Question
The area of an isosceles triangle ABC is 8√5 cm². In
this triangle, sides AB and BC are equal in length, and the base AC has a length of 8 cm. Determine the perimeter of triangle ABCSolution
Let AB = BC = 'x' cm Area of isosceles triangle = {(c/4) X √(4a2 - c2)} [Where 'a' is the length of two equal sides and 'c' is the length of third (unequal) side] 8√5 = (8/4) X {√(4 X x2 - 82)} Or, (4√5) = √(4x2 - 64) Squaring both sides. (16 X 5) = 4x2 - 64 Or, 144 = 4x2 Or, x2 = 36 So, x = 6 cm (Since, length cannot be negative therefore, we will take the positive root only) So, required perimeter = 6 + 6 + 8 = 20 cm
(22 × 52 ) + 4 × 6 = ? - √324
What should come in place of (?) question mark in the given expression.
 (25% of 320) + (3/8 of 400) − 30 = ?
(5832)1/3  × 10.11 × 11.97 ÷ 16.32 = ? + 45.022
82% of 400 + √(?) = 130% of 600 - 85% of 400
If (x + 1/x) = 5, then value of x3 + 1/x3 is:
Simplify: (1 ÷ 0.08)
What should come in place of (?) question mark in the given expression.
{ (144 ÷ 12) × 5 } − (18 ÷ 3) = ?
Simplify the following expressions and choose the correct option.
(3/4 of 256) + (2/5 of 150) - (72 ÷ 7)
464 + 181 +? = (154 × 25) - (15) 2 Â
15% of 1800 + 22 = ?Â
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