Live practice

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:

Numbers 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:

Length = \(20\,\mathrm{m}\) and breadth = \(16\,\mathrm{m}\).

The bag contains only lemon-flavoured candies. Therefore drawing orange is impossible and drawing lemon is certain:

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:

₹ 211

The radius is \(r=2.5\) mm. The two hemispheres together have axial length \(2r=5\) mm, so the cylindrical length is \[\begin{aligned}h &= 14-5\\&= 9\end{aligned}\] mm.

Only curved surfaces are exposed:

Using \(\pi=\frac{22}{7}\), \(A=220\text{ mm}^2\).

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:

20.4375 cm2 ; 686.0625 cm2

Radii are perpendicular to tangents, so \[\begin{aligned}\angle OPT &= \angle OQT\\&= 90^\circ\end{aligned}\]. In quadrilateral \(OPTQ\),

Therefore \(\angle PTQ=70^\circ\).

Formula or theorem used: Draw and label a right triangle for each line of sight. Use \(\tan\theta=\frac{\text{vertical height}}{\text{horizontal distance}}\).

Working: Define the unknown height or distance, write a tangent equation for each observation, solve the equations, and include the observer's height when the question requires it.

After substitution and simplification:

19 \(3\,\mathrm{m}\)

For acute angles, cosine decreases continuously from 1 to 0 as the angle increases from \(0^\circ\) to \(90^\circ\). It therefore cannot take the same value at two different acute angles.

Since \(\cos A=\cos B\) and both angles are acute, the angles must be equal:

Formula or theorem used: Use \(PQ=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\) or the section formula \(\left(\frac{mx_2+nx_1}{m+n},\frac{my_2+ny_1}{m+n}\right)\).

Working: Substitute the coordinates before simplifying. For a shape, compare the required side lengths or diagonals and state the conclusion.

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) Square (ii) No quadrilateral (iii) Parallelogram