 cm

Now, we have to find the sum of areas of semicircles with AB and AC as diameters.

Let the semicircles with AB as diameter be S1 and the semicircles with AC as diameter be S2.

We know that the area of a semicircle is 0.5πr²

radius of semicircle S1 = AB/2 = 8/2 = 4 cm

radius of semicircle S2 = AC/2 = (8√2)/2 = 4√2 cm

area of semicircle S1 = 0.5 × π × (4)² = 0.5 × 3.14 × 16 = 25.12 sq. cm

area of semicircle S2 = 0.5 × π × (4√2)² = 0.5 × 3.14 × 16 × 2 = 50.24 sq. cm

sum of the areas of semicircles drawn on AB and AC as diameters is given by:

⇒ 25.12 + 50.24

⇒ 25.36

Therefore, the sum of the areas of semicircles drawn on AB and AC as diameters is 25.36 sq. cm.

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