Latur School Exam Papers 2026
Raja Narayanlal Lahoti English School
Class 9 | Term Test-I
26 Circle Geometry (Internal Choice)
Draw \(\triangle ABC\) with \(AB=5\text{ cm}\), \(\angle A=70^\circ\) and \(\angle B=60^\circ\). Draw the circumcircle of \(\triangle ABC\). Is the centre inside or outside the triangle?[cite: 49]
OR
Prove that Chords of a circle that subtend equal angles at the centre are equal.[cite: 49]
Concept:
1
Main: The circumcircle passes through all three vertices. Its center (circumcenter) is the intersection of the perpendicular bisectors of the sides. For acute triangles, it lies inside; for obtuse, outside; for right-angled, on the hypotenuse.
2
OR: Use the Side-Angle-Side (SAS) congruence criterion between the two triangles formed by the radii and the chords at the center.
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Hint:
1
Main: Find the third angle \(\angle C\). Since the sum of angles is \(180^\circ\), \(180^\circ - (70^\circ + 60^\circ) = 50^\circ\). Are all angles acute (\(< 90^\circ\))? If yes, the center is inside!
2
OR: Draw a circle with center O. Take chords AB and CD. Given \(\angle AOB = \angle COD\). You know that OA, OB, OC, and OD are all radii of the same circle. Match the parts!
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Mistakes to Avoid:
!
Construction Error: For the main question, students often draw angle bisectors instead of perpendicular bisectors. Angle bisectors find the INCENTER, not the CIRCUMCENTER!
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MAIN QUESTION SOLUTION:
Steps of Construction:
1. Draw a line segment \(AB = 5\text{ cm}\).
2. Construct \( \angle A = 70^\circ \) and \( \angle B = 60^\circ \). Let the rays intersect at C. [1 M]
3. Draw the perpendicular bisectors of sides AB and AC. Let them intersect at point O.
4. With O as centre and radius OA, draw a circle. This is the required circumcircle. [1 M]
To check the centre position, calculate \(\angle C\):
\( \angle C = 180^\circ - (70^\circ + 60^\circ) = 50^\circ \)
Since all three angles (\(70^\circ, 60^\circ, 50^\circ\)) are acute (less than \(90^\circ\)), the triangle is an acute-angled triangle.
Therefore, the circumcentre lies inside the triangle. [1 M]
OR
OR QUESTION SOLUTION:
Given: A circle with centre O. Two chords AB and CD such that they subtend equal angles at the centre, i.e., \( \angle AOB = \angle COD \).
To Prove: \( AB = CD \) [½ M]
Proof:
In \( \Delta AOB \) and \( \Delta COD \),
\( OA = OD \) (Radii of the same circle) [½ M]
\( \angle AOB = \angle DOC \) (Given) [½ M]
\( OB = OC \) (Radii of the same circle) [½ M]
By SAS (Side-Angle-Side) Congruence Criterion:
\( \Delta AOB \cong \Delta DOC \) [½ M]
Since the triangles are congruent, their corresponding parts are equal.
Therefore, \( AB = CD \) (by CPCT). [½ M]
Hence Proved.
27 Coordinate Geometry (Midpoint & Distance)
The coordinates of the vertices of \(\triangle ABC\) are A(7,2), B(9,10) and C(1,4). If E and F are the midpoints of AB and AC respectively, prove that \(EF = \frac{1}{2}BC\).[cite: 49]
Concept:
1
Midpoint Formula: The midpoint \(M\) of a line segment joining \((x_1, y_1)\) and \((x_2, y_2)\) is \( M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) \).
2
Distance Formula: The distance \(d\) between two points is \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \).
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Hint:
1
Step 1: Find coordinates of E using A(7,2) and B(9,10).
Step 2: Find coordinates of F using A(7,2) and C(1,4).
Step 3: Calculate length of EF using the distance formula.
Step 4: Calculate length of BC using the distance formula. Compare the two!
Step 2: Find coordinates of F using A(7,2) and C(1,4).
Step 3: Calculate length of EF using the distance formula.
Step 4: Calculate length of BC using the distance formula. Compare the two!
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Mistakes to Avoid:
!
Just citing the theorem: Stating "By Midpoint Theorem, \(EF = \frac{1}{2}BC\), Hence proved" will yield zero marks. In coordinate geometry problems where specific coordinates are provided, you MUST prove it algebraically by calculating the actual distances.
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Given vertices: \( A(7,2), B(9,10), C(1,4) \).
Step 1: Find coordinates of E (midpoint of AB)
\( E = \left( \frac{7+9}{2}, \frac{2+10}{2} \right) = \left( \frac{16}{2}, \frac{12}{2} \right) \)
\( E = (8, 6) \) [1 M]
Step 2: Find coordinates of F (midpoint of AC)
\( F = \left( \frac{7+1}{2}, \frac{2+4}{2} \right) = \left( \frac{8}{2}, \frac{6}{2} \right) \)
\( F = (4, 3) \) [½ M]
Step 3: Calculate distance EF
\( EF = \sqrt{(4 - 8)^2 + (3 - 6)^2} \)
\( EF = \sqrt{(-4)^2 + (-3)^2} \)
\( EF = \sqrt{16 + 9} = \sqrt{25} \)
\( EF = 5 \text{ units} \) [1 M]
Step 4: Calculate distance BC
\( BC = \sqrt{(1 - 9)^2 + (4 - 10)^2} \)
\( BC = \sqrt{(-8)^2 + (-6)^2} \)
\( BC = \sqrt{64 + 36} = \sqrt{100} \)
\( BC = 10 \text{ units} \) [½ M]
From our calculations, \( EF = 5 \) and \( BC = 10 \).
Therefore, \( 5 = \frac{1}{2}(10) \implies EF = \frac{1}{2}BC \).
Hence Proved.
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