Mixed Practice & Chapter Challenge — Class 9 Mathematics Chapter 1

🖼 Image PendingA wide cinematic vista: a vast workshop floor at dawn, with a coordinate grid faintly etched into the polished surface and a craftsperson assembling a small mechanical sculpture using rulers, compasses, and protractors — every tool the chapter has provided

AI Generation Prompt

Ultra-wide cinematic banner (16:5 ratio). A vast craftsperson's workshop at dawn, with morning light streaming in through tall windows. The polished wooden floor is faintly etched with a coordinate grid that fades softly into the wood grain. On a long workbench centred in the frame, a craftsperson is assembling a small mechanical sculpture from precise components — measuring with a ruler, marking with a compass, comparing angles with a protractor. The image conveys: every tool of this chapter — coordinates, distance, midpoints, reflections, circles — comes together in one craft. Painterly cinematic illustration. Dark background. No text, no labels.

✦Think About It

You have now built a complete toolkit:

Coordinates to name any point in a plane
Quadrants to read a point at a glance
Distance formula to measure between any two points
Midpoint formula to find the halfway point of a segment
Reflections to mirror shapes
Circles & loci to describe sets of equidistant points

The questions on this page don't tell you which tool to use. That is exactly the point. The hardest part of mathematics is not applying a formula — it is recognising which formula applies.

Ready?

If a question mentions 'a square' — what tools can verify a square? (Equal sides via distance; right angles via... we'll see.)

✦भगवद्गीता — १८.७८

The Closing Verse of the Gītā

यत्र योगेश्वरः कृष्णो यत्र पार्थो धनुर्धरः।

तत्र श्रीर्विजयो भूतिर्ध्रुवा नीतिर्मतिर्मम॥

'जहाँ ज्ञान (कृष्ण) और कार्य (अर्जुन) मिलते हैं, वहीं सच्ची विजय, समृद्धि और स्थिर बुद्धि होती है।'

'Where knowledge meets action, there is sure victory, prosperity, and steady judgement.'

The Gītā ends by joining knowledge (the theory you've learned across this chapter) and action (the practice on this page). Mathematics works the same way. Knowing a formula and applying it under uncertainty are two different skills; the second is what this page builds.

NCERT Q3 — Reading a rectangle from coordinates

NCERT Q8 — Building special triangles by hand

NCERT Q16 — When is a quadrilateral a square?

Three fresh integration problems

Analogical ReasoningLevel 5 · Pattern & Analogy

Across all 14 problems on this page (including the worked examples), the same three formulas keep showing up: distance, midpoint, and the equation of a circle. Why are these three so often enough?

Mixed Practice — Six More Problems

Three vertices of a parallelogram are $(2, 1)$ , $(5, 4)$ , $(8, 1)$ . Find the fourth vertex. (Two answers possible — find both.)
Show that the points $(0, 0)$ , $(3, 0)$ , $(3, 4)$ , $(0, 4)$ form a rectangle, and find its perimeter and area.
Find the equation of the circle that has $A(2, 3)$ and $B(8, 11)$ as endpoints of a diameter.
Triangle $PQR$ has vertices $P(0, 0)$ , $Q(6, 0)$ , $R(3, h)$ . For what value of $h$ is the triangle equilateral? Justify.
Reflect the triangle from Problem 4 (with $h = 3\sqrt{3}$ ) across the x-axis. Find the area of the resulting triangle. Compare to the original.
A circle has equation $(x - 3)^2 + (y - 4)^2 = 25$ . The point $P(8, 4)$ lies on the circle (check). Find the coordinates of the point diametrically opposite $P$ on the circle.

Answers:

Three possibilities depending on which pair are opposite: $(5, -2)$ , $(-1, 4)$ , or $(11, 4)$ . The most natural choice (when the given vertices form three corners in order) is $(5, -2)$ .
Sides: $3, 4, 3, 4$ . Diagonals: both $5$ . Equal diagonals + equal opposite sides + right angle at $(0, 0)$ ⇒ rectangle. Perimeter = 14, area = 12.
Centre = midpoint of $AB$ = $(5, 7)$ . Radius = $|AB|/2 = \sqrt{36+64}/2 = 5$ . Equation: $(x - 5)^2 + (y - 7)^2 = 25$ .
Equilateral ⇒ all sides equal. $|PQ| = 6$ , so we need $|PR| = |QR| = 6$ . $|PR| = \sqrt{9 + h^2}$ . Setting $\sqrt{9 + h^2} = 6$ : $h^2 = 27$ , $h = 3\sqrt{3}$ (taking the positive root for the triangle above the x-axis).
Reflected vertices: $(0, 0), (6, 0), (3, -3\sqrt{3})$ . Side lengths unchanged ⇒ same area as original = $\frac{\sqrt{3}}{4} \cdot 6^2 = 9\sqrt{3}$ . Reflection preserves area exactly.
Centre $(3, 4)$ . The diametrically opposite point of $P(8, 4)$ is reflection of $P$ through the centre: $(2 \cdot 3 - 8, 2 \cdot 4 - 4) = (-2, 4)$ .

What if…

What if every map you have ever used had been built without coordinates?

Quick Check

Q1.A quadrilateral has vertices $A(0, 0), B(4, 0), C(4, 4), D(0, 4)$ . What kind of quadrilateral is it?