The proofs the syllabus asks for, grouped by type, each written out step by step with the reasons examiners look for.
What the syllabus asks you to prove
The Class 10 Mathematics syllabus for 2026-27 lists, among others: proofs of the irrationality of root 2, root 3 and root 5; the proof of the Basic Proportionality Theorem (its converse and the similarity criteria are stated without proof); the proofs that a tangent is perpendicular to the radius at the point of contact and that tangents from an external point are equal; and the proof of sin squared A + cos squared A = 1 with simple identities. (CBSE Class 10 Mathematics syllabus 2026-27)
The proofs below are written by us, in our own words and with our own figures and letters. Learn the reasoning, not a form of words: an examiner gives marks for each correct step with its reason.
Proof by contradiction: irrational numbers
Assume the number is rational, follow the algebra, and reach something impossible.
Show that there are no integers \(p\) and \(q\), with \(q \neq 0\), for which \(\sqrt{5} = \dfrac{p}{q}\); that is, \(\sqrt{5}\) is irrational.
Suppose, for contradiction, that \(\sqrt{5}\) is rational. Then \(\sqrt{5} = \dfrac{p}{q}\) for integers \(p, q\) with \(q \neq 0\) and no common factor other than \(1\).
Squaring: \(5q^2 = p^2\). So \(5\) divides \(p^2\), and because \(5\) is prime, \(5\) divides \(p\). Write \(p = 5k\).
Then \(5q^2 = 25k^2\), so \(q^2 = 5k^2\). So \(5\) divides \(q^2\), and again \(5\) divides \(q\).
Now \(5\) divides both \(p\) and \(q\), which contradicts "no common factor other than \(1\)". So \(\sqrt{5}\) is irrational.
Where marks slip: The step examiners look for is the reason \(5 \mid p^2 \Rightarrow 5 \mid p\): say that \(5\) is prime.
Given that \(\sqrt{3}\) is irrational, prove that \(7 - 2\sqrt{3}\) is irrational.
Suppose \(7 - 2\sqrt{3} = r\), where \(r\) is rational.
Then \(\sqrt{3} = \dfrac{7 - r}{2}\). The right-hand side is rational, because rational numbers are closed under subtraction and division by a non-zero rational.
That would make \(\sqrt{3}\) rational, which contradicts the given fact. So \(7 - 2\sqrt{3}\) is irrational.
Where marks slip: Name the closure fact. "So it is rational" with no reason loses the reasoning mark.
Geometric proofs: triangles
Given, to prove, construction, then numbered steps with a reason for each.
In triangle \(PQR\), a line parallel to \(QR\) meets \(PQ\) at \(S\) and \(PR\) at \(T\). Prove that \(\dfrac{PS}{SQ} = \dfrac{PT}{TR}\).
Join \(QT\) and \(RS\). Let \(h_1\) be the perpendicular distance from \(T\) to line \(PQ\), and \(h_2\) the perpendicular distance from \(S\) to line \(PR\).
Triangles \(PST\) and \(QST\) have bases \(PS\) and \(SQ\) on the same line and the same height \(h_1\), so \(\dfrac{\text{ar}(PST)}{\text{ar}(QST)}\) \[= \dfrac{\tfrac12 \cdot PS \cdot h_1}{\tfrac12 \cdot SQ \cdot h_1}\] \(= \dfrac{PS}{SQ}\).
In the same way, using \(h_2\): \[\dfrac{\text{ar}(PST)}{\text{ar}(RST)} = \dfrac{PT}{TR}\]
Triangles \(QST\) and \(RST\) stand on the same base \(ST\) between the same parallels \(ST\) and \(QR\), so \(\text{ar}(QST) = \text{ar}(RST)\).
The two ratios have the same numerator and equal denominators, so \(\dfrac{PS}{SQ} = \dfrac{PT}{TR}\).
Where marks slip: Draw and label the figure, write "Given", "To prove" and "Construction", and give the reason for equal areas.
From a point \(P\) outside a circle with centre \(O\), two tangents touch the circle at \(A\) and \(B\). Prove that \(PA = PB\).
Join \(OA\), \(OB\) and \(OP\).
\(\angle OAP = \angle OBP = 90^\circ\) (a tangent is perpendicular to the radius at the point of contact).
\(OA = OB\) (radii) and \(OP\) is common to both triangles.
So \(\triangle OAP \cong \triangle OBP\) (RHS), and therefore \(PA = PB\) (corresponding parts of congruent triangles).
Where marks slip: Give a reason in brackets on every line; the RHS reason is the key step.
Trigonometric identities
Start from one side, usually the more complicated one, and work to the other. Every identity rests on \(\sin^2 A + \cos^2 A = 1\), which follows from Pythagoras' theorem: divide \(a^2 + b^2 = c^2\) by \(c^2\) in a right triangle.
An identity: expand and use \(\sin^2 A + \cos^2 A = 1\)
Which theorems have to be proved in Class 10 Maths?
The 2026-27 syllabus asks for proofs that root 2, root 3 and root 5 are irrational, the Basic Proportionality Theorem, that a tangent is perpendicular to the radius at the point of contact, that tangents from an external point are equal, and sin squared A + cos squared A = 1 with simple identities. Other results, such as the converse of the Basic Proportionality Theorem and the similarity criteria, are stated without proof.
Do I have to write a proof word for word?
No. Marks are for correct steps with reasons. Any correct, complete proof in your own words earns the marks.
How do I prove that a number like 7 − 2√3 is irrational?
Assume it equals a rational number r, rearrange to make √3 the subject, and note that the other side is rational. That contradicts √3 being irrational.
Official facts on this page were last checked against CBSE's own documents on 5 October 2026. CBSE Math Revision is independent and not affiliated with CBSE or NCERT. If a newer CBSE notice or your school says something different, follow CBSE and the school. Worked examples are written by CBSE Math Revision and checked twice (by computer algebra and by hand); the mark splits are our illustration, not CBSE's scheme for any paper.