Samacheer Class 7 Maths - Geometry Intext Questions

\(\overleftrightarrow { GH } \) is parallel to \(\overleftrightarrow { AB } \)
(ii) \(\overleftrightarrow { IJ } \) and \(\overleftrightarrow { KL } \) intersect \(\overleftrightarrow { CD } \)
(iii) \(\overleftrightarrow { IJ } \) and \(\overleftrightarrow { KL } \) are perpendicular to \(\overleftrightarrow { GH } \)
(iv) Only one line \(\overleftrightarrow { KL } \) is parallel to \(\overleftrightarrow { IJ } \)
(v) Yes, \(\overleftrightarrow { EF } \) will intersect \(\overleftrightarrow { AB } \) at some point.
Try These (Text Book Page No. 85)
Choose the correct answer

(i) Angles between leaf veins. [ ∠1 and ∠2],
(ii) Angles between adjacent pages of a book, when it is open [ ∠1 and ∠2 ].
(iii) Adjacent angles of scissors [ ∠1 and ∠2 ]

Four pairs of adjacent angles are
1. ∠A and ∠B
2. ∠B and ∠C
3. ∠C and ∠D
4. ∠D and ∠E

Four pairs of non adjacent angles are.
1. ∠A and ∠C
2. ∠C and ∠F
3. ∠E and ∠D
4. ∠A and ∠F

Given ∠PQR =150°
∠QPS = 30°
They are supplementary angles,
But they are not adjacent angles as they don’t have common vertex or common arm.
∴ They are not a linear pair.
Try this (Text book Page No. 90)

New adjacent angles are formed.
The new angles become smaller in measure. But their sum is 180° as it is a linear angle.
Try this (Text book Page No. 90)

We know that the sum of angles at a point is 360°
∠AOB + ∠BOC + ∠COD + ∠DOE + ∠EOF + ∠FOA = 360° as they are the sum of angles at the point ‘O’
Try These (Text book Page No. 91)

(i) a, b, c are transversal to l, m and n.
(ii) a, b, c are transversal to l, m, n and p. More transversals can be drawn.

If the radius of the arc is less than half of AB, then both the arcs will not cut at a point
and we can’t draw perpendicular bisector.
About Us
Privacy Policy
Disclaimer
Contact Us
Facebook
Twitter
Pinterest
LinkedIn
Copyright © 2026 Samacheer Kalvi Guru

(i) ∠ABG and ∠GBC are adjacent angles.
(ii) ∠BCF and ∠FCD are adjacent angles.
(iii) ∠BCF and ∠FCE are adjacent angles.
(iv) ∠FCE and ∠ECD are adjacent angles.

If the angle are linear pair, then their sum is 180°.
Given one angle is right angle ie 90°.
∴ The other angle = 180° -90° = 90°
∴ The other angle also a right angle

We know that the sum of angles at a point is 360°.
Given the three angles are in the ratio 1:4:7.
Let the three angles be 1x, 4x, 7x.

∴ The three angles are 30°, 120° and 210°.

We know that the sum of angles at a point is 360°.
One angle = 45°
Let the equal angles be x° each
∴ x° + x° + x° + x° + x° + 45° = 360°
5x° + 45° – 45° = 360° – 45°

5x° = 315°

∴ Measure of all 5 equal angles = 63°.

Lines \(\overleftrightarrow { RP }\) and \(\overleftrightarrow { SQ }\) are interesting at ‘O’

∴ Vertically opposite angles are equal.
∴ x = 105°
∴ Missing angle = 105°

(i) m and n are parallel lines. l is the transversal.
Interior angles on the same side of the transversal are supplementary. But here it is 75 + 75 ≠ 180°
105 + 105 ≠ 180°
∴ Angles marked are not correct
(ii) m and n are parallel lines. l is the transversal.
Corresponding angles must be equal. So here the marking is wrong.

When two parallel lines are intersected by a transversal, we need a minimum of a single angle to find the remaining angle.
Using the concept of linear pair of angles, we can final one more angle.
By the concepts of corresponding angles, alternate interior angles and alternate , exterior angles we could find all other angles.
Objective Type Questions

1. A. 8 cm
2. B. 7 cm
3. C. 5.6 cm
4. D. 10.4 cm

(a) 8 cm
Construction :

Step 1: Drawn a line. Marked two points A and B on it so that AB = 8 cm
Step 2: Using compass with A as centre and radius more than half of the length of AB, drawn two arcs of the same length one above AB and one below AB
Step 3: With the same radius and B as centre drawn two arcs to cut the arcs drawn in step 2. Marked the points of intersection of the arcs as C and D.
step 4: Joined C and D, CD intersect AB. Marked the point of intersection as ‘O’.
CD is the required perpendicular bisector of AB.

(b) 7 cm
Construction :

step 1: Drawn a line and marked points A and B on it so that AB = 7 cm.
step 2: Using compass with A as centre and radius more than half of the length of AB drawn two arcs of same length one above AB and one below AB.
step 3: With the same radius and B as centre drawn two arcs to cut the already drawn arcs in step 2. Marked the intersection of the arcs as C and D
step 4: Joined C and D, CD is the required perpendicular bisector of AB.


(c) 5.6 cm.
Construction :

Step 1: Drawn a line and marked two points A and B on it so that AB = 5.6cm
Step 2: Using compass with A as centre and radius more than half of the length of AB, drawn two arcs of the same length, one above AB and one below AB
Step 3: With the same radius and B as centre drawn two arcs to cut the arcs drawn in step 2 and marked the points of intersection of the arcs as C and D
Step 4: Joined C and D. CD intersects AB. Marked the point of intersection as ‘O’CD is the required perpendicular bisector of AB.
Now ∠AOC = 90° AO = BO = 2.8 cm

(d) 10.4 cm
Construction :

Step 1: Drawn a line and marked two points A and B on it so that AB = 10.4 cm.
Step 2: Using compass with A as centre and radius more than half of the length of AB, drawn two arcs of same length one above AB and one below AB.
Step 3: With the same radius and B as centre drawn two arcs to cut the arcs drawn in step 2 and marked the points of intersection of the arcs as C and D.
Step 4: Joined C and D. CD intersects AB. Marked the points of intersection as O. CD is the required perpendicular bisector.
Now ∠AOC = 90° ; AO = BO = 5.2 cm

(e) 58 mm

Construction :

Step 1: Drawn a line. Marked two points A and B on it so that
AB = 5.8 cm = 58 mm.
Step 2: Using compass with A as centre and radius more than half of the length of AB, drawn two arcs of the same length one above AB and one below AB.
Step 3: With the same radius and B as centre drawn two arcs to cut the arcs of drawn in step 2. Marked the points of intersection of the arcs as C and D.
Step 4: Joined C and D. CD intersects AB. Marked the point of intersection as O. CD is the required perpendicular bisector. ∠AOC = 90°
AO = BO = 2.9 cm
About Us
Privacy Policy
Disclaimer
Contact Us
Facebook
Twitter
Pinterest
LinkedIn
Copyright © 2026 Samacheer Kalvi Guru

1. A. 60°
2. B. 100°
3. C. 90°
4. D. 48°

(a) 60°
Construction:

Step 1: Drawn the given angle ∠ABC with the measure 60° using protractor.
Step 2: With B as centre and convenient radius, drawn an arc to cut BA and BC. Marked the points of intersection as E on BA and F on BC.
Step 3: With the same radius and E as centre drawn an arc in the interior of ∠ABC and another arc of same measure with centre at F to cut the previous arc.
Step 4: Marked the point of intersection as G. Drawn a ray BX through G. BG is the required bisector of the given ∠ABC

(b) 100°

Construction :
Step 1: Drawn the given angle ∠ABC with the measure 100° c using protractor.
Step 2: With B as centre and convenient radius, drawn an arc to cut BA and BC. Marked the points of intersection as E on BA and F on BC.
Step 3: With the same radius and E as centre drawn an arc in the interior of ∠ABC and another arc of the same measure with centre at F to cut the previous arc.
Step 4: Marked the point of intersection at G. Drawn a ray BX through G.
BG is the required bisector of angle ∠ABC
∠ABG = ∠GBC = 50°

(c) 90°
Construction :

Step 1: Drawn the given angle ∠ABC with the measure 90° using protractor.
Step 2: With B as center and convenient radius, drawn an arc to cut BA and BC. Marked the points of intersection as E on BA and F on BC.
Step 3: With the same radius and E as center drawn an arc in the interior of ∠ABC and another arc of same measure with center at F to cut the previous arc.
Step 4: Mark the point of interaction as G. Drawn a ray BX through G. BG is the required bisector of the given angle ∠ABC
∠ABG = ∠GBC = 45°
(d) 48°

Construction :
Step 1: Drawn the given angle ∠ABC with the measure 48° using protractor.
Step 2: With B as center and convenient radius, drawn an arc to cut BA and to cut BA and BC. Marked the points of intersection as E on BA and F on BC.
Step 3: With the same radius and E as center drawn an arc in the interior of ∠ABC and another arc of the same measure with center at F to cut the previous arc.
Step 4: Marked the point of intersection as G. Drawn a ray BX through G. BG is the required bisector of the given angle ∠ABC
Now ∠ABC = ∠GBC = 24°

(e) 110°
Construction:

Step 1: Drawn the given angle ∠ABC with the measure 110° using protractor.
Step 2: With B as center and convenient radius, drawn an arc to cut BA and BC. Marked points of intersection as E on BA and F BC.
Step 3: With the same radius and E as center, drawn an arc in the interior of ∠ABC and another arc of same measure with center at F to cut the previous arc.
Step 4: Mark the point of intersection as G. Drawn a ray BX through G. BG is the
required bisector of the given angle ∠ABC
∠ABG = ∠GBC = 55°
About Us
Privacy Policy
Disclaimer
Contact Us
Facebook
Twitter
Pinterest
LinkedIn
Copyright © 2026 Samacheer Kalvi Guru

(i) 60°
Construction :
Step 1: Drawn a line and marked a point ‘A’ on it.
Step 2: With A as center drawn an arc of convenient radius to meet the line at a point B.

Step 3: With the same radius and B as center drawn an arc to cut the previous arc at C.
Step 4: Joined AC. The ∠ABC is the required angle with the measure 60°.
(ii) 120°
Construction :

We know that there are two 60° angles in 120°.
∴ We can construct two 60° angles consecutively construct 120°
Step 1: Drawn a line and marked a point ‘A’ on it.
Step 2: With ‘A’ as center, drawn an arc of convenient radius to the line at a point B.
Step 3: With the same radius and B as center, drawn an arc to cut the previous arc at C.
Step 4: With the same radius and C as center, drawn an arc to cut the arc drawn in step 2 at D.
Step 5: Joined AD. Then ∠BAD is the required angle with measure 120°.

(iii) 30°
Constructions :

Since 30° is half of 60°, we can construct 30° by bisecting the angle 60°.
Step 1: Drawn a line and marked a point A on it.
Step 2: With A as center drawn an arc of convenient radius to the line to meet at a point B.
Step 3: With the same radius and B as center drawn an arc to cut the previous arc at C.
Step 4: Joined AC to get ∠BAC = 60°
Step 5: With B as center drawn an arc of convenient radius in the interior of ∠BAC
Step 6: With the same radius and C as center drawn an arc to cut the previous arc at D.
Step 7: Joined AD.
∴ ∠BAD is the required angle of measure 30°.

(iv) 90°
Construction :

Step 1: Drawn a line and marked a point ‘A’ on it.
Step 2: With ‘A’ as center, drawn an arc of convenient radius to the line at a point B.
Step 3: With the same radius and B as center drawn an arc to cut the previous arc at ‘C’.
Step 4: With the same radius and C as center, drawn an arc to cut the arc drawn in step 2 at D.
Step 5: Joined AD. ∠BAD = 120°.
Step 6: With C as center, drawn an arc of convenient radius in the interior of ∠CAD.
Step 7: With the same radius and D as center, drawn an arc to cut the arc at E.
Step 8: Joined AF ∠BAE = 90°.
(v) 45°
Construction :
Step 1: Drawn a line and marked a point A on it

Step 2: With A as center, drawn an arc of convenient radius to the line at a point B.
Step 3: With the same radius and B as center drawn an arc to cut the previous arc at C.
Step 4: With the same radius and C as center, drawn an arc to cut the arc drawn in step 2 at D.
Step 5: Joined AD. ∠BAD = 120°.
Step 6: With G as center and any convenient radius drawn an arc in the interior of ∠GAB
Step 7: With the same radius and B as center drawn an arc to cut the arc at F.
Step 8: Joined AF. ∠BAF = 45°

(vi) 150°
Construction :

Since 150° = 60° + 60° + 30°; we construct as follows
Step 1: Drawn a line and marked a point A on it.
Step 2: With ‘A’ as center, drawn a full arc of convenient radius to the line at a point B and at E the other end.
Step 3: With the same radius and B as center, drawn an arc to cut the previous arc at C.
Step 4: With the same radius and C as center drawn an arc to cut the already drawn arc at D.
Step 5: With D as center, drawn an arc of convenient radius in the interior of ∠DAE
Step 6: With E as center and with the same radius drawn an arc to cut the previous arc at F.
Step 7: Joined AF, ∠FAB = 150°.
(vii) 135°
Construction :

Step 1: Drawn a line and marked a point A on it.
Step 2: With ‘A’ as center, drawn an arc of convenient radius to the line at a point B.
Step 3: With the same radius and B as center drawn an arc to cut the previous arc at C.
Step 4: With the same radius and C as center, drawn an arc to cut the arc at D.
Step 5: With C and D as centers drawn arcs of convenient (same) radius in the interior of ∠CAD. Marked the point of intersection as E.
Step 6: Joined AE, through G. ∠BAE = 90°.
Step 7: Drawn angle bisector to ∠GAH through F.
Now ∠BAF = 135°.
About Us
Privacy Policy
Disclaimer
Contact Us
Facebook
Twitter
Pinterest
LinkedIn
Copyright © 2026 Samacheer Kalvi Guru

Since ∠BOC and ∠AOC are linear pair, their sum = 180°
2x + 23 + 3x – 48 = 180°
5x – 25 = 180°
5x – 25 + 25 = 180° + 25

∠DOB and ∠BOC are linear pair
∴ ∠DOB + ∠BOC = 180°
x + 3x + 40 = 180°
4x + 40 = 180°
4x + 40 – 40 = 180° – 40°
4x = 140°

Also ∠BOD and ∠AOC are vertically opposite angles.
∴ ∠BOD = ∠AOC
x = z + 10
35° = z + 10
z + 10 – 10 = 35 – 10
z = 25°
Again ∠AOD and ∠AOC are linear pair.
∴ ∠AOD + ∠AOC = 180°
y + 30 + z + 10 = 180°
y + 30 + 25 + 10 = 180°
y + 65 = 180°
y + 65 – 65 = 180° – 65
y = 115°
∴ x = 35°,
y = 115°,
z = 25°

Given two angles are in the ratio 11 : 25.
Let the angles be 11x and 25x.
They are also linear pair
∴ 11x + 25x = 180°.

∴ The angles 11x = 11 × 5° = 55° and 25x = 25 × 5 = 125°.
∴ The angles are 55° and 125°.

(i) Yes, ∠1 is adjacent to ∠2.
Because they both have the common vertex ‘O’ and the common arm OA . Also their interiors do not overlap.
(ii) No, ∠AOB and ∠BOB are not adjacent angles because they have overlapping interiors.
(iii) No, ∠BOC and ∠BOD does not form a linear pair.
Because ∠BOC itself a straight angle, so the sum of ∠BOC and ∠BOD exceed 180°.
(iv) Yes, the angles ∠COD and ∠BOD are supplementary ∠COD + ∠BOD = 180°, [∵ linear pair of angles]
∴ ∠COD and ∠BOD are supplementary.
(v) No. ∠3 and ∠1 are not formed by intersecting lines. So they are not vertically opposite angles.

Given TOU is a straight line.
∴ The sum of all angles formed at a point on a straight line is 180°
∠TOR + ∠ROP + ∠POV + ∠VOU = 180°.
36° + 47°+ 45° + x° = 180°.
128° + x° = 180°
128° + x° – 128° = 180° – 128°
x = 52°

Given AB || DC
AB and CD are parallel lines Taking CE as transversal we have.
∠1 = 30°, [∵ alternate interior angles]
Taking DE as transversal
∠2 = 80°.[∵ alternate interior angles]
∠1 = 30° and ∠2 = 80°
Justification:
CDE is a triangle

l and m are parallel lines and n is the transversal.
∠G and ∠H are alternate interior angles.
∠G = ∠H …… (1)
Given ∠G and ∠G are Suplementary

Given ∠1 = 53°

Clearly ∠1 and ∠2 are interior angles on the same side of the transversal and so they are supplementary.
∠1 + ∠2 = 180°
53° + ∠2 = 180°
53° + ∠2 – 53° = 180° – 53°
∠2 = 127°
Challenge Problems

Let the two parallel lines be m and n and l be the transversal
Let one of the interior angles on the same side of the transversal be x°
Then the other will be 2x + 48.
We know that they are supplementary.

Given GH || IZ
∠1 = 108°
∠2 = 123°
∠1 + ∠KGH = 180 [linear pair]
108° + ∠KGH = 180°

108° + ∠KGH – 108° = 180° – 108°
∠KGH = 72°
∠KGH = x° (corresponding angles if KG is a transversal)
∴ x° = 72°
Similarly
∠2 + ∠GHK = 180° (∵ linear pair)
123° + ∠GHK = 180°
123° + ∠GHK – 123° = 180° – 123°
∠GHK = 57°
Again ∠GHK = y° (corresponding angles if KH is a transversal)
y = 57°
x° +y° + z° = 180° (sum of three angles of a triangle is 180°)
72° + 57° + z° = 180°
129° + z° = 180°
129° + z° – 129° = 180° – 129°
z = 51°
x = 72°,
y = 57°,
z = 51°

From the picture
∠2 + 65° = 180° [Sum of interior angles on the same side of a transversal]
x + 39° + 65° = 180°
x + 104° = 180°
x + 104° – 104° = 180° – 104°
x = 76°
Also from the picture
∠1 = 65° [alternate exterior angles]
2x – 3y = 65°
2 (76) – 3y = 65°
152° – 3y = 65°
152° – 3y – 152° = 65 – 152°
-3y = -87

Given AB||CD
Then AD and BC are transversals.
x = 48°, alternate interior angles; AD is transversal y = 60°, alternate interior angles; BC is transversal
∠AEB + 48° + y° = 180°, (sum of angles of a triangle is 180°)

Given ∠8 = 107°
∠2 = 107°
[∵ ∠8 and ∠2 are alternate exterior angles, ∵ ∠8 = ∠2]


About Us
Privacy Policy
Disclaimer
Contact Us
Facebook
Twitter
Pinterest
LinkedIn
Copyright © 2026 Samacheer Kalvi Guru