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Each side of an equilateral triand 12 is \( 24 cm \) the mid - point of its sides are joined to form another tranpla this brress is doins centinuously infinite 46. The perimeter of 7 th triangle is \( ( \) in \( cm ) \) a) \( \frac{3}{4} \) b) \( \frac{5}{8} \) c) \( \frac{9}{8} \) d) \( \frac{7}{8} \) 47. The sum of perimeter of all triangle is (in \( cm \) ) a) 144 b) 625 c) 400 d) 169 48. The area of all the triangle is (in sq cm) a) 576 b) \( 144 \sqrt{3} \) c) \( 169 \sqrt{3} \) d) \( 192 \sqrt{3} \) 49. The sum of perimeter of first 6 triangle is (in \( cm \) ) a) 120 b) \( \frac{567}{4} \) c) \( \frac{569}{4} \) d) 144 50. The side of the 5th triangle is (in \( cm \) ) a) 6 b) \( 1.5 \) c) \( 0.75 \) d) 3

Monjur Alahi

Each side of an equilateral triand 12 is \( 24 cm \) the mid - point of its sides are joined to form another tranpla this brress is doins centinuously infinite 46. The perimeter of 7 th triangle is \( ( \) in \( cm ) \) a) \( \frac{3}{4} \) b) \( \frac{5}{8} \) c) \( \frac{9}{8} \) d) \( \frac{7}{8} \) 47. The sum of perimeter of all triangle is (in \( cm \) ) a) 144 b) 625 c) 400 d) 169 48. The area of all the triangle is (in sq cm) a) 576 b) \( 144 \sqrt{3} \) c) \( 169 \sqrt{3} \) d) \( 192 \sqrt{3} \) 49. The sum of perimeter of first 6 triangle is (in \( cm \) ) a) 120 b) \( \frac{567}{4} \) c) \( \frac{569}{4} \) d) 144 50. The side of the 5th triangle is (in \( cm \) ) a) 6 b) \( 1.5 \) c) \( 0.75 \) d) 3


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The longest side of a triangle is three times the shortest side and the third side is 2 cm shorter than the longest side if the perimeter of the trian

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2 Answers 1 views
The sides of a triangle are in A.P. and its area is `(3)/(5)` th of an equilateral triangle of the same perimeter. Find the greatest angle of the tria

Correct Answer - `120^(@)` Let the sides be `x - d, x, x + d`. Then `s = (3x)/(2), (s -a) = (x)/(2) + d`, `(s-b) = (x)/(2), (s-c) = (x)/(2)...

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(i) `(a^(p+q))^(p-q) (a^(q+r))^(q-r)(a^(r+p))^(r-p)` (ii) `[(a/b)^(sqrt(p)+sqrt(q))]^(sqrt(p)-sqrt(q))[(a/b)^(sqrt(q)+sqrt(r))]^(sqrt(q)-sqrt(r))` `[(

Correct Answer - (i) 1 (ii) 1

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One side of an equilateral triangle is 36 cm. The mid-points of its sides are joined to form another triangle. Again another triangle is formed by joi

Correct Answer - `432sqrt3 cm^(2)`

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4. The value of \( \left(\frac{1}{2}\right)^{-2}+\left(\frac{1}{3}\right)^{-2}+\left(\frac{1}{4}\right)^{-2} \) is (a) \( \frac{61}{144} \) (b) \( \frac{144}{61} \) (c) 29 (d) none of these

Correct answer is:- (c) 29 Explanation:-  1/2-2 + 1/3-2 + 1/4-2 = 22 + 32 + 42 (x-1 = 1/x) = 4 + 9 + 16  = 29 

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The perimeter of an isosceles triangle is 32 cm. The ratio of the equal side to its base is 3 : 2. Find the area of the triangle. Hint Ratio of sides

Correct Answer - `32 sqrt(3) cm^(2)`

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12 In a plane electromagnetic wave, the directions of electric field and magnetic field are represented by \( \hat{ k } \) and \( 2 \hat{ i }-2 \hat{ j } \), respectively. What is the unit vector along direction of propagation of the wave? (a) \( \frac{1}{\sqrt{2}}(\hat{ i }+\hat{ j }) \) (b) \( \frac{1}{\sqrt{2}}(\hat{ j }+\hat{ k }) \) (c) \( \frac{1}{\sqrt{5}}(\hat{ i }+2 \hat{ j }) \) (d) \( \frac{1}{\sqrt{5}}(2 \hat{ i }+\hat{ j }) \)

Correct answer is (a) We know that propagation wave vector ∵ \(\vec{E} = \hat{k}\) \(\vec{B} = 2\hat{i} - 2\hat{j}\) \(\vec{C} = \vec{E} \times \vec{B}\) \(\vec{C} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 0 & 1...

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Consider a particle of mass \( m \) having linear momentum \( P \) at position \( r \) relative to the origin \( O \). Which of the following equations correctly relates \( r , P , L \) ? a) \( \left[\left(\frac{d L}{d t}\right)+r x\left(\frac{d P}{d t}\right)\right]=0 \) b) \( \left[\left(\frac{d L}{d t}\right)-r x\left(\frac{d P}{d t}\right)\right]=0 \) c) \( \left[\left(\frac{d L}{d t}\right) \times\left(\frac{d r}{d t}\right) \times P\right]=0 \) d) \( \left[\left(\frac{d L}{d t}\right)-\left(\frac{d r}{d t}\right) \times P\right]=0 \)

Correct answer is (b) We know that \(\tau\) = r × F ∵ \(\tau\) \(= \frac{dL}{dt},\) \(F = \frac{dP}{dt}\) \(\left(\frac{dL}{dt}\right) = r \times \left(\frac{dP}{dt}\right)\) \(\left(\frac{dL}{dt}\right) - r \times \left(\frac{dP}{dt}\right) = 0\)

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If the roots of the equation \( a x^{2}+b x+c=0 \) are equal then \( c=? \) \( \quad \) 1) \( -\frac{b}{2 a} \) 2) \( \frac{b}{2 a} \) 3) \( -\frac{b^{2}}{4 a} \) 4) \( \frac{b^{2}}{4 a} \)

b^2 -4ac = 0 (as roots are equal ) b^2 =4ac b^2/4a = c

2 Answers 1 views
If the roots of the equation \( a x^{2}+b x+c=0 \) are equal then \( c=? \) \( \quad \) 1) \( -\frac{b}{2 a} \) 2) \( \frac{b}{2 a} \) 3) \( -\frac{b^{2}}{4 a} \) 4) \( \frac{b^{2}}{4 a} \) 2 Answers 1 views

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