CLASS 7 · GANITA PRAKASH PART II · CHAPTER 1
Learn when figures and triangles are congruent and use matching parts to find angles and prove results.
34 worked answers · short steps first, explanations when you need them
Textbook pages 1–4
Question 1
The two lengths alone allow the angle at B to change, so they are not enough.
Draw AB=4 cm. At B, construct an 80° angle.
On the new ray, mark C so BC=8 cm.
Two side lengths alone do not fix the opening between the arms.
Question 2
Imagine placing one figure over the other.
A rotation or reflection may be used before matching.
They have the same shape and size and can be superimposed exactly; turning or flipping is allowed.
Question 3
A circle is determined by its radius (or equivalently its diameter).
Measure the radius; draw a circle with the same radius.
Question 4
Measure two adjacent side lengths.
All angles in a rectangle are right angles.
Its length and width are sufficient.
Question 5
Compare their radii for circles.
For rectangles, compare the two side lengths, allowing a rotation that exchanges length and width.
Circles are congruent exactly when their radii match. Rectangles are congruent when their pairs of side lengths match.
Question 6
Match corresponding side lengths and angles, then check that the full outlines can overlap after a turn or flip.
Use corresponding side and angle measurements and verify exact superposition; a visual guess alone cannot prove congruence.
Textbook pages 5–17
Question 1
Fix one side as a base.
The third vertex must lie at the intersection of circles centered at the base endpoints, with radii equal to the other two sides.
The two possible intersections are mirror images across the base.
The triangles are congruent by SSS; the two constructions are reflections of one another.
Question 2
Read the correspondence in order: H↔B, E↔I, N↔G.
Keep these pairings when writing another order.
ΔHEN≅ΔBIG; ΔHNE≅ΔBGI; ΔEHN≅ΔIBG; ΔENH≅ΔIGB; ΔNHE≅ΔGBI; ΔNEH≅ΔGIB.
Question 3
AB matches CD; AD matches CB; BD is shared.
Thus A↔C, B↔D, and D↔B.
ΔABD≅ΔCDB by SSS. The order ΔABD≅ΔCBD would mismatch the side lengths in general.
Question 4
Compare ΔABC with ΔADC.
AB=AD and CB=CD are given; AC is common.
Apply SSS, then use corresponding angles.
ΔABC≅ΔADC by SSS. Therefore AC bisects ∠BAD and ∠BCD.
Question 5
DF=DG and FE=GE are given.
DE is common to both triangles.
Yes. ΔDFE≅ΔDGE by SSS.
Question 6
The equal angle lies between the two equal corresponding sides.
Yes, by SAS.
Question 7
Draw the known base and angle ray.
A circle for the other side can intersect that ray twice.
No. SSA may produce two different triangles, so SSA is not a general congruence condition.
Question 8
Draw the known side.
At its endpoints, construct the two known angles.
Their rays meet at one determined third vertex.
The triangles are congruent by ASA.
Question 9
The third angle is fixed because the angle sum is 180°.
The data then give two angles and a side.
Yes, by AAS (equivalently, find the third angle and use ASA).
Question 10
Both have a right angle.
The hypotenuse and one leg are equal.
Yes, by RHS.
Question 11
AO=OD and BO=OC because O is each segment’s midpoint.
∠AOB=∠DOC as vertically opposite angles.
Use SAS on ΔAOB and ΔDOC.
ΔAOB≅ΔDOC by SAS; hence AB=CD as corresponding sides.
Question 12
Angles opposite equal sides are equal, so ∠B=∠C.
The angles sum to 180°: ∠B+∠C=100°.
∠B=∠C=50°.
Question 13
All three sides are equal, so the three opposite angles are equal.
Their total is 180°.
180°÷3=60°.
Each angle is 60°.
Textbook pages 3, 8–21
Question 1
Compare the lengths of the matching arms and the angle at their meeting point.
The pictured arms do not have matching measurements.
No. The arm lengths and opening angle do not both match, so one outline cannot be placed exactly over the other.
Question 2
Compare size and outline, allowing a rotation or flip.
The two droplet shapes match after turning; the two-petal shapes match after turning.
The cloud outlines and starbursts differ in size or shape.
Circle the droplet pair and the two-petal pair.
Question 3
Match letters in the same position.
Vertices: A↔F, I↔L, R↔Y. Sides: AI↔FL, IR↔LY, AR↔FY. Angles: ∠A↔∠F, ∠I↔∠L, ∠R↔∠Y.
Question 4
(a) AB=DE, BC=EF, CA=DF gives SSS.
(b) AB=EF, ∠A=∠E, AC=ED gives SAS.
(c) AB=DF, ∠B=∠D=90°, AC=FE gives RHS.
(d) ∠A=∠D, ∠B=∠E, AC=DF gives AAS.
(e) AB=DF, ∠B=∠F, AC=DE gives SSA.
(a) Yes, ΔABC≅ΔDEF by SSS. (b) Yes, ΔABC≅ΔEFD by SAS. (c) Yes, ΔABC≅ΔFDE by RHS. (d) Yes, ΔABC≅ΔDEF by AAS. (e) The measurements do not guarantee congruence: SSA can form two different triangles.
Question 5
With AC as a transversal, ∠BAO=∠DCO by alternate interior angles.
With BD as a transversal, ∠ABO=∠CDO.
AB=CD is the included side between these angle pairs.
The parallel sides create two pairs of equal alternate interior angles.
Question 6
Both triangles have two equal angle pairs.
Subtract the equal pairs from 180° to see the remaining angles match.
Yes, ∠BAC=∠BDC. The triangles need not be congruent: the information gives AAA, which fixes shape but not size.
Question 7
All square sides are equal and AC is shared.
Apply SSS to ΔABC and ΔADC.
For the second naming, match A↔C, B↔D, C↔A.
ΔABC≅ΔADC by SSS. Also ΔABC≅ΔCDA; the vertex order C-D-A corresponds to A-B-C.
Question 8
A non-equilateral isosceles triangle has a reflection symmetry that exchanges its equal-side vertices.
An equilateral triangle has every permutation of its three vertices as a symmetry correspondence.
An isosceles non-equilateral triangle gives two correspondences; two equilateral triangles give six.
Question 9
AB=AC because both are radii, so the base angles are equal.
The remaining 60° is shared equally.
∠B=∠C=30°.
Question 10
At V, RV=VO and ∠RVO=68°, so the base angles of triangle RVO are (180°−68°)÷2=56°.
The diagram gives ∠PRO=34°. Therefore ∠VRP=56°+34°=90°, so RP is perpendicular to the top edge.
Set U=(0,0). Since CU=CR=UP and the corner at C is a right angle, we can set C=(0,s), R=(s,s), P=(s,y), with CU=CR=UP=s. The perpendicular mark makes RP vertical.
UP=s gives s²+y²=s², so y=0. Thus UP is horizontal and UA is vertical, making ∠AUP=90°.
Since ∠PUQ=34°, ∠AUQ=90°−34°=56°.
But AU=UQ, so triangle AUQ is isosceles. Also ∠UAK=90° and ∠QAK=34°, hence ∠UAQ=56° and ∠AQU=56°. Its third angle must be 180°−56°−56°=68°.
No. As printed, the marked diagram is inconsistent: it forces ∠AUQ to be both 56° and 68°. At least one printed angle or equal-side mark must be corrected before a consistent list of missing angles can be given.
Question 11
OA=OD and OB=OC by the midpoint information.
∠AOB=∠DOC because they are vertically opposite.
Use SAS to prove ΔAOB≅ΔDOC.
Then ∠ABO=∠DCO as corresponding angles; these are alternate interior angles on transversal BC.
ΔAOB≅ΔDOC by SAS. Therefore ∠ABO=∠DCO, so AB∥CD by the converse of the alternate-interior-angle rule.
Question 12
Match the vertex where the 3.5 cm and 5 cm sides meet: E↔A.
Then R↔J and D↔M.
All three matching sides have equal lengths.
Yes. ΔRED≅ΔJAM by SSS.
Question 13
The equal angle is between the two equal sides in each triangle.
Match B↔Z, A↔X and C↔Y.
Yes. ΔABC≅ΔXZY by SAS.
Question 14
At B, ∠ABC=∠ABD+∠DBC.
At C, ∠DCB=∠DCA+∠ACB.
The given equal pairs make these two sums equal.
Use each triangle’s angle sum to compare the remaining angles.
Then ∠BAC=∠BDC. The triangles have three equal angle pairs (AAA), which proves the same shape but does not by itself prove congruence.
Question 15
Look for repeated triangular panels with the same side lengths and angles.
For example, repeated matching glass or metal panels in a dome or bridge can be congruent.
In a rangoli, repeated triangles may match after a turn or flip; a pyramid picture may show matching triangular faces.
The exact answer depends on the visible design: compare the sides and angles rather than assuming every triangle in the picture is equal.
The pictures show repeated triangular shapes in architecture and art. A valid example is any pair of panels whose three side lengths match (SSS), even if one is turned or flipped. Check each pictured pair before calling it congruent.