Question

The sides of certain triangles are given below. Determine which option can form a right-angled triangle?

• A. (a - 1) units, 2√a units, (a + 1) units
• B. Both Option (b) and Option (c)
• C. 7 units, 24 units and 25 units.
• D. 9 units, 16 units and 18 units.

Answer 1

The correct option is B Both Option (b) and Option (c)

Applying the Pythagorean theorem:
$\text{We know}{a}^2+b^2=c^2$

For a given triangle to be a right-angled, the sum of the squares of the two sides must be equal to the square of the largest side.

Option 'a':
Let a = 9 units, b = 16 units and c = 18 units.
Here, 9 < 16 < 18.

$\text{So,}(a^2+b^2)=[9^2+(16{)}^2]$
$=(81+256)sq.unit$
$=337\text{sq. unit}$
$\text{And}{c}^2=(18{)}^2\text{sq. unit}=324\text{sq. unit}$

$\therefore (a^2+b^2)\ne c^2$

Hence, the given triangle is not right-angled.

Option 'b':
Let a = 7 units, b = 24 units and c = 25 units.
Here, 7 < 24 < 25.

$\text{So,}(a^2+b^2)=[7^2+(24{)}^2]$
$=(49+576)sq.unit$
$=625\text{sq. unit}$
$\text{And}{c}^2=(25{)}^2\text{sq. unit}=625\text{sq. unit}$

$\therefore (a^2+b^2)=c^2$

Hence, the given triangle is a right-angled triangle.

Option 'c':
Let p = (a - 1) unit, q = 2√a unit and r = (a + 1) unit

$(p^2+q^2)=\left[\right(a-1{)}^2+\left(2\sqrt{a}{)}^2\right]$
$=(a^2+1-2a+4a)\text{sq. unit}$
$=(a+1{)}^2\text{sq. unit}$
$\text{And}{r}^2=(a+1{)}^2\text{sq. unit}$

Therefore,
$(p^2+q^2)=r^2$

Hence, the given triangle is a right-angled triangle.

So, option 'd' is correct.