NCERT Solutions Of Ch Triangles Class 10

Similarity Criteria

All congruent figures are similar, but it does not mean that all similar figures are congruent.

Two polygons of the same number of sides are similar, if:

• Their corresponding angles are equal.
• Their corresponding sides are in the same ratio.

Two triangles are similar, if:

• Their corresponding angles are equal.
• Their corresponding sides are in the same ratio.

According to Greek mathematician Thales, “The ratio of any two corresponding sides in two equiangular triangles is always the same.”

• If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.
• If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.
• If in two triangles, corresponding angles are equal, then their corresponding sides are in the same ratio (or proportion) and hence the two triangles are similar.
• If two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar.
• If in two triangles, sides of one triangle are proportional to the sides of the other triangle, then their corresponding angles are equal and hence the two triangles are similar.
• If one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, then the two triangles are similar.
• The ratio of areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
• If a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse then triangles on both sides of the perpendicular are similar to the whole triangle and to each other.
• In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

According to the Indian mathematician Budhayan, “The diagonal of a rectangle produces by itself the same area as produced by its both sides (i.e., length and breadth).”
In a triangle, if square of one side is equal to the sum of the squares of the other two sides, then the angle opposite the first side is a right angle.
If in a triangle, square of one side is equal to the sum of the squares of the other two sides, then the angle opposite the first side is a right angle.

THEOREM 1:

If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.

Construction: ABC is a triangle. DE || BC and DE intersects AB at D and AC at E.
Join B to E and C to D. Draw DN ⊥ AB and EM ⊥ AC.
To prove:
Proof:
ar (AEM)
Similarly;

Hence;

Similarly;

Triangles BDE and DEC are on the same base, i.e. DE and between same parallels, i.e. DE and BC.
Hence, ar(BDE) = ar(DEC)
From above equations, it is clear that;

THEOREM 2:

If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

Construction: ABC is a triangle in which line DE divides AB and AC in the same ratio. This means:
To Prove: DE || BC
Let us assume that DE is not parallel to BC. Let us draw another line DE’ which is parallel to BC.
Proof:
If DE’ || BC, then we have;

According to the theorem;

Then according to the first theorem; E and E’ must be coincident.
This proves: DE || BC

THEOREM 3:

If in two triangles, corresponding angles are equal, then their corresponding sides are in the same ratio (or proportion) and hence the two triangles are similar. This is also called AAA (Angle-Angle-Angle) criterion.

Construction: Two triangles ABC and DEF are drawn so that their corresponding angles are equal. This means:
∠ A =∠ D, ∠ B = ∠ E and ∠ C = ∠ F
To prove:

Draw a line PQ in the second triangle so that DP = AB and PQ = AC
Proof:

Because corresponding sides of these two triangles are equal
This means; ∠ B = ∠ P = ∠ E and PQ || EF
This means;

Hence;

Hence;

THEOREM 4:

If in two triangles, sides of one triangle are proportional to the sides of the other triangle, then their corresponding angles are equal and hence the two triangles are similar. This is also called SSS (Side-Side-Side) criterion.

Construction: Two triangles ABC and DEF are drawn so that their corresponding sides are proportional. This means:

To Prove: ∠ A = ∠ D, ∠ B = ∠ E and ∠ C = ∠ F
And hence; Δ ABC ~ Δ DEF
In triangle DEF, draw a line PQ so that DP = AB and DQ = AC
Proof:

Because corresponding sides of these two triangles are equal
This means;

This also means; ∠ P = ∠ E and ∠ Q = ∠ F
We have taken; ∠ A = ∠ D, ∠ B = ∠ P and ∠ C = ∠ Q
Hence; ∠ A = ∠ D, ∠ B = ∠ E and ∠ C = ∠ F
From AAA criterion;
Δ ABC ~ Δ DEF proved

THEOREM 5:

If one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, then the two triangles are similar. This is also called SAS (Side-Angle-Side) criterion.
Construction: Two triangles ABC and DEF are drawn so that one of the angles of one triangle is equal to one of the angles of another triangle. Moreover, two sides included in that angle of one triangle are proportional to two sides included in that angle of another triangle. This means;
∠ A = ∠ D and

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To Prove: Δ ABC ~ Δ DEF
Draw PQ in triangle DEF so that, AB = DP and AC = DF
Proof:

Because corresponding sides of these two triangles are equal
given
∠ A = ∠ D
Hence; from SSS criterion
Hence;

Hence; Δ ABC ~ Δ DEF proved

THEOREM 6:

The ratio of areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

Construction: Two triangles ABC and PQR are drawn so that, ΔABC ~ Δ PQR.
To Prove:

Draw AD ⊥ BC and PM ⊥ PR
Proof:

Hence;

Now, in Δ ABD and Δ PQM;
∠ A = ∠ P, ∠ B = ∠ Q and ∠ D = ∠ M (because Δ ABC ~ Δ PQR)
Hence; Δ ABD ~ Δ PQM
Hence;

Since, Δ ABC ~ Δ PQR
So,

Hence;

Similarly, following can be proven:

THEOREM 7:

If a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse then triangles on both sides of the perpendicular are similar to the whole triangle and to each other.

Construction: Triangle ABC is drawn which is right-angled at B. From vertex B, perpendicular BD is drawn on hypotenuse AC.
To Prove: Δ ABC ~ Δ ADB ~ Δ BDC
Proof: In Δ ABC and Δ ADB;
∠ ABC = ∠ ADB
∠ BAC = ∠ DAB
∠ ACB = ∠ DBA
From AAA criterion; Δ ABC ~ Δ ADB
In Δ ABC and Δ BDC;
∠ ABC = ∠ BDC
∠ BAC = ∠ DBC
∠ ACB = ∠ DBC
From AAA criterion; Δ ABC ~ Δ BDC
Hence; Δ ABC ~ Δ ADB ~ Δ BDC proved.

THEOREM 8:

Pythagoras Theorem: In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Construction: Triangle ABC is drawn which is right angled at B. From vertex B, perpendicular BD is drawn on hypotenuse AC.
To Prove: AC² = AB² + BC²
Proof: In Δ ABC and Δ ADB;

Because these are similar triangles (as per previous theorem)
In Δ ABC and Δ BDC;

Adding equations (1) and (2), we get;
AC x AD + AC x CD = AB² + BC²
Or, AC(AD + CD) = AB² + BC²
Or, AC x AC = AB² + BC²
Or, AC² = AB² + BC²
Proved

Exercise 6.1 (NCERT Book)

Question 1: Fill in the blanks using the correct word given in brackets:
(a) All circles are ………… (congruent, similar).
Answer: similar
Explanation: All circles are similar irrespective of difference in their radii. A smaller circle can be enlarged to the size of a larger circle and vice-versa is also true.
(b) All squares are ……………(similar, congruent).
Answer: similar
Explanation: All squares are similar because all the angles in a square are right angles and all the sides are equal. Hence, a smaller square can be enlarged to the size of a larger square and vice-versa is also true.

c) All …………..triangles are similar. (isosceles, equilateral).Answer: equilateral
Explanation: All equilateral triangles are similar because all the angles in an equilateral triangle are 60°. Moreover, all the sides of an equilateral triangle are equal. Hence, a smaller equilateral triangle can be enlarged to the size of a larger equilateral triangle and vice-versa is also true.
(d) Two polygons of the same number of sides are similar, if their corresponding angles are …………and their corresponding sides are ……………(equal, proportional).
Answer: equal, proportional
Explanation: This is similar to AAA criterion in case of triangles. If corresponding angles are same and the number of sides are similar in two polygons then the polygons are similar.
Question 2: Give two different examples of pair of
(a) Similar figures
Answer: Two equilateral triangles of different sides.
Explanation: Two equilateral triangles of different sides would be similar because all the angles are 60°. Hence, AAA criterion applies in this case.
(b) Non-similar figures
Answer: A rhombus and a rectangle
Explanation: In case of a rhombus, all the sides are equal and the angles can either be right angles or a combination of acute and obtuse angles. In case of rectangle; opposite sides are equal and all the angles are right angles. Hence, a rhombus and a rectangle are non-similar figures. There can be some exceptions; like in case of two squares of same sides. A square is a rhombus and a rectangle too.


Question 3: State whether the following quadrilaterals are similar or not.

Answer: These quadrilaterals are not similar.
Explanation: In the given quadrilaterals, the sides are in the same ration, i.e. all the sides are same in respective quadrilaterals. But angles in the smaller quadrilateral are not right angles but all the angles in the larger quadrilateral are right angle. Hence, these quadrilaterals are not similar.

Exercise 6.2 (NCERT Book) Part 1

Question 1: In the given figures, DE || BC. Find EC in first figure and AD in second figure.

Solution: In the first figure;
Δ ADE ~ Δ ABC (Because DE || BC)
Hence;

Similarly, in the second figure;
Δ ADE ~ Δ ABC (Because DE || BC)
Hence;

Question 2: E and F are points on the sides PQ and PR respectively of a Δ PQR. For each of the following cases, state whether EF || QR.

(a) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm
Solution: For EF || QR, the figure should fulfill following criterion;

In this case;

It is clear that;

Hence; EF and QR are not parallel.
(b) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm
Solution: In this case;

It is clear that;

Hence; EF || QR
(c) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm
Solution: In this case;

It is clear that;

Hence; EF || QR






Question 3: In the given figure, if LM || CB and LN || CD, prove that

Solution: In Δ ABC and Δ AML;
Δ ABC ~ Δ AML (because ML || BC)
Hence;

Similarly, in Δ ADC and Δ ANL;
Δ ADC ~ Δ ANL (because NL || DC)
Hence;

From above two equations;

Question 4: In the given figure, DE || AC and DF || AE. Prove that

Solution: In Δ ABC and ΔDBE;

Because Δ ABC ~ Δ DBE
Similarly, in Δ ABE and Δ DBF;

From above two equations, it is clear;

Exercise 6.2 (NCERT Book) Part 2


Question 5: In the given figure, DE || OQ and DF || OR. Show that EF || QR.

Solution: In Δ PQO and Δ PED;

(Because these are similar triangles, as per Basic Proportionality theorem.)
Similarly, in Δ PRO and Δ PFD;

From above two equations, it is clear;

Hence;

Hence, EF || QR proved.






Question 6: In the given figure, A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR. Show that BC || QR.

Solution: In Δ OPQ and Δ OAB;

(Because these are similar triangles as per BPT.)
Similarly, in Δ OPR and Δ OAC;

From above two equations, it is clear;

Hence; BC || QR proved
Question 7: Using theorem 6.1, prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side.

Solution: PQR is a triangle in which DE || QR. Line DE intersects PQ at D so that PD = DQ
To Prove: PE = ER
In Δ PQR and Δ PDE;

(Because as per BPT, these are similar triangles.)

Hence; E is the midpoint of PR proved.

Question 8: Using theorem 6.2, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side.
Solution: The figure from the previous question can be used to solve this question.
ABC is a triangle in which D and E are mid points of PQ and PR respectively.
To Prove: DE || QR
In Δ PQR and Δ PDE;

Hence, as per BPT these triangles are similar.
Hence; DE || QR proved.
Question 9: ABCD is a trapezium in which AB || CD and its diagonals intersect each other at the point O. show that;

Solution: Draw a line EF || CD which is passing through O.
In Δ ABC and Δ EOC;
These are similar triangles as per BPT.

Similarly, in Δ BOD and Δ FOD;

In Δ ABC and Δ BAD;

(Because diagonals of a trapezium divide each other in same ratio)
From above three equations, it is clear;

Hence, Δ ABC ~ Δ BAD
Using the third equation;

Question 10: The diagonals of a quadrilateral ABCD intersect each other at point O such that Show that ABCD is a trapezium.




Solution: This question can be proved by using the figure in previous question. Since diagonals are dividing each other in the same ratio hence, ABCD is a trapezium; proved.

Exercise 6.3 (NCERT Book) Part 1

Question 1: State which pairs of triangles in the given figure are similar? Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form:

Solution: (i) Δ ABC ~ Δ PQR (AAA criterion)

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Solution: (ii) Δ ABC ~ Δ QRP (SSS criterion)

Solution: (iii) Not similar

Solution: (iv) Δ LMN ~ Δ PQR (SAS criterion)

Solution: (v) Not similar

Solution: (vi) Δ DEF ~ Δ PQR (AAA criterion)
Question 2: In the given figure, Δ ODC ~ Δ OBA, ∠BOC = 125° and ∠CDO = 70°. Find ∠DOC, ∠DCO and ∠OAB.

Solution: ∠DOC + ∠COB = 180° (Linear pair of angles)
Or, ∠DOC + 120° = 180°
Hence, ∠DOC = 180° - 120° = 60°
In Δ DOC;
∠DCO + ∠CDO + ∠DOC = 180° (Angle sum of triangle)
Or, ∠DCO + 70° + 60° = 180°
Or, ∠DCO = 180° - 130° = 50°
∠OCD = ∠OAB = 50° (Because Δ ODC ~ Δ OBA; given)
Question 3: Diagonals AC and BD of a trapezium ABCD with AB || CD intersect each other at the point O. Using a similarity criterion for two triangles, show that

Solution: Draw a line EF || CD which is passing through O.
In Δ ABC and Δ EOC;

These are similar triangles as per BPT.

Similarly, in Δ BOD and Δ FOD;

In Δ ABC and Δ BAD;

Because diagonals of a trapezium divide each other in same ratio
From above three equations, it is clear;

Hence, Δ ABC ~ Δ BAD
Using the third equation;

Or, Proved
Question 4: In the given figure, and ∠1 = ∠2. Show that Δ PQS ~ Δ TQR.

Solution: Proof:
∠1 = ∠2 (given)
Hence, PQ = PR (Sides opposite to equal angles are equal in isosceles triangle)

Hence, PS || TR
And; Δ PQS ~ Δ TQR proved
Question 5: S and T are points on sides PR and QR of Δ PQR such that ∠P = ∠RTS. Show that Δ RPQ ~ Δ RTS.

Solution: In Δ RPQ and Δ RTS;
∠RPQ = ∠RTS (given)
∠PRQ = ∠TRS (common)
Hence, Δ RPQ ~ Δ RTS (AAA Criterion)



Exercise 6.3 (NCERT Book) Part 2


Question 6: In the given figure, if Δ ABE ≈ Δ ACD, show that Δ ADE ~ Δ ABC.

Solution: Since Δ ABE ≈ Δ ACD
Hence, BE = CD
And ∠DBE = ∠ECD



In Δ DBE and Δ ECD
BE = CD (proved earlier)
∠DBE = ∠ECD (proved earlier)
DE = DE (common)
Hence, Δ DBE ≈ Δ ECD
This means; DB = EC
This also means;

Hence; DE || BC
Thus, Δ ADE ~ Δ ABC proved.
Question 7: In the given figure, altitudes AD and CE of Δ ABC intersect each other at point P. Show that:

(a) Δ AEP ~ Δ CDP
Solution: In Δ AEP and Δ CDP
∠AEP = ∠CDP (Right angle)
∠APE = ∠CPD (Opposite angles)
Hence; Δ AEP ~ Δ CDP proved (AAA criterion)


(b) Δ ABD ~ Δ CBE
Solution: In Δ ABD and Δ CBE
∠ADB = ∠CEB (Right angle)
∠DBA = ∠EBC (Common angle)
Hence; Δ ABD ~ Δ CBE proved (AAA criterion)


(c) Δ AEP ~ Δ ADB
Solution: In Δ AEP and Δ ADB
∠AEP = ∠ADB (Right angle)
∠EAP = ∠DAB (Common angle)
Hence; Δ AEP ~ Δ ADB proved (AAA criterion)


(d) Δ PDC ~ Δ BEC
Solution: In Δ PDC and Δ BEC
∠PDC = ∠BEC (Right angle)
∠PCD = ∠BCE (Common angle)
Hence; Δ PDC ~ Δ BEC proved (AAA criterion)





Question 8: E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. show that Δ ABE ~ Δ CFB.

Solution: In Δ ABE and Δ CFB
∠ABE = ∠CFB (Alternate angles)
∠AEB = ∠CBF (Alternate angles)
Hence; Δ ABE ~ Δ CFB proved (AAA criterion)
Question 9: In the given figure, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that:

(a) Δ ABC ~ Δ AMP
Solution: In Δ ABC and Δ AMP
∠ABC = ∠AMP (Right angle)
∠CAB = ∠PAM (Common angle)
Hence; Δ ABC ~ Δ AMP proved (AAA criterion)
(a)
Solution: Since Δ ABC ~ Δ AMP;
Hence;

Proved

Exercise 6.4 (NCERT Book)

Question 1: Let Δ ABC ~ Δ DEF and their areas be, respectively, 64 sq cm and 121 sq m. If EF = 15.4 cm, find BC.
Solution: In Δ ABC ~ Δ DEF;

   

Question 2: Diagonals of a trapezium ABCD with AB || CD intersect each other at point O. if AB = 2 CD, find the ratio of the areas of triangles AOB and COD.

Solution: In triangles AOB and COD;
∠ AOB = ∠ COD (Opposite angles)
∠ OAC = ∠ OCD (Alternate angles)
Hence; Δ AOB ~Δ COD
Hence;

Question 3: In the given figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that

Solution: Let us draw altitudes AM and DN on BC; respectively from A and D

In ΔAMO and ΔDNO;
∠ AMO = ∠ DNO (Right angle)
∠ AOM = ∠ DON (Opposite angles)
Hence; ΔAMO ~ ΔDNO
Hence;

Question 4: If the areas of two similar triangles are equal, prove that they are congruent.
Solution: Let us take two triangles ABC and PQR with equal areas.
Then, we have;

In this case;

Hence; the triangles are congruent.

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Question 5: D, E and F are respectively the mid-points of sides AB, BC and CA of Δ ABC. Find the ratio of the areas of Δ DEF and Δ ABC.

Solution: Since D, E and F are mid points of AB, BC and AC
Hence; ΔBAC ~ΔDFE
So,

So,

Question 6: Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.
Solution: In case of two similar triangles ABC and PQR;

Let us assume AD and PM are the medians of these two triangles.
Then;

Question 7: Prove that the area of an equilateral triangle described on one side of a square is equal to half the area of the equilateral triangle described on one of its diagonals.
Solution: Let us take a square with side ‘a’
Then the diagonal of square will be a√2
Area of equilateral triangle with side ‘a’

Area of equilateral triangle with side a√2

Ratio of two areas can be given as follows:

Proved
Question 8: ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Ratio of the areas of triangles ABC and BDE is

• 2 : 1
• 1 : 2
• 4 : 1
• 1: 4

Solution: (c) 4 : 1
For explanation; refer to question 5
Question 9: Sides of two similar triangles are in the ratio 4 : 9. Areas of these triangles are in the ratio

• 2 : 3
• 4 : 9
• 81 : 16
• 16: 81

Solution: (d) 16 : 91
Note: Areas are in duplicate ratio of corresponding sides in case of similar figures.

Exercise 6.5 (NCERT Book) Part 1

Question 1: Sides of triangles are given below. Determine which of them are right triangles. In case of a right triangle, write the length of its hypotenuse.
(a) 7 cm, 24 cm, 25 cm
Solution: In a right triangle, the longest side is the hypotenuse. We also know that according to Pythagoras theorem:
Hypotenuse² = Base² + Perpendicular²
Let us check if the given three sides fulfill the criterion of Pythagoras theorem.
25² = 24² + 7²
Or, 625 = 576 + 49
Or, 625 = 625
Here; LHS = RHS
Hence; this is a right triangle.

(b) 3 cm, 8 cm, 6 cm
Solution: Let us check if the given sides fulfill the criterion of Pythagoras theorem.
8² = 6² + 3²
Or, 64 = 36 + 9
Or, 64 ≠ 45
Here; LHS is not equal to RHS
Hence; this is not a right triangle.
(c) 50 cm, 80 cm, 100 cm
Solution: Let us check if the given sides fulfill the criterion of Pythagoras theorem.
100² = 80² + 50²
Or, 10000 = 6400 + 2500
Or, 10000 ≠ 8900
Here; LHS is not equal to RHS
Hence; this is not a right triangle.
(d) 13 cm, 12 cm, 5 cm
Solution: Let us check if the given sides fulfill the criterion of Pythagoras theorem.
13² = 12² + 5²
Or, 169 = 144 + 25
Or, 169 = 169
Here; LHS = RHS
Hence; this is a right triangle.
Question 2: PQR is a triangle right angled at P and M is a point on QR such that PM ⊥ QR. Show that PM² = QM. MR.

Solution: In triangles PMQ and RMP
∠ PMQ = ∠ RMP (Right angle)
∠ PQM = ∠ RPM (90 – MRP)
Hence; PMQ ~ RMP (AAA criterion)

Question 3: In the given figure, ABD is a triangle right angled at A and AC ⊥ BD. Show that

(a) AB² = BC. BD
Solution: In triangles ACB and DAB
∠ ACB = ∠ DAB (Right angle)
∠ CBA = ∠ ABD (common angle)
Hence; ACB ~ DAB

(b) AC² = BC. DC
Solution: In triangles ACB and DCA
∠ ACB = ∠ DCA (right angle)
∠ CBA = ∠ CAD
Hence; ACB ~ DCA

(c) AD² = BD. CD
Solution: In triangles DAB and DCA
∠ DAB = ∠ DCA (right angle)
∠ ABD = ∠ CAD
Hence; DAB ~ DCA

Question 4: ABC is an isosceles triangle right angled at C. Prove that AB² = 2AC².
Solution: In this case; AB is hypotenuse and AC = BC are the other two sides
According to Pythagoras theorem:
AB² = AC² + BC²
Or, AB² = AC² + AC²
Or, AB² = 2AC² proved
Question 5: ABC is an isosceles triangle with AC = BC. If AB² = 2AC², prove that ABC is a right triangle.
Solution: This question will be sold in the same way as the earlier question.
In this case; square of the longest side = sum of squares of other two sides
Hence, this is a right triangle

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Question 6: ABC is an equilateral triangle of side 2a. Find each of its altitudes.
Solution: In case of an equilateral triangle, an altitude will divide the triangle into two congruent right triangles. In the right triangle thus formed, we have;
Hypotenuse = One of the sides of the equilateral triangle = 2a
Perpendicular = altitude of the equilateral triangle = p
Base = half of the side of the equilateral triangle = a
Using Pythagoras theorem, the perpendicular can be calculated as follows:
p² = h² – b²
Or, p² = (2a)² – a²
Or, p² = 4a² – a² = 3a²
Or, p = a√3
Question 7: Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.

Solution: ABCD is a rhombus in which diagonals AC and BD intersect at point O.
To Prove: AB² + BC² + CD² + AD² = AC² + BD²
In ∆ AOB; AB² = AO² + BO²
In ∆ BOC; BC² = CO² + BO²
In ∆ COD; CD² = CO² + DO²
In ∆ AOD; AD² = DO² + AO²
Adding the above four equations, we get;
AB² + BC² + CD² + AD²
= AO² + BO² + CO² + BO² + CO² + DO² + DO² + AO²
Or, AB² + BC² + CD² + AD² = 2(AO² + BO² + CO² + DO²)
Or, AB² + BC² + CD² + AD² = 2(2AO² + 2BO²)
(Because AO = CO and BO = DO)
Or, AB² + BC² + CD² + AD² = 4(AO² + BO²) ………(1)
Now, let us take the sum of squares of diagonals;
AC² + BD² = (AO + CO)² + (BO + DO)²
= (2AO)² + (2BO)²
= 4AO² + 4BO² ……(2)
From equations (1) and (2), it is clear;
AB² + BC² + CD² + AD² = AC² + BD² proved
Question 8: In the given figure, O is a point in the interior of a triangle ABC, OD ⊥ BC, OE ⊥ AC and OF ⊥ AB. Show that

(a) OA² + OB² + OC² – OD² – OE² – OF² = AF² + BD² + CE²
Solution: In ∆ AFO; AF² = OA² – OF²
In ∆ BDO; BD² = OB² – OD²
In ∆ CEO; CE² = OC² – OE²
Adding the above three equations, we get;
AF² + BD² + CE² = OA² + OB² + OC² – OD² – OE² – OF² proved
(b) AF² + BD² + CE² = AE² + CD² + BF²
Solution: In ∆ AEO; AE² = OA² – OE²
In ∆ CDO; CD² = OC² – OD²
In ∆ BFO: BF² = OB² – OF²
Adding the above three equations, we get;
AE² + CD² + BF² = OA² + OB² + OC² – OD² – OE² – OF²
From the previous solution, we also have;
AF² + BD² + CE² = OA² + OB² + OC² – OD² – OE² – OF²
Comparing the RHS of the above two equations, we get;
AF² + BD² + CE² = AE² + CD² + BF² proved

Exercise 6.5 (NCERT Book) Part 2

Question 9: A ladder 10 m long reaches a window 8 m above the ground. Find the distance of the foot of the ladder from the base of the wall.
Solution: In this case, the ladder makes the hypotenuse, the height of top of the ladder is perpendicular and the distance of foot of the ladder from the base of the wall is the base.
So, h = 10 m, p = 8 m and b = ?
From Pythagoras theorem;
b² = h² – p²
Or, b² = 10² – 8²
= 100 – 64 = 36
Or, b = 6 m






Question 10: A guy wire attached to a vertical pole of height 18 m is 24 m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut?
Solution: Here; h = 24 m, p = 18 m and b = ?
From Pythagoras theorem;
b² = h² – p²
Or, b² = 24² – 18²
= 576 – 324 = 252
Or, b = √252 = 6√7 m
Question 11: An aeroplane leaves and airport and flies due north at a speed of 1000 km per hour. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1200 km per hour. How far apart will be the two planes after 1.5 hours?
Solution: Distance covered by the first plane in 1.5 hours = 1500 km
Distance covered by the second plane in 1.5 hours = 1800 km
The position of the two planes after 1.5 hour journey can be shown by a right triangle and we need to find the hypotenuse to know the aerial distance between them.
Here; h = ? p = 1800 km and b = 1500 km
From Pythagoras theorem;
h² = p² + b²
Or, h² = 1800² + 1500²
= 3240000 + 2250000 = 5490000
Or, h = 300√61 km
Question 12: Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between the feet of the poles is 12 m, find the distance between their tops.

Solution: In this figure; AB = 6 m, CD = 11 m and BC = AE = 12 m
In right triangle DEA;
DE = CD – AB = 11 – 6 = 5 m, AE = 12 m and AD = ?
Distance between the tops of the two poles can be calculated by using Pythagoras theorem;
AD² = DE² + AE²
= 5² + 12²
= 25 + 144 = 169
Or, AD = 13 m
Question 13: D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that AE² + BD² = AB² + DE².

Solution: In ∆ ACE; AE² = AC² + CE² ……… (1)
In ∆ DCB; BD² = DC² + CB² ……… (2)
In ∆ ACB; AB² = AC² + CB² ……… (3)
In ∆ DCE; DE² = DC² + CE² ……… (4)
Adding equations (1) and (2), we get;
AE² + BD² = AC² + CE² + DC² + CB² …….. (5)
Adding equations (3) and (4), we get;
AB² + DE² = AC² + CB² + DC² + CE² ………. (6)
On comparing the RHS of equations (5) and(6), we get;
AE² + BD² = AB² + DE² proved




Question 14: The perpendiculars from A on side BC of a Δ ABC intersects BC at D such that DB = 3CD. Prove that 2AB² = 2AC² + BC².

Solution: AB² = AD² + DB²
Or, AB² = AD² + 9CD² (because DB = 3CD) …….. (1)
In ∆ ADC; AD² = AC² – CD²
Substituting the value of AD in equation (1), we get;
AB² = AC² – CD² + 9CD² = AC² + 8CD²
Or, 2AB² = 2AC² + 16CD² ……… (2)
Since BC = CD + 3CD = 4CD
Hence, BC² = 16CD²
Substituting the value from above equation in equation (2), we get;
2AB² = 2AC² + BC² proved
Question 15: In an equilateral triangle, ABC, D is a point on side BC such that BD = 1/3 BC. Prove that 9AD² = 7AB².

Solution: Let us assume that each side of triangle ABC is ‘a’
Then, BD = a/3, MC = a/2 and

According to Pythagoras theorem, in triangle ADM;

Question 16: In an equilateral triangle, prove that three times the square of one side is equal to four times the square of one of its altitudes.
Solution: Let us assume that each side of the triangle is ‘a’, then its altitude AM is as follows:

Four times of square of altitude can be calculates as follows:

Hence; three times of square of a side = four times of square of altitude proved
Question 17: Tick the correct answer and justify: In Δ ABC, AB = 6√ 3 cm, AC = 12 cm and BC = 6 cm. The angle B is:

• 120°
• 60°
• 90°
• 45°

Solution: Here; the longest side is 12 cm. Let us check if the sides of the given triangle fulfill the criterion of Pythagoras triplet.

Here; LHS = RHS and hence the given triangle is a right triangle.
Hence; angle B = 90°, i.e. option (c) is the correct answer.

Exercise 6.6 (NCERT Book) Part 1

Question 1: In the given figure, PS is the bisector of ∠ QPR or Δ PQR. Prove that

Solution: Draw a line RT || SP; which meets QP extended up to QT.






∠ QPS = ∠ SPQ (given)
∠ SPQ = ∠ PRT (Alternate angles)
∠ QPS = ∠ PTR (Corresponding angles)
From these three equations, we have;
∠ PRT = ∠ PTR
Hence, in triangle PRT;
PT = PR (Sides opposite to equal angles) -----------(1)
Now; in triangles SQP and RQT;
∠ QPS = ∠ QTR (Corresponding angles)
∠ QSP = ∠ QRT (Corresponding angles)
Hence; Δ SQP ~ ΔRQT (AAA criterion)

Because PT = PR (from equation 1)
Question 2: In the given figure, D is a point on hypotenuse AC of Δ ABC, such that BD ⊥ AC, DM⊥ BC and DN ⊥ AN. Prove that:

(a) DM² = DN.MC
Solution: DN || BC and DM || AB
DNMB is a rectangle because all the four angles are right angles.
Hence; DN = MB and DM = NB
In triangles DMB and CMD;
∠ DMB = ∠ CMD (Right angle)
∠ DBM = ∠ CDM
DM = DM
Hence; Δ DMB ~ Δ CMD

(Because DN = MB)
(b) DN² = DM.AN
Solution: In triangles DNB and AND;
∠ DNB = ∠ AND (Right angles)
∠ NDB = ∠ NAD
DN = DN
Hence; Δ DNB ~ Δ AND

(Because DM = NB)
Question 3: In the given figure, ABC is a triangle in which ∠ ABC > 90° and AD ⊥ CB produced. Prove that AC² = AB² + BC² + 2 BC.BD

Solution: In triangle ADB;
AB² = AD² + BD² …….. (1)
In triangle ADC;
AC² = AD² + DC²
Or, AC² = AD² + (BD + BC)²
= AD² + BD² + BC² + 2BD.BC …….. (2)
Substituting the value of AB² from equation (1) into equation (2), we get;
AC² = AB² + BC² + 2BC.BD proved






Question 4: In the given figure, ABC is triangle in which ∠ ABC < 90° and AD ⊥ BC. Prove that AC² = AB² + BC² - 2BC.BD.

Solution: In triangle ABD;
AB² = AD² + BD² …….. (1)
In triangle ADC;
AC² = AD² + DC²
Or, AC² = AD² + (BC – BD)²
= AD² + BD² + BC² – 2BC.BD ……… (2)
Substituting the value of AB² from equation (1) in equation (2), we get;
AC² = AB² + BC² – 2BC.BD proved
Question 5: In the given figure, AD is a median of a triangle ABC and AM ⊥ BC. Prove that:

(a)
Solution: In triangle AMD;

Substituting the value of AD² from equation (1) in equation (2), we get;

Proved
(b)
Solution: In triangle ABM;

Substituting the value of AD2 from equation (1) in equation (2), we get;

(c)
Solution: From question (a), we have;

And from question (b), we have;

Adding these two equations, we get;

Proved

Exercise 6.6 (NCERT Book) Part 2

Question 6: Prove that the sum of the squares of the diagonals of parallelogram is equal to the sum of the squares of its sides.

Solution: ABCD is a parallelogram in which AB = CD and AD = BC
Perpendicular AN is drawn on DC and perpendicular DM is drawn on AB extended up to M.
In ∆ AMD; AD² = DM² + AM² ……… (1)
In ∆ BMD; BD² = DM² + (AM + AB)²
Or, BD² = DM² + AM² + AB² + 2AM.AB ………. (2)




Substituting the value of AM² from equation (1) in equation (2), we get;
BD² = AD² + AB² + 2AM.AB ………. (3)
In triangle AND;
AD² = AN² + DN² ………. (4)
In triangle ANC;
AC² = AN² + (DC – DN)²
Or, AC² = AN² + DN² + DC² – 2DC.DN ……… (5)
Substituting the value of AD² from equation (4) in equation (5), we get;
AC² = AD² + DC² – 2DC.DN ……….. (6)
We also have; AM = DN and AB = CD.
Substituting these values in equation (6), we get;
AC² = AD² + DC² – 2AM.AB ……… (7)
Adding equations (3) and (7), we get;
AC² + BD² = AD² + AB² + 2AM.AB + AD² + DC² – 2AM.AB
Or, AC² + BD² = AB² + BC² + DC² + AD² proved
Question 7: In the given figure, two chords AB and CD intersect each other at the point P. Prove that:

(a) Δ APC ~ Δ DPB
Solution: In Δ APC and Δ DPB;
∠ CAP = ∠ BDP (Angles on the same side of a chord are equal)
∠ APC = ∠ DPB (Opposite angles)
Hence; Δ APC ~ Δ DPB (AAA Criterion)
(b) AP.PB = CP. DP
Solution: Since the two triangles are similar, hence;

Question 8: In the given figure, two chords AB and CD of a circle intersect each other at the point P (when produced) outside the circle. Prove that:

(a) Δ PAC ~ Δ PDB
Solution: In triangle Δ PAC ~ Δ PDB;
∠ PAC + ∠ CAB = 180° (Linear pair of angles)
∠ CAB + ∠ BDC = 180° (Opposite angles of cyclic quadrilateral are supplementary)
Hence; ∠ PAC = ∠ PDB
Similarly; ∠ PCA = ∠ PBD can be proven
Hence; Δ PAC ~ Δ PDB
(b) PA.PB = PC.PD
Solution: Since the two triangles are similar, so;

Question 9: In the given figure, D is a point on side BC of Δ ABC such that Prove that AD is the bisector of ∠ BAC.

Solution: Draw a line CM which meets BA extended up to AM so that AM = AC
This means; ∠ AMC = ∠ ACM (Angles opposite to equal sides)

Hence; Δ ABD ~ Δ MBC
Hence; AD || MC
This means;
∠ DAC = ∠ ACM (Alternate angles)
∠ BAD = ∠ AMC (Corresponding angles)
Since, ∠ AMC = ∠ ACM
Hence; ∠ DAC = ∠ BDA
Hence; AD is the bisector of ∠ BAC
Question 10: Nazima is fly fishing in a stream. The tip of her fishing rod is 1.8 m above the surface of the water and the fly at the end of the string rests on the water 3.6 m away and 2.4 m from a point directly under the tip of the rod. Assuming that her string (from the tip of her rod to the fly) is taut, how much string does she have out? If she pulls in the string at the rate of 5 cm per second, what will be the horizontal distance of the fly from her after 12 seconds?

Solution: Here; AD = 1.8 m, BD = 2.4 m and CD = 1.2 m
Retracting speed of string = 5 cm per second
In triangle ABD; length of string, i.e. AB can be calculated as follows:
AB² = AD² + BD²
= (1.8)² + (2.4)²
= 3.24 + 5.76 = 9
Or, AB = 3
Let us assume that the string reaches at point M after 12 seconds
Length of retracted string in 12 seconds = 5 x 12 = 60 cm
Remaining length of string = 3 m – 0.6 m = 2.4 m
In triangle AMD, we can find MD by using Pythagoras Theorem.
MD² = AM² – AD²
= 2.4² – 1.8²
= 5.76 – 3.24
= 2.52
Or, MD = 1.58
Hence; horizontal distance between the girl and the fly = CD + MD = 1.2 + 1.58 = 2.78 m