Live practice

In \(\triangle ABC\), \(LM\parallel CB\). Therefore, corresponding sides of \(\triangle AML\) and \(\triangle ABC\) are proportional:

In \(\triangle ADC\), \(LN\parallel CD\). Thus \(\triangle ALN\sim\triangle ACD\), and

The right-hand sides of \(1\) and \(2\) are equal. Hence,

Formula or theorem used: For an AP use \(a_n=a+(n-1)d\) and \(S_n=\frac n2[2a+(n-1)d]\).

Working: Identify \(a\), \(d\), and the required term or sum; substitute them in the appropriate formula and simplify line by line.

Write each reason beside the corresponding equality. The conclusion follows only after the stated theorem or congruence/similarity criterion has been established.

After substitution and simplification:

(i) C (ii) B

Formula or theorem used: Write the equation in the form \(ax^2+bx+c=0\). Factorise where possible; otherwise use \(x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\).

Working: Display the standard form, the factorisation or discriminant calculation, both roots, and reject any value that is impossible in the stated context.

After substitution and simplification:

Positive integers are 13 and 14.

Formula or theorem used: Translate the information into two linear equations. Use the method named in the exercise, keeping equivalent equations aligned.

Working: Eliminate or substitute one variable, obtain the other variable, substitute back, and verify the ordered pair in both original equations.

After substitution and simplification:

One possible answer for the three parts:
(i) 3x + 2y – 7 = 0 (ii) 2x + 3y – 12 = 0 (iii) 4x + 6y – 16 = 0

Formula or theorem used: For grouped data use class marks and frequencies systematically. Apply the direct, assumed-mean, or step-deviation formula and show the working table totals.

Working: Write the relevant mean, median, or mode formula, identify the modal/median class, substitute the table totals, and interpret the numerical result in the question's context.

After substitution and simplification:

0.099 ppm

Formula or theorem used: Write the surface-area or volume formula for every solid before combining them. Hidden joining faces are excluded from exposed surface area.

Working: Identify every radius, height, and slant height; calculate the component measures, add or subtract them as the model requires, and attach cubic or square units.

After substitution and simplification:

44 m2, ₹ 22000

Formula or theorem used: Use \(C=2\pi r\), \(A=\pi r^2\), sector area \(=\frac{\theta}{360^\circ}\pi r^2\), and arc length \(=\frac{\theta}{360^\circ}2\pi r\).

Working: Split the figure into sectors, segments, triangles, or rectangles; calculate each labelled area and combine them with the correct addition or subtraction.

After substitution and simplification:

88.44 cm2

\(OA=OB\) (radii), \(PA=PB\) (tangents from \(P\)), and \(OP\) is common. Thus \(\triangle OAP\cong\triangle OBP\), so \(OP\) bisects \(\angle APB\).

Also \(OA\perp PA\). In right triangle \(OAP\),

Let the horizontal distance be \(x\) and the total height be \(H\).

Therefore the tower alone is \(H-20=20(\sqrt3-1)\text{ m}\).