Live practice

For every tangential quadrilateral, the sums of opposite sides are equal:

In a parallelogram, \(AB=CD\) and \(AD=BC\). Substituting in \(1\),

Thus adjacent sides are equal; together with opposite-side equality, all four sides are equal. Hence the parallelogram is a rhombus.

Let tower height be \(h\). The initial and later distances from the foot are

In 6 seconds the car covers

The remaining distance \(y\) is half the distance covered in 6 seconds, so the remaining time is \(3\) seconds.

Formula or theorem used: In a right triangle use \(\sin\theta=\frac{P}{H}\), \(\cos\theta=\frac{B}{H}\), and \(\tan\theta=\frac{P}{B}\), together with the standard-angle values and identities.

Working: Write the relevant ratio or identity, substitute the known value, simplify one equality at a time, and state restrictions when a denominator could be zero.

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) False (ii) True (iii) False (iv) False (v) True

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.

After substitution and simplification:

Ratio \(1:1\); point \(\left(-\frac32,0\right)\)

\(\angle DOC\) and \(\angle BOC\) form a linear pair. Therefore,

In \(\triangle ODC\),

Since \(\triangle ODC\sim\triangle OBA\), vertex \(C\) corresponds to \(A\). 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.

After substitution and simplification:

65th term

Good pens (=132), total pens (=132+12=144).

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:

Median = 137 units, Mean = 137.05 units, Mode = 135.76 units.
The three measures are approximately the same in this case.

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:

1.131 m3 (approx.)

Let the equal tangent lengths from \(A\) be \(x\). Equal tangents from \(B\) and \(C\) give

The semiperimeter is \(s=x+14\). Since the inradius is \(4\,\mathrm{cm}\), the area is (\Delta=rs=4\(x+14\)).

By Heron's formula,

Equating and squaring, (16\(x+14\)^2=48x\(x+14\)), so \(x=7\). Therefore \[\begin{aligned}AB &= 7+8\\&= 15\end{aligned}\] cm and \[\begin{aligned}AC &= 7+6\\&= 13\end{aligned}\] cm.