Question

In a triangle ABC, angle A 90° and area of triangle is 3, where AB and AC are having integral lengths , then BC can be

Answer 1

BC can be ✓13 or ✓37 units.

Given - Angle A and area

Find - Length of BC

Solution - As per the given information, AB and AC will be base and perpendicular.

Area of right angled triangle = 1/2*base*perpendicular

Keep the values in formula

3 = 1/2*B*P

B*P = 3*2

B*P = 6

Now, the multiples of 6 will be the base and perpendicular of given triangle. So,

B and P = (2*3), (3*2), (1*6) or (6*1).

So, base and perpendicular will be either combination of 2 and 3 or 1 and 6. Now, calculating BC which is hypotenuse. Using Pythagoras theorem.

Case I -

BC² = 2² + 3²

BC² = 4 + 9

BC² = 13

BC = ✓13 units

Case II -

BC² = 1² + 6²

BC² = 1 + 36

BC² = 37

BC = ✓37 units

Thus, BC can be ✓13 or ✓37 units.

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Answer 2

The MaIn Answer is:

                           The Possible Values Of BC are  √37 and √13

GIVEN:  ΔABC,     ∠A = 90°,    area of ΔABC = 3 units,    AB and AC are integral lengths (they are integers)

TO FIND: possible length(s) of BC

SOLUTION:

• We know, for any triangle, its area = $\frac{1}{2}$×base×height.
• In this case, the base and height are AB and AC, in any order.
• Appling the formula for the area of a triangle--
• 3 = $\frac{1}{2}$×AB×AC
• ⇒ AB×AC = 6
• Now, the only integral positive values possible for AB and AC, to give a product of six are-- a pair of (1 and 6) or (2 and 3).
• Also, for the given right-angled ΔABC, we can apply the 'Pythagoras theorem' which states:

        (Hypotenuse)² = (Base)² + (Height)²

• For given ΔABC,

            (BC)² = 1² + 6²       or     (BC)² = 2² + 3²

        ⇒ BC = $\sqrt{1 + 36}$        or       BC = $\sqrt{4 + 9}$

        ⇒ BC = √37              or       BC = √13

Therefore, the possible values of BC are √37 and √13.