Are you looking for solution of Class 10 ICSE Important Maths Questions for 2026 Exams? Here you’ll find complete, step-by-step solutions for every question. Designed for both students and teachers, these answers make board exam revision easier with clear explanations, formula highlights, and downloadable PDF options.
Q 1: In the adjoining diagram, O is the centre of the circle and PT is the tangent. Find the value of x.
Solution
Step 1: Identify the angle between the radius and the tangent.
Since OP is the radius and PT is the tangent at point P, the angle formed between them is a right angle.
\angle OPT = 90^{\circ}
Step 2: Calculate the adjacent angle inside the triangle.
The line segment QOT is a straight line passing through the centre O.
Therefore, \angle POQ and \angle POT form a linear pair.
We are given that \angle POQ = 110^{\circ}.
\angle POT = 180^{\circ} - 110^{\circ} = 70^{\circ}
Step 3: Apply the Angle Sum Property.
Now, look at the right-angled triangle \Delta OPT.
We know two of its angles (\angle OPT = 90^{\circ} and \angle POT = 70^{\circ}).
The sum of angles in \Delta OPT is 180^{\circ}.
x + \angle OPT + \angle POT = 180^{\circ}
x + 90^{\circ} + 70^{\circ} = 180^{\circ}
x + 160^{\circ} = 180^{\circ}
x = 20^{\circ}
Final Answer: The value of \boxed {x = 20} .
Solution to Question 2: Perpendicular Lines and Slopes
Problem Statement: If the lines kx-y+4=0 and 2y=6x+7 are perpendicular to each other, find the value of k.
Concept Explanation:
For coordinate geometry questions involving perpendicular lines, the easiest approach is to convert the equations into the slope-intercept form (y = mx + c), where m represents the slope. If two lines are perpendicular, the product of their slopes is always -1 (i.e., m_1 \times m_2 = -1).
Step-by-Step Solution:
- Step 1: Find the slope of the first line. Equation of the first line: kx - y + 4 = 0 Rearranging to make y the subject:
y = kx + 4
Comparing this with y = mx + c, the slope of the first line (m_1) is:
m_1 = k
- Step 2: Find the slope of the second line. Equation of the second line: 2y = 6x + 7 Divide the entire equation by 2 to isolate y:
y = 3x + \frac{7}{2}
Comparing this with y = mx + c, the slope of the second line (m_2) is:
m_2 = 3
- Step 3: Apply the condition for perpendicular lines.
Since the two lines are perpendicular, the product of their slopes must equal -1.
m_1 \times m_2 = -1
Substitute the values we found:
k \times 3 = -1
k = -\frac{1}{3}
Final Answer:
The value of k is -1/3.
Q 3: The roots of the equation (q-r)x^{2}+(r-p)x+(p-q)=0 are equal. Prove that: 2q=p+r, that is p, q & r are in A.P.
Solution
Step 1: Identify the coefficients of the quadratic equation.
Comparing the given equation with the standard quadratic form ax^2 + bx + c = 0, we get:
a = (q-r)
b = (r-p)
c = (p-q)
Step 2: Apply the condition for equal roots.
For a quadratic equation to have equal roots, its discriminant (D) must be equal to zero.
D = b^2 - 4ac = 0
Step 3: Substitute the coefficients into the discriminant formula and expand.
(r-p)^2 - 4(q-r)(p-q) = 0
Using the algebraic identity (a-b)^2 = a^2 - 2ab + b^2, we expand the terms:
r^2 - 2rp + p^2 - 4(qp - q^2 - rp + rq) = 0
r^2 - 2rp + p^2 - 4qp + 4q^2 + 4rp - 4qr = 0
Step 4: Simplify the expression by combining like terms.
Grouping the squared terms and the product terms together:
p^2 + 4q^2 + r^2 - 4pq - 4qr + 2rp = 0
Step 5: Recognize the algebraic identity.
The expression matches the expansion of (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca.
Here, we can rewrite our equation as:
(p)^2 + (-2q)^2 + (r)^2 + 2(p)(-2q) + 2(-2q)(r) + 2(p)(r) = 0
This simplifies to a perfect square:
(p - 2q + r)^2 = 0
Step 6: Solve for the final relation.
Taking the square root of both sides gives:
p - 2q + r = 0
p + r = 2q
Rearranging gives:
2q = p + r
Final Answer: We have successfully proved that
\boxed {2q = p + r}
, which is the necessary condition for p, q, and r to be in Arithmetic Progression (A.P.).