1
For the ratio 3:2 the equal points on the ray are —
💡 Hint:
3 plus 2.
Explanation:
5.
Class 10 · Maths · Chapter 11 · बिहार बोर्ड (BSEB)CBSE · NCERT 2026-27
Constructions
Constructions
How to use this page:
1. Read — Division of a segment · similar triangles · external tangents · Exercises 11.1 and 11.2, diagram, worked example, board tip
2. Check — each lesson has its own questions; the number follows the lesson
3. Mastery ★ — all of that lesson correct. Redo the wrong ones
4. The memory figure shows equal points on a ray and a parallel line dividing a segment in the ratio 3:2.
This is Chapter 11 of the Bihar book and is not in the NCERT 14-chapter book; NCERT Chapter 11 is Areas Related to Circles. Progress stays in this browser.
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- 1 A construction and its justification
- 2 The ratio with one ray
- 3 Two parallel rays
- 4 A smaller similar triangle
- 5 A larger similar triangle
- 6 Two tangents from an outside point
- 7 The remaining tangents
- Chapter winner — every lesson at mastery ★
🎬 Shorts
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📄 PDF download — notes + answer key
A construction and its justification
रचना और उसका औचित्य · Bihar 11.1 · Class 9 constructions continue
A compass and a straight edge, not a measuring scaleNotes
Class 9 already has the angle bisector, the perpendicular bisector and the constructions of triangles. This chapter adds three new jobs with the same tools.
With every construction you also write why it is correct. The book asks for a justification in both Exercise 11.1 and Exercise 11.2. The NCERT 2026-27 book of 14 chapters does not print this chapter.
Three jobs, in this orderActivity
First divide a segment in a given ratio. Then build a triangle similar to a given triangle, with a scale factor smaller or larger than 1. At the end draw the two tangents to a circle from an outside point. A similar quadrilateral or polygon can follow the same steps, but it has no exercise of its own.
Worked exampleExample
Question: How many equal points go on the ray for the ratio 3:2?
Formula: number of points = m + n.
Substitution: m = 3, n = 2. 3 + 2 = 5.
Answer: 5 points. The parallel passes through the third point.
10-second revision
- The tools are a compass and a straight edge
- The ratio m:n needs m+n points
- Write a justification in every answer
Board tip · BSEBBoard tip
On BSEB a figure alone is not enough. Name the theorem in one sentence.
Board tip · CBSEBoard tip
The CBSE 2026-27 book does not have this chapter. On the Bihar paper this is Chapter 11.
Check your understandingall correct = mastery ★
2
The exercises of this chapter do not ask for a justification.
Explanation:
False — both exercises ask for a justification.
3
In the ratio m:n the number of points is ______.
💡 Hint:
Add the two whole numbers.
Explanation:
m+n.
4
NCERT 2026-27 Chapter 11 is —
💡 Hint:
Constructions is not in that book.
Explanation:
Areas Related to Circles.
5
Write the three jobs of this chapter in order.
Model answer (board format)
Divide a segment in a ratio, construct a similar triangle for a given scale factor, and draw two tangents from an outside point.
Check your answer:
The ratio with one ray
एक किरण से अनुपात · Construction 11.1 · Exercise 11.1 question 1
The third point, out of fiveNotes
Draw AB. From A take a ray AX at an acute angle. On it cut m+n equal arcs. Join the last point to B. Through the m-th point draw a parallel in the same direction. It meets AB at C.
For 3:2 there are five points and the parallel passes through A3. AC:CB = 3:2. Justification: by the basic proportionality theorem a parallel cuts the ray and the segment in the same ratio. Equal arcs give AA3:A3A5 = 3:2.
Thirteen arcs on question 1Activity
The length is 7.6 cm and the ratio is 5:8. Cut 13 equal arcs on the ray. Join the thirteenth to B. Draw the parallel through the fifth. Check with a measurement, and also write the calculation of the parts.
Worked exampleExample
Question: Divide a segment of 7.6 cm in the ratio 5:8. Both parts.
Formula: part = (share / m+n) × length.
Substitution: (5/13)×7.6 = 38/13 cm. (8/13)×7.6 = 60.8/13 cm.
Answer: 38/13 cm and 60.8/13 cm. The sum is 7.6 cm.
10-second revision
- The ratio 5:8 needs 13 points
- The parallel starts at the fifth point
- 38/13 cm and 60.8/13 cm
Board tip · BSEBBoard tip
On BSEB the 13 arcs should look equal. Keep the compass at one opening.
Board tip · CBSEBoard tip
The current CBSE book does not ask this question. On the Bihar paper write both the measured parts and the justification.
Check your understandingall correct = mastery ★
1
To divide 7.6 cm in the ratio 5:8, the points on the ray are —
💡 Hint:
5 plus 8.
Explanation:
13.
2
In the ratio 5:8 the parallel is drawn through the eighth point.
Explanation:
False — through the fifth point.
3
In the 5:8 division of 7.6 cm the larger part is ______ cm.
💡 Hint:
(8/13)×7.6.
Explanation:
60.8/13 cm.
4
Write the justification of the 3:2 construction in two lines.
Model answer (board format)
A3C is parallel to A5B. By basic proportionality, AC:CB = AA3:A3A5 = 3:2.
Check your answer:
Two parallel rays
दो समांतर किरणें · The second way of Construction 11.1
m points at A, n points at BNotes
The second way uses one join. Draw AX at an acute angle with AB. From B draw BY parallel to AX, by copying the corresponding angle.
Cut m equal arcs on AX and n equal arcs on BY. Join the last points. The join meets AB at C. For 3:2 there are three arcs on AX and two on BY. Triangles AA3C and BB2C are similar, because the rays are parallel and the vertically opposite angles at C are equal. So AC:CB = 3:2.
Both ways give the same CActivity
The first way puts m+n points on one ray. The second way splits the points across two rays. The ratio stays the same. In the answer say which way you chose.
Worked exampleExample
Question: Divide AB in the ratio 3:2 by the second way. How many arcs on each ray?
Formula: m arcs on AX, n arcs on BY.
Substitution: m = 3, n = 2.
Answer: 3 on AX and 2 on BY. A3B2 cuts AB in the ratio 3:2.
10-second revision
- BY is parallel to AX
- The arcs are 3 and 2
- Similar triangles give the ratio
Board tip · BSEBBoard tip
On BSEB show the copied corresponding angle, or the rays will not be accepted as parallel.
Board tip · CBSEBoard tip
The current CBSE book does not keep this construction. On the Bihar paper write one reason for the similarity.
Check your understandingall correct = mastery ★
1
In the second way, the arcs on BY for 3:2 are —
💡 Hint:
n = 2.
Explanation:
2.
2
In the second way BY is drawn parallel to AX.
Explanation:
True.
3
In the second way for 3:2, the arcs on AX are ______.
💡 Hint:
m = 3.
Explanation:
3.
4
A3B2 gives the ratio because the triangles are —
💡 Hint:
The sides are not equal, the ratios are.
Explanation:
They are similar.
5
The second way gives a different ratio from the first way.
Explanation:
False — both give 3:2.
6
Write one reason for the similarity in the second way.
Model answer (board format)
Because AX is parallel to BY, corresponding angles are equal, and the vertically opposite angles at C are equal. So triangle AA3C is similar to triangle BB2C.
Check your answer:
A smaller similar triangle
छोटा समरूप त्रिभुज · Construction 11.2 · factor below 1 · questions 2 and 5
As many points as the greater number, the parallel at the smallerNotes
The scale factor is the ratio of a new side to the given side. 3/4 means the new triangle is smaller. On the side of BC opposite A, take a ray BX at an acute angle.
Mark 4 points, because 4 is the greater number. Join B4 to C. Through B3 draw a parallel and take C′ on BC. Through C′ draw a parallel to CA and take A′ on BA. Each side of triangle A′BC′ is 3/4 of the given triangle. BC′:BC = 3:4, and the parallel to CA makes the angles equal, so the triangles are similar.
Questions 2 and 5 are both smallerActivity
In question 2 the sides are 4, 5 and 6 cm and the factor is 2/3. Mark three points and draw the parallel through the second. In question 5, BC = 6 cm, AB = 5 cm, angle B = 60 degrees, and the factor is 3/4. Four points, parallel through the third. The cosine rule gives the third side as √31 cm, and that side is also multiplied by 3/4.
Worked exampleExample
Question: Sides 4, 5, 6 cm. New sides at 2/3.
Formula: new side = factor × given side.
Substitution: (2/3)×4 = 8/3 cm, (2/3)×5 = 10/3 cm, (2/3)×6 = 4 cm.
Answer: 8/3 cm, 10/3 cm and 4 cm.
10-second revision
- The factor 3/4 needs 4 points
- The parallel is from the third point, inside the side
- At 2/3 the new sides are 8/3, 10/3, 4 cm
Board tip · BSEBBoard tip
On BSEB show both A′ and C′. One parallel alone is an unfinished construction.
Board tip · CBSEBoard tip
The current CBSE book does not ask this construction. On the Bihar paper write the factor as a ratio of sides.
Check your understandingall correct = mastery ★
1
For the factor 3/4 the points on the ray are —
💡 Hint:
The greater number.
Explanation:
4.
2
Two-thirds of a 6 cm side is ______ cm.
💡 Hint:
(2/3)×6.
Explanation:
4 cm.
3
The new triangle of factor 3/4 is larger than the given triangle.
Explanation:
False — 3/4 is less than 1.
4
In question 5, angle B is —
💡 Hint:
BC = 6 cm, AB = 5 cm.
Explanation:
60°.
5
Sides 5 cm and 6 cm with an included angle of 60 degrees. Write the third side and its 3/4.
Model answer (board format)
Cosine rule: AC² = 25+36−2×5×6×1/2 = 31. AC = √31 cm. The new part is (3/4)√31 cm.
Check your answer:
A larger similar triangle
बड़ा समरूप त्रिभुज · Second case of Construction 11.2 · questions 3, 4, 6, 7
Join the smaller number to CNotes
When the factor is greater than 1, the new triangle sticks out past the given triangle. For 5/3, mark five points. The smaller number is 3, so join B3 to C.
Through B5 draw a parallel to B3C. It meets BC extended at C′. Through C′ draw a parallel to CA, up to A′ on BA extended. BC′:BC = 5:3. Question 3 has factor 7/5, question 4 has 3/2, question 6 has 4/3 and question 7 has 5/3. All four use this outer step.
Find the third angle before you drawActivity
In question 6, angle B = 45 degrees and angle A = 105 degrees. The third angle is 180−105−45 = 30 degrees. Side BC = 7 cm. For the factor 4/3, mark four points, join the third to C, and take the parallel from the fourth onto the extension. In question 4 the base is 8 cm and the altitude is 4 cm. The perpendicular bisector of the base carries the altitude.
Worked exampleExample
Question: A right triangle has legs 4 cm and 3 cm. The new triangle is at 5/3.
Formula: hypotenuse = √(base² + height²). New side = (5/3) × old side.
Substitution: hypotenuse = √(16+9) = 5 cm. New sides (5/3)×4 = 20/3 cm, (5/3)×3 = 5 cm, (5/3)×5 = 25/3 cm.
Answer: 20/3 cm, 5 cm and 25/3 cm.
10-second revision
- For 5/3 join B3 to C
- The parallel meets the extension
- The third angle in question 6 is 30 degrees
Board tip · BSEBBoard tip
On BSEB draw the extension as a light line, so that C′ is visibly outside the given triangle.
Board tip · CBSEBoard tip
The current CBSE book does not ask this construction. On the Bihar paper do not swap the steps of a factor above 1 with those of a factor below 1.
Check your understandingall correct = mastery ★
1
For the factor 5/3, the point joined to C is —
💡 Hint:
The smaller number.
Explanation:
The third point.
2
In question 6, angle C is 30 degrees.
Explanation:
True — 180−105−45.
3
In question 7 the new hypotenuse is ______ cm.
💡 Hint:
(5/3)×5.
Explanation:
25/3 cm.
4
Write the original equal side of question 4 and the new base.
Model answer (board format)
Half the base is 4 cm and the altitude is 4 cm. The equal side is √(16+16) = 4√2 cm. At the factor 3/2 the new base is (3/2)×8 = 12 cm.
Check your answer:
Two tangents from an outside point
बाह्य बिंदु से दो स्पर्श रेखाएँ · Construction 11.3 · Exercise 11.2 questions 1 and 2
An angle in a semicircle is 90 degreesNotes
No tangent starts at an inside point. A point on the circle has one tangent, and it is perpendicular to the radius. An outside point has two tangents.
Join O to P. Take the midpoint M of OP. Draw the circle with centre M and radius MO. It cuts the given circle at Q and R. PQ and PR are the tangents. Angle PQO is 90 degrees in a semicircle, so PQ is perpendicular to the radius OQ.
The length from the difference of squares, before you measureActivity
The length of the tangent is √(OP² − r²). In question 1, r = 6 cm and the point is 10 cm from the centre. In question 2 the point lies on a concentric circle of radius 6 cm, and the given circle has radius 4 cm. Check the measurement against this calculation.
Worked exampleExample
Question: Radius 6 cm. The point is 10 cm from the centre. Length of the tangent.
Formula: length = √(OP² − r²).
Substitution: √(10² − 6²) = √(100 − 36) = √64 = 8 cm.
Answer: 8 cm. The two tangents are equal.
10-second revision
- None from inside, one from on the circle, two from outside
- Question 1 has length 8 cm
- Question 2 has length 2√5 cm
Board tip · BSEBBoard tip
On BSEB label M as the midpoint of OP. The semicircle sentence is the justification.
Board tip · CBSEBoard tip
The current CBSE book does not ask this construction. On the Bihar paper write √(OP² − r²) along with the measurement.
Check your understandingall correct = mastery ★
1
With radius 6 cm and distance 10 cm, the tangent is —
💡 Hint:
√(100−36).
Explanation:
8 cm.
2
Two tangents can be drawn from a point inside the circle.
Explanation:
False — not even one from inside.
3
For concentric circles of 4 cm and 6 cm the tangent is ______ cm.
💡 Hint:
√(36−16).
Explanation:
2√5 cm.
4
In Construction 11.3, angle PQO is —
💡 Hint:
A semicircle.
Explanation:
90°.
5
A point on the circle has exactly one tangent.
Explanation:
True.
6
Find the length in question 2 with formula, substitution and unit.
Model answer (board format)
Length = √(OP² − r²). OP = 6 cm, r = 4 cm. √(36−16) = √20 = 2√5 cm.
Check your answer:
The remaining tangents
शेष स्पर्श रेखाएँ · Exercise 11.2 questions 3 to 7 · summary 11.4
If the centre is not given, find it firstNotes
In question 3 the radius is 3 cm. On an extended diameter, P and Q are each 7 cm from the centre. Draw two tangents from each point. In question 4 the two tangents meet at 60 degrees and the radius is 5 cm. Half of that angle is 30 degrees.
In question 7 the circle is drawn with a bangle, and the centre is not given. The perpendicular bisectors of two non-parallel chords meet at the centre. After that, use the same outside-point construction. The three jobs in the summary are the same: a ratio, a similar triangle, and a pair of tangents.
In questions 5 and 6, fix the distance firstActivity
In question 5, AB = 8 cm. The circle at A has radius 4 cm and the circle at B has radius 3 cm. The tangent from B to the first circle is √(8²−4²). The tangent from A to the second circle is √(8²−3²). In question 6, AB = 6, BC = 8 and angle B = 90 degrees. The circle through B, C and D has diameter BC, because the angle at D is 90 degrees.
Worked exampleExample
Question: Radius 5 cm. Tangents at 60 degrees. The length.
Formula: Half the angle is 30 degrees. tan 30° = 1/√3 = r / length.
Substitution: length = 5 × √3 = 5√3 cm. OP = r / sin 30° = 10 cm. Check: √(10²−5²) = √75 = 5√3 cm.
Answer: 5√3 cm.
10-second revision
- Question 3 has length 2√10 cm
- At 60 degrees the length is 5√3 cm
- In question 6 one tangent is AB itself, of length 6 cm
Board tip · BSEBBoard tip
On BSEB, for the bangle question, first show two chords and their perpendicular bisectors.
Board tip · CBSEBoard tip
The current CBSE book does not keep this exercise. On the Bihar paper write that the two tangents have equal length.
Check your understandingall correct = mastery ★
1
With radius 3 cm and distance 7 cm, the tangent is —
💡 Hint:
√(49−9).
Explanation:
√40 = 2√10 cm.
2
The tangent inclined at 60 degrees has length ______ cm when the radius is 5 cm.
💡 Hint:
tan 30° = 5 / length.
Explanation:
5√3 cm.
3
The centre of a bangle circle comes from two parallel chords.
Explanation:
False — the perpendicular bisectors of two non-parallel chords.
4
In question 5 the tangent from B to the circle of radius 4 cm is —
💡 Hint:
√(64−16).
Explanation:
√48 = 4√3 cm.
5
Why is the tangent in question 6 equal to 6 cm? Write the formula and the substitution.
Model answer (board format)
BC is a diameter, the centre is the midpoint of BC, and the radius is 4 cm. With B(0,0), C(8,0) and A(0,6), AM = √(16+36) = √52. The length is √(52−16) = √36 = 6 cm. AB itself is one tangent.
Check your answer:
❓ Full question bank — with answers and explanations — 65 questions
Pick a type. The 35 lesson checks are separate — each lesson has as many as its topic needs. All correct earns mastery ★.
BSEB modelpractice · not annual60 tried · 0 correctMultiple choiceMCQ120 tried · 0 correctTrue / falseTrue / false80 tried · 0 correctFill in the blanksFill80 tried · 0 correctMatchMatch20 tried · 0 correctAssertion–reasonAssertion–reason50 tried · 0 correctBalancingBalancing40 tried · 0 correctIdentifyIdentify20 tried · 0 correctShort and longVSA, SA, LA120 tried · 0 correctCBSE-stylenot a PYQ60 tried · 0 correct
No question is marked as a verified past paper. The BSEB set is a model for practice. CBSE items are CBSE-style, not a copy of any year’s paper.
Multiple choice
0/12Pick one option. A wrong try brings a hint.
1
The points for 3:2 are —
💡 Hint:
m+n.
Explanation:
5.
2
In 5:8 the parallel passes through —
💡 Hint:
m = 5.
Explanation:
The fifth point.
3
The larger part of 7.6 cm is —
💡 Hint:
8/13.
Explanation:
60.8/13 cm.
4
The points for factor 3/4 are —
💡 Hint:
The greater number.
Explanation:
4.
5
For factor 5/3, the join to C is —
💡 Hint:
The smaller number.
Explanation:
B3.
6
10 cm away, radius 6 cm. The length is —
💡 Hint:
√64.
Explanation:
8 cm.
7
On concentric circles 6 and 4 the length is —
💡 Hint:
√20.
Explanation:
2√5 cm.
8
At 60 degrees and radius 5 cm the length is —
💡 Hint:
tan 30°.
Explanation:
5√3 cm.
9
From an inside point the tangents are —
💡 Hint:
No line meets the circle at exactly one point.
Explanation:
0.
10
The angle in a semicircle is —
💡 Hint:
Construction 11.3.
Explanation:
90°.
11
The tangent in question 6 is —
💡 Hint:
√(52−16).
Explanation:
6 cm.
12
In question 7 the centre is found by —
💡 Hint:
The centre is not given.
Explanation:
The perpendicular bisectors.
True or false
0/81
Exercise 11.1 asks for a justification.
Explanation:
True.
2
In 3:2 the parallel passes through the fifth point.
Explanation:
False — through the third.
3
The factor 3/4 makes a smaller triangle.
Explanation:
True.
4
When the factor is greater than 1, the parallel meets the side inside it.
Explanation:
False — on the extension.
5
An outside point has exactly one tangent.
Explanation:
False — two.
6
The perpendicular bisectors of two non-parallel chords meet at the centre.
Explanation:
True.
7
The new 6 cm side in question 2 is 4 cm.
Explanation:
True — (2/3)×6.
8
In NCERT 2026-27, Constructions is Chapter 11.
Explanation:
False — that chapter is not in that book.
Fill in the blanks
0/81
The points for m:n are ______.
💡 Hint:
m+n.
Explanation:
The sum.
2
The share of 5 in 7.6 cm is ______ cm.
💡 Hint:
38/13 cm.
Explanation:
(5/13)×7.6.
3
The tangent in question 1 is ______ cm.
💡 Hint:
8 cm.
Explanation:
√64.
4
The tangent in question 2 is ______ cm.
💡 Hint:
2√5 cm.
Explanation:
√20.
5
The third angle in question 6 is ______ degrees.
💡 Hint:
30 degrees.
Explanation:
180−105−45.
6
The new hypotenuse in question 7 is ______ cm.
💡 Hint:
25/3 cm.
Explanation:
(5/3)×5.
7
The tangent in question 3 is ______ cm.
💡 Hint:
2√10 cm.
Explanation:
√(49−9).
8
The new base in question 4 is ______ cm.
💡 Hint:
12 cm.
Explanation:
(3/2)×8.
Match
0/21
Match the factor with the number of points.
Column B: A. 7 · B. 4 · C. 5 · D. 3
1. 3/4
2. 5/3
3. 2/3
4. 7/5
Explanation:
3/4 has 4, 5/3 has 5, 2/3 has 3, and 7/5 has 7.
2
Match the question with its length.
Column B: A. 6 cm · B. 8 cm · C. 2√5 cm · D. 5√3 cm
1. Tangent 6 and 10
2. Tangent 4 and 6
3. Angle 60, radius 5
4. Question 6
Explanation:
6 and 10 give 8 cm, 4 and 6 give 2√5 cm, 60 degrees gives 5√3 cm, and question 6 gives 6 cm.
Assertion–reason
0/5Check both statements, then see whether the reason explains the assertion.
1
Assertion (A): When 7.6 cm is divided in the ratio 5:8, the larger part is 60.8/13 cm.
Reason (R): The larger share is 8/13 of the whole.
Explanation:
Both are true and R explains A.
2
Assertion (A): For the factor 3/4, four points are marked on the ray.
Reason (R): An outside point has two tangents.
Explanation:
Both are true, but R does not explain the points.
3
Assertion (A): With radius 6 cm and distance 10 cm, the tangent is 8 cm.
Reason (R): The length is the sum of OP and r.
Explanation:
A is true. R is false, the length is the square root of the difference of squares.
4
Assertion (A): For the factor 5/3 the parallel meets the side inside it.
Reason (R): For a factor greater than 1, the parallel meets the side extended.
Explanation:
A is false. R is true.
5
Assertion (A): In Construction 11.3, PQ is a tangent.
Reason (R): The angle in a semicircle is 90 degrees, so PQ is perpendicular to the radius.
Explanation:
Both are true and R explains A.
Coefficient practice
0/4These chemistry lines are not a result of this maths chapter. They are only practice in filling coefficients. A blank means 1.
1
This is coefficient practice and not a result of this chapter. In the water line, Fill coefficients (blank = 1).
H2 + O2 → H2O
Explanation: 2H2 + O2 → 2H2O
2
This is coefficient practice and not a result of this chapter. In the line of H2 and O2, Fill coefficients (blank = 1).
H2 + O2 → H2O
Explanation: 2H2 + O2 → 2H2O
3
This is coefficient practice and not a result of this chapter. In this coefficient line, Fill coefficients (blank = 1).
H2 + O2 → H2O
Explanation: 2H2 + O2 → 2H2O
4
This is coefficient practice and not a result of this chapter. In the line with blank coefficients, Fill coefficients (blank = 1).
H2 + O2 → H2O
Explanation: 2H2 + O2 → 2H2O
Classify
0/21
Place each factor as a smaller or a larger triangle.
3/4
2/3
5/3
7/5
Explanation:
3/4 and 2/3 are smaller. 5/3 and 7/5 are larger.
2
Place the number of tangents by where the point lies.
Inside
On the circle
Outside
The centre
Explanation:
Inside and the centre give none, a point on the circle gives one, and an outside point gives two.
Very short answer
0/51
Write the rule for the number of points in m:n.
Model answer (board format)
m+n.
Check your answer:
2
Write the formula for the length of a tangent.
Model answer (board format)
√(OP² − r²).
Check your answer:
3
Through which point does the parallel pass in 3:2?
Model answer (board format)
Through the third point.
Check your answer:
4
Write the third angle of question 6.
Model answer (board format)
30 degrees.
Check your answer:
5
How is the centre of a bangle circle found?
Model answer (board format)
At the intersection of the perpendicular bisectors of two non-parallel chords.
Check your answer:
Short answer
0/41
Divide 7.6 cm in the ratio 5:8. Both parts, with units.
Model answer (board format)
(5/13)×7.6 = 38/13 cm. (8/13)×7.6 = 60.8/13 cm.
Check your answer:
2
A 2/3 triangle on sides 4, 5, 6 cm. The three new sides.
Model answer (board format)
8/3 cm, 10/3 cm, 4 cm.
Check your answer:
3
Find the tangent lengths of questions 1 and 2.
Model answer (board format)
√(100−36) = 8 cm. √(36−16) = 2√5 cm.
Check your answer:
4
Right legs 4 cm and 3 cm. The three sides at factor 5/3.
Model answer (board format)
The hypotenuse is 5 cm. The new sides are 20/3 cm, 5 cm and 25/3 cm.
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Long answer
0/31
Write the first way of Construction 11.1 and its justification, for the ratio 3:2.
Model answer (board format)
Five equal points on the ray. Join A5 to B. The parallel through A3 meets AB at C. By basic proportionality, AC:CB = AA3:A3A5 = 3:2.
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2
Write the steps for the factor 5/3 and say why the ratio 5/3 appears.
Model answer (board format)
Five points. Join B3 to C. The parallel through B5 meets BC extended at C′. Then draw a parallel to CA. BB5:BB3 = 5:3, so BC′:BC = 5:3. The parallel makes the triangles similar.
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3
Write the steps and the justification of Construction 11.3. Also find the length in question 1.
Model answer (board format)
M is the midpoint of OP. The circle of radius MO cuts at Q and R. PQ and PR are tangents, because the angle in a semicircle is 90 degrees. The length is √(100−36) = 8 cm.
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BSEB model paper · practice
0/6This model set is for practice. It is not a question from any year’s annual examination.
1
In question 5 the tangent from A to the circle of radius 3 cm is —
💡 Hint:
√(64−9).
Explanation:
√55 cm.
2
In question 4 the new altitude is —
💡 Hint:
(3/2)×4.
Explanation:
6 cm.
3
The points for the factor 7/5 are ______.
💡 Hint:
7.
Explanation:
The greater number.
4
Find the three new sides of question 3. The original sides are 5, 6 and 7 cm.
Model answer (board format)
(7/5)×5 = 7 cm. (7/5)×6 = 42/5 cm. (7/5)×7 = 49/5 cm.
Check your answer:
5
Write the justification of the second way for 3:2, using similar triangles.
Model answer (board format)
AX is parallel to BY. Triangle AA3C is similar to triangle BB2C. AC:CB = AA3:BB2 = 3:2, because the arcs are equal.
Check your answer:
6
Assertion (A): The factor in question 4 is greater than 1.
Reason (R): For a factor greater than 1, the parallel stops inside the side.
Explanation:
A is true, 3/2 is greater. R is false.
CBSE-style questions
0/6These are competency-based practice questions. They are not a copy of any year’s paper.
1
A student marks 8 points for 5:8. The correct number is —
💡 Hint:
5+8.
Explanation:
13.
2
A student joins B5 to C for the factor 5/3. The join should be —
💡 Hint:
The smaller number.
Explanation:
B3.
3
Assertion (A): Both tangents in question 1 are 8 cm.
Reason (R): Tangents from an outside point are equal, and the length is √(OP² − r²).
Explanation:
Both are true and R explains A.
4
Assertion (A): This chapter is printed in the NCERT 2026-27 book of 14 chapters.
Reason (R): Chapter 11 of that book is Areas Related to Circles.
Explanation:
A is false. R is true.
5
A student takes the parallel for the factor 3/4 onto the extension of the side. What is the mistake?
Model answer (board format)
3/4 is less than 1. Mark four points, join the fourth to C, and draw the parallel from the third inside the side. The extension is the step for a factor greater than 1.
Check your answer:
6
A student draws the tangent on a bangle circle without a centre. What should be done first?
Model answer (board format)
Take two non-parallel chords. Their perpendicular bisectors meet at the centre. Then take the midpoint of OP and follow Construction 11.3.
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🏛️ Board exam corner — Bihar Board (BSEB)— CBSE
Switch board with BSEB | CBSE above. The lessons follow the same NCERT chapter.
BSEB model · practiceBoard tip
This page has no verified annual-exam question, because no source page has been added. The model set below is practice in the board pattern.
🏛️ Model questions on one page →
The verified label will be used only when a source page for the question is available.
CBSE-style · competency-based
These are case and assertion-reason practice items. Do not treat them as past CBSE questions.
🔁 Spaced review — today’s questions
Wrong questions return soon; correct ones return after a few days.
🧠 What you learned + equation sheet
What you learned
- For the ratio m:n, mark m+n equal points on the ray.
- Through the m-th point draw a parallel to the join of the last point.
- The justification comes from the basic proportionality theorem.
- When the scale factor is less than 1, the new triangle is smaller.
- When the scale factor is greater than 1, the parallel meets the side extended.
- An outside point gives two tangents, and an angle in a semicircle is 90 degrees.
| What | Keep this |
|---|---|
| Ratio 3:2 | पाँच बिंदु, तीसरे से समांतर |
| 7.6 cm, 5:8 | 38/13 सेमी और 60.8/13 सेमी |
| Factor 3/4 | चार बिंदु, तीसरे से समांतर |
| Factor 5/3 | पाँच बिंदु, तीसरे को जोड़ो |
| Tangent length | √(OP² − r²) |
| Angle 60°, r = 5 | लंबाई 5√3 सेमी |
The notes are original writing. The textbook was used only for activity order and numbers. “Verified” will be used only on a question that has a source page.