Question
The area of an isosceles triangle ABC is 8√5 cm
2 . If AB = BC and AC = 8 cm, then find the perimeter of triangle ABC.Solution
Let AB = BC = 'x' cm
Area of isosceles triangle = {(c/4) X √(4a 2 - c 2 )} [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 x 2 - 8 2 )}
Or, (4√5) = √(4x 2 - 64)
Squaring both sides.
(16 X 5) = 4x 2 - 64
Or, 144 = 4x 2
Or, x 2 = 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
?% of (168 ÷ 8 × 20) = 126
20% of 1500 – 75% of 200 = 125% of ?
Find the value of 16 X [(8 - 5) of 12 ÷ 4].
√196 + (0.25 × 144) + 19 = ? + 72
22 * 6 + 45% of 90 + 65% of 180 = ?
52% of 400 + √(?) = 60% of 600 - 25% of 400
(25 × 12 + 30 × 8 – 22 × 10) = ?
What will come in the place of question mark (?) in the given expression?
(240% of 175 ÷ √16) X 6 + 80% of 400 = ?3 + 179 + 42
What will come in the place of question mark (?) in the given expression?
(144 × 16 ÷ 12) × 6 = ?
808 ÷ (128)1/7 + 482 = 4 × ? + 846