Mathematics - MCQ Practice Questions

The diagonals of a rhombus bisect each other at right angles. So half-diagonals are $8\text{ cm}$ and $6\text{ cm}$. Each side is:

Perimeter $=4\times 10=\boxed{40\text{ cm}}$

First, find $BC$ using the Pythagorean theorem:

A key property states that the median from the right angle to the hypotenuse equals half the hypotenuse:

When two circles touch externally, the distance between their centres equals the sum of their radii:

The arc length formula is:

By the Midpoint Theorem, the line segment joining the midpoints of two sides of a triangle is parallel to the third side and equals half its length:

Volume of a cone:

First find the other leg:

Area of right triangle:

The sum of all exterior angles of any regular polygon is $360°$. The number of sides is:

The perpendicular from the centre bisects the chord, so half the chord $=8\text{ cm}$. Using the Pythagorean theorem:

Check: $7^2+24^2=49+576=625=25^2$. So this is a right triangle with legs $7$ and $24$.

Understanding:

We must find the perimeter of a rhombus whose diagonals are given.

Formula:

The diagonals of a rhombus bisect each other at right angles. The side $a$ of the rhombus is:

and the perimeter is $P=4a$.

Step 1: Find the half-diagonals.

Step 2: Find the side using Pythagoras.

Step 3: Find the perimeter.

Answer:

The perimeter of the rhombus is 40 cm.

Quick Tip:

The half-diagonals $6$ and $8$ form a classic $3$-$4$-$5$ right triangle (scaled by $2$), so the hypotenuse $=10$ can be spotted instantly without full calculation.

Understanding:

We must find the length of a tangent from an external point to a circle.

Formula:

The tangent from an external point is perpendicular to the radius at the point of contact, so by Pythagoras:

Step 1: Substitute the given values.

Answer:

The length of the tangent is 12 cm.

Quick Tip:

The numbers $5$, $12$, $13$ form a Pythagorean triple — recognising it saves computation time in exams.

Understanding:

We must find the height of a trapezium given its area and parallel sides.

Formula:

Step 1: Substitute the known values and solve for $h$.

Answer:

The height of the trapezium is 10 cm.

Understanding:

Two chords intersect inside a circle. We must find the length of one segment using the intersecting chords theorem.

Formula:

When two chords intersect inside a circle at point $P$:

Step 1: Substitute and solve for $PD$.

Answer:

The length of $PD$ is 12 cm.

Quick Tip:

The intersecting chords theorem states that the products of the two segments of each chord are equal. A common error is to add instead of multiply the segments.

Understanding:

We must find the area of a triangle with given side lengths.

Formula:

First check whether the triangle is right-angled:

If yes, use:

Step 1: Check for right angle.

So the triangle is right-angled with legs $7$ cm and $24$ cm.

Step 2: Compute the area.

Answer:

The area of the triangle is 84 square centimetres.

Quick Tip:

The triple $7$-$24$-$25$ is a standard Pythagorean triple worth memorising alongside $3$-$4$-$5$ and $5$-$12$-$13$.

Understanding:

We must find the length of a chord given the radius and its distance from the centre.

Formula:

The perpendicular from the centre bisects the chord. Using Pythagoras on the right triangle formed:

Step 1: Substitute the values.

Answer:

The length of the chord is 16 cm.

Quick Tip:

The numbers $6$, $8$, $10$ form a scaled $3$-$4$-$5$ triple, making the half-chord $=8$ easy to spot.

Understanding:

We must find the area of a right triangle given the hypotenuse and one leg.

Formula:

First find the missing leg, then use:

where $a$ and $b$ are the two legs.

Step 1: Find the other leg $b$.

Step 2: Compute the area.

Answer:

The area of the right triangle is 120 square centimetres.

Quick Tip:

The triple $10$-$24$-$26$ is simply $2\times (5\text{-}12\text{-}13)$. Recognising such scaled triples avoids lengthy square-root computation.

Understanding:

We must identify the measure of an angle inscribed in a semicircle.

Formula:

The inscribed angle theorem states:

Step 1: Identify the central angle for a semicircle.

A semicircle subtends an arc of $180^{\circ }$ at the centre.

Step 2: Apply the inscribed angle theorem.

Answer:

The angle inscribed in a semicircle is always $90^{\circ }$.

Quick Tip:

This is Thales' theorem — any angle inscribed in a semicircle (i.e., the angle at the circumference subtended by the diameter) is a right angle. It is one of the most frequently tested circle theorems.

Understanding:

We must find the area of a rectangle given its perimeter and diagonal.

Formula:

Let the length and width be $l$ and $w$. Then:

Step 1: Find $l+w$.

Step 2: Find $l^2+w^2$.

Step 3: Compute the area.

Verification:

$l$ and $w$ satisfy $t^2-35t+300=0$.

So $l=20\text{ cm}$ and $w=15\text{ cm}$. Check: $20+15=35$ ✓, $\sqrt{20^2+15^2}=\sqrt{625}=25$ ✓, Area $=20\times 15=300$ ✓.

Answer:

The area of the rectangle is 300 square centimetres.

Quick Tip:

The identity $\text{Area}=\frac{(l+w{)}^2-(l^2+w^2)}{2}$ converts the perimeter and diagonal directly into the area without solving for $l$ and $w$ individually.