NCERT Solutions Class 7 Mathematics Ganita Prakash 2 Chapter 01 Geometric Twins

Question 1. Check if the two figures are congruent.
Answer: To determine if two figures are congruent, you must measure the angles at their corners using a protractor. When you compare the measurements, you discover that angle ABC does not match angle DEF. This tells you the figures are not congruent.
In simple words: Two shapes are only congruent if all their angles and side lengths are the same. Here, the angles are different, so the shapes do not match.

Exam Tip: Always measure all corresponding angles and sides carefully - even one mismatch means the figures are not congruent.

Question 2. Circle the pairs that appear congruent.
Answer: When you look at the given shapes, pairs (a) and (d) are the ones that match perfectly. You can confirm this because when you place one shape directly over the other, they fit exactly on top of each other without any parts sticking out or gaps left behind.
In simple words: Congruent shapes look the same and fit exactly when stacked on top of each other. Shapes (a) and (d) do this perfectly.

Exam Tip: The key test for congruence is superposition - if one shape covers another shape completely with no overlap or gaps, they are congruent.

Question 3. What measurements would you take to create a figure congruent to a given: (a) Circle (b) Rectangle Using this, state how you would check if two (a) Are circles congruent? (b) Rectangles are congruent?
Answer:
(a) For a circle, the key measurement to take is the radius or diameter of the circle. To check if two circles match, place one circle on top of the other. If they sit exactly on top of each other with no parts sticking out, they are congruent - this happens because they will have the same radius.
(b) For a rectangle, you need to measure both the length and the breadth of the shape. To check if two rectangles match, place one rectangle over the other. If they fit perfectly with edges aligned and no gaps or overlaps, they are congruent - this means both shapes have the same length and breadth.
In simple words: Circles match if they have the same size radius. Rectangles match if they have the same length and width.

Exam Tip: Remember that for circles, only the radius (or diameter) matters - all other circles with that same radius are congruent to it.

Question 4. How would we check if two figures like the one below are congruent? Use this to identify whether each of the following pairs is congruent.
Answer: To check if two figures made of line segments are congruent, you need to measure the lengths of all the line segments in both figures and also find the angle between them. Then compare these measurements - if the measurements match, the figures are congruent. Looking at each pair given here, both figures have a horizontal line that is 3.3 cm long and a vertical line that is 2.3 cm long, with an angle of 82 degrees between them. Since both figures have these exact same measurements, each pair shown is congruent.
In simple words: Two line-segment figures match if they have the same line lengths and the same angles between them.

Exam Tip: Always measure all corresponding segments and angles - congruent figures must have every single measurement the same.

Question 1. Suppose ∆HEN is congruent to ∆BIG. List all the other correct ways of expressing this congruence.
Answer: When ∆HEN ≅ ∆BIG, the vertices H, E, and N match with B, I, and G in that order. A congruence statement can be written in six different ways for two matching triangles. Besides the one given, the five other correct ways are:
(i) ∆HNE ≅ ∆BGI
(ii) ∆EHN ≅ ∆IBG
(iii) ∆ENH ≅ ∆IGB
(iv) ∆NHE ≅ ∆GBI
(v) ∆NEH ≅ ∆GIB
In simple words: You can name a pair of matching triangles in six ways - what matters is that the letters stay in the same matching order.

Exam Tip: When writing congruence statements, the order of vertices must show which corners match - the first letter matches the first letter in the other triangle, and so on.

Question 2. Determine whether the triangles are congruent. If yes, express the congruence.
Answer: Looking at the side lengths of triangle RED, we have RE = 3.5 cm, ED = 5 cm, and RD = 6 cm. For triangle JAM, we have JA = 3.5 cm, AM = 5 cm, and JM = 6 cm. When you compare them, all three sides of the first triangle equal the matching sides of the second triangle: RE equals JA at 3.5 cm, ED equals AM at 5 cm, and RD equals JM at 6 cm. Because all three pairs of matching sides are equal, the triangles are congruent by the side-side-side rule. Therefore, ∆RED ≅ ∆JAM.
In simple words: If all three sides of one triangle match all three sides of another triangle, the triangles are congruent.

Exam Tip: The SSS (side-side-side) rule is the most straightforward way to prove triangles match - measure all three sides of each triangle.

Question 3. In the figure below, AB = AD, CB = CD. Can you identify any pair of congruent triangles? If yes, explain why they are congruent. Does AC divide ∠BAD and ∠BCD into two equal parts? Give reasons.
Answer: First, look at triangles ABC and ADC. We know AB equals AD (given), and CB equals CD (given). Both triangles also share the side AC. Since all three sides of triangle ABC match all three sides of triangle ADC, the triangles are congruent by the side-side-side rule. Therefore, ∆ABC ≅ ∆ADC.

Yes, AC does divide both angles into equal parts. When ∆ABC ≅ ∆ADC, the matching angles must be equal. This means angle BAC equals angle DAC, and angle BCA equals angle DCA. When a line divides an angle into two equal parts, it is called the angle bisector. So AC acts as the bisector for both angle BAD and angle BCD.
In simple words: AC is a line that cuts both angles exactly in half. When two triangles are congruent, their matching angles are always equal.

Exam Tip: When you prove triangles congruent, you can then use CPCT (Corresponding Parts of Congruent Triangles) to find equal angles and sides.

Question 4. In the figure below, are ∆DFE and ∆GED congruent to each other? It is given that DF = DG and FE = GE.
Answer: You are given that DF = DG and FE = GE. Notice that side DE is shared by both triangles ∆DFE and ∆DGE. Since all three corresponding sides are equal - DF matches DG, FE matches EG, and DE is common to both - the triangles are congruent by the side-side-side rule. Therefore, ∆DFE ≅ ∆DGE.

However, the question asks if ∆DFE and ∆GED are congruent. This is a different order of vertices. In the congruence statement ∆DFE and ∆DGE, vertex F in the first triangle matches G in the second, and E matches E. But in ∆GED, the vertices are in a different sequence, and the matching does not work correctly. The corresponding sides would not match the way they need to. So ∆DFE and ∆GED are not congruent because the vertex order does not create proper matching of corresponding sides.
In simple words: The order of the letters in a congruence statement matters - it shows which corner matches which corner. A different letter order means a different matching.

Exam Tip: Always pay attention to the order of vertices in congruence statements - ∆ABC ≅ ∆DEF means A matches D, B matches E, and C matches F.

Question 1. Identify whether the triangles below are congruent. What conditions did you use to establish their congruence? Express the congruence.
Answer: Looking at the triangles, side CB equals ZY (both are 5 cm), side BA equals ZX (both are 7 cm), and the angle between these two sides is identical in both triangles at 47 degrees. When two sides and the angle formed between them match in one triangle with two sides and the included angle in another triangle, the triangles are congruent using the side-angle-side method. The triangles are therefore congruent and can be expressed as ∆ABC ≅ ∆XZY.
In simple words: If two sides and the angle between them in one triangle match two sides and the angle between them in another, the triangles must be congruent.

Exam Tip: The SAS (side-angle-side) rule requires the angle to be between the two equal sides - an angle at either end of both sides will work.

Question 2. Given that CD and AB are parallel, and AB = CD, what are the other equal parts in this figure? (Hint: When the lines are parallel, the alternate angles are equal. Are the two resulting triangles congruent? If so, express the congruence.)
Answer: Since DC is parallel to AB, we can use the property of alternate angles. When a transversal line crosses two parallel lines, the alternate angles are equal. This gives us angle OCD equals angle OAB, and angle OBA equals angle ODC. We already know AB equals CD. Looking at triangles DOC and BOA, we find that the angle at O is shared by both triangles (it is the same angle). With angle OCD equal to angle OAB, angle OBA equal to angle ODC, and these being alternate angles formed by the parallel lines, we can use the angle-side-angle rule. From the congruence of these triangles, the other equal parts are: OA equals OC, and OB equals OD. The triangles ∆DOC ≅ ∆BOA.
In simple words: Parallel lines create equal angles. When you have equal angles and parallel lines, you can find many other equal parts using congruence.

Exam Tip: Remember the properties: alternate angles are equal when lines are parallel, and vertically opposite angles are always equal.

Question 3. Given that ∠ABC = ∠DBC and ∠ACB = ∠DCB, show that ∠BAC = ∠BDC. Are the two triangles congruent?
Answer: Let angle ABC equal angle DBC, and call this measure a. Similarly, let angle ACB equal angle DCB, and call this measure b. Side BC is shared by both triangles ABC and DBC - it is a common side. Now we have two angles in each triangle that are equal to their corresponding angles in the other triangle, plus a shared side between those angles. Using the angle-side-angle rule, ∆ABC ≅ ∆DBC. When two triangles are congruent, their matching parts are equal - this is known as CPCT (Corresponding Parts of Congruent Triangles). Since the triangles are congruent, angle BAC must equal angle BDC.
In simple words: When two angles and the side between them are the same in two triangles, the triangles are congruent. Then all other matching parts are equal too.

Exam Tip: The ASA (angle-side-angle) rule is powerful - you only need two angles and the side between them to prove congruence.

Question 4. Identify the equal parts in the following figure, given that ∠ABD = ∠DCA and ∠ACB = ∠DBC.
Answer: You are given that angle ABD equals angle DCA, and angle ACB equals angle DBC. The angles at point O are vertically opposite - angles AOB and DOC are vertically opposite angles, and vertically opposite angles are always equal. Looking at triangles COD and BOA, we have two pairs of equal angles and the angle at O is equal in both. By the angle-side-angle rule, the triangles are congruent: ∆COD ≅ ∆BOA. From this congruence, using CPCT, we can identify the other equal parts: AO equals DO, and CO equals BO.
In simple words: Vertically opposite angles are always equal. When you find equal angles, you can often prove triangles congruent and find many other equal parts.

Exam Tip: Vertically opposite angles are a reliable tool - they are always equal, even before you prove anything else about the figure.

Question 1. ∆AIR ≅ ∆FLY. Identify the corresponding vertices, sides, and angles.
Answer: When ∆AIR ≅ ∆FLY, the vertices match in the order they are written. The matching parts are:

Corresponding Vertices:
A matches with F
I matches with L
R matches with Y

Corresponding Sides:
AI matches with FL
IR matches with LY
AR matches with FY

Corresponding Angles:
Angle A matches with angle F
Angle I matches with angle L
Angle R matches with angle Y
In simple words: In a congruence statement, letters in the same position match with each other - first letter with first, second with second, third with third.

Exam Tip: Always write matching parts in the same order as the triangle names to avoid mistakes.

Question 2. Each of the following cases contains certain measurements taken from two triangles. Identify the pairs in which the triangles are congruent to each other, with reason. Express the congruence whenever they are congruent. (a) AB = DE BC = EF CA = DF (b) AB = EF ∠A = ∠E AC = ED (c) AB = DF ∠B = ∠D = 90° AC = FE (d) ∠A = ∠D ∠B = ∠E AC = DF (e) AB = DF ∠B = ∠F AC = DE
Answer:
(a) Here, AB = DE, BC = EF, and CA = FD. All three matching sides are equal between the two triangles. By the side-side-side congruence rule, ∆ABC ≅ ∆DEF.

(b) Given AB = EF, angle A = angle E, and AC = ED. Two sides and the angle between them in triangle ABC match two sides and the angle between them in triangle EFD. By the side-angle-side rule, ∆ABC ≅ ∆EFD.

(c) Here, AB = FD, angle B = angle D = 90 degrees, and AC = FE. Both triangles have a right angle. In a right triangle, the side opposite the right angle is called the hypotenuse. Here, AC and FE are the hypotenuses and they are equal. Also, one other side is equal. When two right triangles have equal hypotenuses and one equal side, they are congruent by the right-angle-hypotenuse-side rule. Therefore, ∆ABC ≅ ∆FDE.

(d) Here, angle A = angle D, angle B = angle E, and AC = DF. Two angles and one side match between the triangles. By the angle-angle-side rule, ∆ABC ≅ ∆DEF.

(e) Here, AB = DF, angle B = angle F, and AC = DE. Two sides and one angle are equal, but the angle is not between the two sides - it is not included. This is called the side-side-angle arrangement, which is not a valid congruence rule. Therefore, ∆ABC does not have to be congruent to ∆DFE.
In simple words: Use SSS, SAS, RHS, and AAS to prove triangles match. SSA does not work - the angle must be between the two sides for SAS to work.

Exam Tip: Remember the valid rules: SSS, SAS, ASA, AAS, and RHS. SSA is not valid unless it is a right triangle (then it becomes RHS).

Question 3. It is given that OB = OC, and OA = OD. Show that AB is parallel to CD. [Hint: AD is a transversal for these two lines. Are there any equal alternate angles?]
Answer: Given OB = OC and OA = OD, look at triangles AOB and COD. The angle at O is shared by both triangles - angle AOB equals angle COD because they are the same angle. With OA = OD, angle AOB = angle COD, and OB = OC, we have a match of side-angle-side between the two triangles. By the side-angle-side rule, ∆AOB ≅ ∆COD. These triangles can be placed exactly on top of each other. From this congruence, matching angles must be equal, so angle A equals angle D, and angle B equals angle C. When line AD crosses lines AB and CD, angles A and D are on opposite sides of the transversal. Since these alternate angles are equal, line AB must be parallel to line CD.
In simple words: If alternate angles are equal when a line crosses two other lines, those two lines are parallel to each other.

Exam Tip: Parallel lines create equal alternate angles, and equal alternate angles prove lines are parallel - this works both ways.

Question 4. ABCD is a square. Show that ∆ABC ≅ ∆ADC. Is ∆ABC also congruent to ∆CDA? Give more examples of two triangles where one triangle is congruent to the other in two different ways, as in the case above. Can you give an example of two triangles where one is congruent to the other in six different ways?
Answer: Since ABCD is a square, all sides are equal: AB = AD = BC = CD. Also, both diagonal segments from A equal each other because in a square, the diagonals are equal. We have CD = CB (all sides of a square are equal), AC = AC (the diagonal is common to both triangles), and AB = AD. By the side-side-side rule, ∆ABC ≅ ∆ADC.

For ∆ABC to be congruent to ∆CDA: In the order CDA, side AB matches CD (equal sides of the square), side BC matches DA (equal sides of the square), and AC matches CA (same diagonal). By the side-side-side rule again, ∆ABC ≅ ∆CDA. So yes, the triangle is congruent in two different ways depending on how you name it.

Now consider two congruent triangles ∆HEN and ∆BIG. There are six ways to write this same congruence, each one showing different vertex matches:
(i) ∆HNE ≅ ∆BGI
(ii) ∆EHN ≅ ∆IBG
(iii) ∆ENH ≅ ∆IGB
(iv) ∆NHE ≅ ∆GBI
(v) ∆NEH ≅ ∆GIB

When two triangles are congruent (all sides equal and all angles equal), you can write their congruence in six different ways by rearranging the order of the vertex letters, as long as the matching is done correctly.
In simple words: The same pair of matching triangles can be written in different ways, depending on which corner you list first - there are six ways total for any pair of congruent triangles.

Exam Tip: All six congruence statements for a single pair of matching triangles are correct if the vertex matching is right - pick whichever naming order is clearest for your answer.

Question 5. Find ∠B and ∠C, if A is the centre of the circle.
Answer: In triangle BAC, both AB and AC are radii of the circle, so they are equal: AB = AC. When a triangle has two equal sides, the angles opposite those sides are also equal. The angle opposite AB is angle C, and the angle opposite AC is angle B. Therefore, angle ABC equals angle ACB. Let's call each of these angles x. The sum of all angles in any triangle is 180 degrees. So x + x + 120 degrees = 180 degrees. Solving: 2x = 180 - 120 = 60 degrees, which means x = 30 degrees. Therefore, angle B = 30 degrees and angle C = 30 degrees.
In simple words: When two sides of a triangle are equal, the angles across from them are also equal. You can use this plus the 180-degree rule to find missing angles.

Exam Tip: In isosceles triangles (two equal sides), the base angles are always equal - this saves time when finding unknown angles.

Question 6. Find the missing angles. As per the convention that we have been following, all line segments marked with a single '|' are equal to each other, and those marked with a double '|' are equal to each other, etc.
Answer: In ∆CUR, we see that CU and CR are marked with the same single tick mark, meaning CU = CR. When two sides are equal, the angles opposite them are equal. The angle opposite CU is angle CRU, and the angle opposite CR is angle CUR. So angle CUR = angle CRU. Let's call each x. The sum of angles in a triangle is 180 degrees: x + x + 90 = 180. Solving: 2x = 90, so x = 45 degrees. Therefore, angle CUR = 45 degrees and angle CRU = 45 degrees.

In ∆VRN, the marks show VR = VN. The angles opposite these equal sides are angle VNR and angle VRN, which must be equal. Call each of these a. The angle at V is 68 degrees. So a + a + 68 = 180. This gives 2a = 112, so a = 56 degrees. Therefore, angle VRN = 56 degrees and angle VNR = 56 degrees.

In ∆AUP, we find that UA = UP (from the equal marks). This makes angle UAP = angle UPA. We determined angle UPA = 56 degrees from the work above. Using 56 + 56 + angle AUP = 180, we get angle AUP = 68 degrees.

Triangle BOF is marked with equal sides on all three sides: OB = OF = BF. A triangle with all three sides equal is called equilateral, and all three angles in an equilateral triangle are 60 degrees. So angle FOB = angle FBO = angle OFB = 60 degrees.

Since angles RVN and DVN form a straight line, they sum to 180 degrees: 68 + angle DVN = 180, so angle DVN = 112 degrees. In ∆DVN, we have VN = VD (from the marks). The angles opposite the equal sides are angle VND and angle VDN, both equal to c. So c + c + 112 = 180. This gives 2c = 68, so c = 34 degrees. Therefore, angle VND = 34 degrees and angle VDN = 34 degrees.

In ∆OLB, angle OBL = 90 - 60 = 30 degrees (since angle OBF is 60 degrees from the equilateral triangle). Also, angle LOB = 60 degrees because LO is parallel to BF and BO is a transversal, creating equal corresponding angles.

In ∆OPN, the angles sum to 180: angle OPN + 56 + 90 = 180, so angle OPN = 34 degrees. On the straight line at P, angles APK, KPO, and OPN sum to 180: 44 + angle KPO + 34 = 180, giving angle KPO = 102 degrees.

In ∆KPO, the angles sum to 180: 102 + 30 + angle PKO = 180, so angle PKO = 48 degrees. In ∆KAP, the angles sum to 180: 34 + 44 + angle AKP = 180, so angle AKP = 102 degrees. On the straight line at K, angles AKP, PKO, and OKL sum to 180: 102 + 48 + angle OKL = 180, so angle OKL = 30 degrees.

In ∆KOL, the angles sum to 180: 30 + 90 + angle KOL = 180, so angle KOL = 60 degrees. We can also show that ∆OKL ≅ ∆OBL by the side-angle-side rule: KL = LB, angle OLK = angle OLB = 90 degrees (both right angles), and OL is common.
In simple words: Use the rule about equal sides giving equal angles. Mark similar sides with the same number of ticks. Find angles step by step using the 180-degree rule and marks showing which sides match.

Exam Tip: Always identify which sides are marked as equal - this immediately tells you which angles are equal. For equilateral triangles (all sides equal), all angles are 60 degrees.