NCERT Solutions Class 10 Mathematics Chapter 6 Triangles

Exercise 6.1

Q.1) Fill in the blanks using correct word given in the brackets: −
(i) All circles are __________. (congruent, similar)
(ii) All squares are __________. (similar, congruent)
(iii) All __________ triangles are similar. (isosceles, equilateral)
(iv) Two polygons of the same number of sides are similar, if (a) their corresponding angles are __________ and (b) their corresponding sides are __________. (equal, proportional)
Sol.1) (i) Similar
(ii) Similar
(iii) Equilateral
(iv) (a) Equal (b) Proportional

Q.2) Give two different examples of pair of
(i) Similar figures (ii)Non-similar figures
Sol.2) (i) Two twenty-rupee notes, Two rupees coins.
(ii) One rupee coin and five rupees coin, One rupee not and ten rupees note

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

Sol.3) The given two figures are not similar because their corresponding angles are not equal.

Exercise 6.2

Q.1) In fig. (i) and (ii), 𝐷𝐸 || 𝐵𝐶. Find 𝐸𝐶 in (i) and 𝐴𝐷 in (ii).

Q.2) E and F are points on the sides PQ and PR respectively of a 𝛥𝑃𝑄𝑅. For each of the following cases, state whether 𝐸𝐹 || 𝑄𝑅
(i) 𝑃𝐸 = 3.9 𝑐𝑚, 𝐸𝑄 = 3 𝑐𝑚, 𝑃𝐹 = 3.6 𝑐𝑚 and 𝐹𝑅 = 2.4 𝑐𝑚
(ii) 𝑃𝐸 = 4 𝑐𝑚, 𝑄𝐸 = 4.5 𝑐𝑚, 𝑃𝐹 = 8 𝑐𝑚 and 𝑅𝐹 = 9 𝑐𝑚
(iii) 𝑃𝑄 = 1.28 𝑐𝑚, 𝑃𝑅 = 2.56 𝑐𝑚, 𝑃𝐸 = 0.18 𝑐𝑚 and 𝑃𝐹 = 0.63 𝑐𝑚
Sol.2) In 𝛥𝑃𝑄𝑅, 𝐸 and 𝐹 are two points on side PQ and PR respectively.
(i) 𝑃𝐸 = 3.9 𝑐𝑚, 𝐸𝑄 = 3 𝑐𝑚 (Given)
𝑃𝐹 = 3.6 𝑐𝑚, 𝐹𝑅 = 2,4 𝑐𝑚 (Given)

Q.3) In the fig., if 𝐿𝑀 || 𝐶𝐵 and 𝐿𝑁 || 𝐶𝐷, prove that 𝐴𝑀/𝑀𝐵 = 𝐴𝑁/𝐴𝐷
Sol.3) In the given figure,
𝐿𝑀 || 𝐶𝐵
By using basic proportionality theorem, we get,

Q.7) Using Basic proportionality theorem, prove that a line drawn through the mid-points of one side of a triangle parallel to another side bisects the third side. (Recall that you have proved it in Class IX).
Sol.7) Given: 𝛥𝐴𝐵𝐶 in which D is the midpoint of AB such that 𝐴𝐷 = 𝐷𝐵.
A line parallel to 𝐵𝐶 intersects 𝐴𝐶 at E as shown in above figure such that 𝐷𝐸 || 𝐵𝐶.
To Prove:
E is the midpoint of AC.
Proof: D is the mid-point of 𝐴𝐵.
∴ 𝐴𝐷 = 𝐷𝐵
⇒ 𝐴𝐷/𝐵𝐷 = 1 ... (i)
In 𝛥𝐴𝐵𝐶, 𝐷𝐸 || 𝐵𝐶,
Therefore,
𝐴𝐷/𝐷𝐵 = 𝐴𝐸/𝐸𝐶
[By using Basic Proportionality Theorem]
⇒ 1 = 𝐴𝐸/𝐸𝐶
[From equation (i)]
∴ 𝐴𝐸 = 𝐸𝐶
Hence, E is the mid point of AC.

Q.8) Using Converse of basic proportionality theorem, prove that the line joining the midpoints of any two sides of a triangle is parallel to the third side. (Recall that you have done it in Class IX).
Sol.8) Given: 𝛥𝐴𝐵𝐶 in which D and E are the mid points of AB and AC respectively such that
𝐴𝐷 = 𝐵𝐷 and 𝐴𝐸 = 𝐸𝐶.
To Prove: 𝐷𝐸 || 𝐵𝐶
Proof: D is the midpoint of AB           (Given)
∴ 𝐴𝐷 = 𝐷𝐵
⇒ 𝐴𝐷/𝐵𝐷 = 1              ... (i)
Also, E is the mid-point of AC           (Given)
∴ 𝐴𝐸 = 𝐸𝐶 ⇒ 𝐴𝐸/𝐸𝐶 = 1                     [From equation (i)]
From equation (i) and (ii), we get
𝐴𝐷/𝐵𝐷 = 𝐴𝐸/𝐸𝐶
Hence, 𝐷𝐸 || 𝐵𝐶 [By converse of Basic Proportionality Theorem]

Q.9) ABCD is a trapezium in which 𝐴𝐵 || 𝐷𝐶 and its diagonals intersect each other at the point O. Show that 𝐴𝑂/𝐵𝑂 = 𝐶𝑂/𝐷𝑂.
Sol.9) Given: ABCD is a trapezium in which 𝐴𝐵 || 𝐷𝐶 in which diagonals AC and BD intersect each other at O.
To Prove:
𝐴𝑂/𝐵𝑂 = 𝐶𝑂/𝐷𝑂
Construction:
Through 𝑂, draw 𝐸𝑂 || 𝐷𝐶 || 𝐴𝐵
Proof: In 𝛥𝐴𝐷𝐶, we have 𝑂𝐸 || 𝐷𝐶 (By Construction)
∴ 𝐴𝐸/𝐸𝐷 = 𝐴𝑂/𝐶𝑂      ...(i)           [By using Basic Proportionality Theorem] In ΔABD, we
have 𝑂𝐸 || 𝐴𝐵 (By Construction)
∴ 𝐷𝐸/𝐸𝐴 = 𝐷𝑂/𝐵𝑂 .    ..(ii)           [By using Basic Proportionality Theorem]
From equation (i) and (ii), we get
𝐴𝑂/𝐶𝑂 = 𝐵𝑂/𝐷𝑂
⇒ 𝐴𝑂/𝐵𝑂 = 𝐶𝑂/𝐷𝑂

Q.10) The diagonals of a quadrilateral ABCD intersect each other at the point O such that 𝐴𝑂/𝐵𝑂 = 𝐶𝑂/𝐷𝑂. Show that 𝐴𝐵𝐶𝐷 is a trapezium.
Sol.10) Given: Quadrilateral ABCD in which diagonals AC and BD intersects each other at O such that 𝐴𝑂/𝐵𝑂 = 𝐶𝑂/𝐷𝑂.
To Prove:
𝐴𝐵𝐶𝐷 is a trapezium

Construction:
Through O, draw line 𝐸𝑂, where 𝐸𝑂 || 𝐴𝐵, which meets 𝐴𝐷 at E.
Proof:
In 𝛥𝐷𝐴𝐵, we have 𝐸𝑂 || 𝐴𝐵
∴ 𝐷𝐸/𝐸𝐴 = 𝐷𝑂/𝑂𝐵 ...(i) [By using Basic Proportionality Theorem]
Also,
𝐴𝑂/𝐵𝑂 = 𝐶𝑂/𝐷𝑂
(Given)
⇒ 𝐴𝑂/𝐶𝑂 = 𝐵𝑂/𝐷𝑂
⇒ 𝐶𝑂/𝐴𝑂 = 𝐵𝑂/𝐷𝑂
⇒ 𝐷𝑂/𝑂𝐵 = 𝐶𝑂/𝐴𝑂 ...(ii)
From equation (i) and (ii), we get
𝐷𝐸/𝐸𝐴 = 𝐶𝑂/𝐴𝑂
Therefore, By using converse of Basic Proportionality Theorem,
𝐸𝑂 || 𝐷𝐶
also 𝐸𝑂 || 𝐴𝐵
⇒ 𝐴𝐵 || 𝐷𝐶.
Hence, quadrilateral ABCD is a trapezium with 𝐴𝐵 || 𝐶𝐷.

Exercise 6.3

Q.1) State which pairs of triangles in Fig. 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:

Sol.1) (i) In 𝛥𝐴𝐵𝐶 and 𝛥𝑃𝑄𝑅, we have
∠𝐴 = ∠𝑃 = 60°       (Given)
∠𝐵 = ∠𝑄 = 80°       (Given)
∠𝐶 = ∠𝑅 = 40°       (Given)
∴ 𝛥𝐴𝐵𝐶 ~ 𝛥𝑃𝑄𝑅 (AAA similarity criterion)
(ii) In 𝛥𝐴𝐵𝐶 and 𝛥𝑃𝑄𝑅, we have
𝐴𝐵/𝑄𝑅 = 𝐵𝐶/𝑅𝑃 = 𝐶𝐴/𝑃𝑄
∴ 𝛥𝐴𝐵𝐶 ~ 𝛥𝑄𝑅𝑃 (SSS similarity criterion)
(iii) In 𝛥𝐿𝑀𝑃 and 𝛥𝐷𝐸𝐹, we have

(vi) In 𝛥𝐷𝐸𝐹, we have
∠𝐷 + ∠𝐸 + ∠𝐹 = 180°    (sum of angles of a triangle)
⇒ 70° + 80° + ∠𝐹 = 180°
⇒ ∠𝐹 = 180° − 70° − 80°
⇒ ∠𝐹 = 30°
In PQR, we have ∠𝑃 + ∠𝑄 + ∠𝑅 = 180 (Sum of angles of 𝛥)
⇒ ∠𝑃 + 80° + 30° = 180°
⇒ ∠𝑃 = 180° − 80° − 30°
⇒ ∠𝑃 = 70°
In 𝛥𝐷𝐸𝐹 and 𝛥𝑃𝑄𝑅, we have
∠𝐷 = ∠𝑃 = 70°
∠𝐹 = ∠𝑄 = 80°
∠𝐹 = ∠𝑅 = 30°
Hence, 𝛥𝐷𝐸𝐹 ~ 𝛥𝑃𝑄𝑅      (AAA similarity criterion)

Q.2) In the fig., 𝛥𝑂𝐷𝐶 ∝ ¼ 𝛥𝑂𝐵𝐴, ∠ 𝐵𝑂𝐶 = 125° and ∠ 𝐶𝐷𝑂 = 70°. Find ∠𝐷𝑂𝐶, ∠𝐷𝐶𝑂 and ∠ 𝑂𝐴𝐵.
Sol.2) 𝐷𝑂𝐵 is a straight line.
Therefore, ∠𝐷𝑂𝐶 + ∠ 𝐶𝑂𝐵 = 180°
⇒ ∠𝐷𝑂𝐶 = 180° − 125° = 55°
In 𝛥𝐷𝑂𝐶,
∠𝐷𝐶𝑂 + ∠ 𝐶𝐷𝑂 + ∠ 𝐷𝑂𝐶 = 180° (Sum of the measures of the angles of a triangle
is 180°.)
⇒ ∠𝐷𝐶𝑂 + 70° + 55° = 180°
⇒ ∠𝐷𝐶𝑂 = 55°
It is given that 𝛥𝑂𝐷𝐶 ~ 𝛥𝑂𝐵𝐴.
∴ ∠𝑂𝐴𝐵 = ∠𝑂𝐶𝐷              [Corresponding angles are equal in similar triangles.]
⇒ ∠ 𝑂𝐴𝐵 = 55°
∴ ∠𝑂𝐴𝐵 = ∠𝑂𝐶𝐷              [Corresponding angles are equal in similar triangles.]
⇒ ∠𝑂𝐴𝐵 = 55°

Q.3) Diagonals AC and BD of a trapezium ABCD with 𝐴𝐵 || 𝐷𝐶 intersect each other at the point O. Using a similarity criterion for two triangles, show that 𝐴𝑂/𝑂𝐶 = 𝑂𝐵/𝑂𝐷
Sol.3) In 𝛥𝐷𝑂𝐶 and 𝛥𝐵𝑂𝐴,
∠𝐶𝐷𝑂 = ∠𝐴𝐵𝑂               [Alternate interior angles as 𝐴𝐵 || 𝐶𝐷]
∠𝐷𝐶𝑂 = ∠𝐵𝐴𝑂               [Alternate interior angles as 𝐴𝐵 || 𝐶𝐷]
∠𝐷𝑂𝐶 = ∠𝐵𝑂𝐴               [Vertically opposite angles]
∴ 𝛥𝐷𝑂𝐶 ~ 𝛥𝐵𝑂𝐴                 [AAA similarity criterion]
∴ 𝐷𝑂/𝐵𝑂 = 𝑂𝐶/𝑂𝐴              [Corresponding sides are proportional]
⇒ 𝑂𝐴/𝑂𝐶 = 𝑂𝐵/𝑂𝐷

Q.4) In the fig., 𝑄𝑅/𝑄𝑆 = 𝑄𝑇/𝑃𝑅 and ∠1 = ∠2. Show that 𝛥𝑃𝑄𝑆 ~ 𝛥𝑇𝑄𝑅.
Sol.4) In 𝛥𝑃𝑄𝑅, ∠𝑃𝑄𝑅 = ∠𝑃𝑅𝑄
∴ 𝑃𝑄 = 𝑃𝑅                       ...(i)
Given,
𝑄𝑅/𝑄𝑆 = 𝑄𝑇/𝑃𝑅
Using (i), we get
𝑄𝑅/𝑄𝑆 = 𝑄𝑇/𝑄𝑃               ...(ii)
In 𝛥𝑃𝑄𝑆 and 𝛥𝑇𝑄𝑅,
𝑄𝑅/𝑄𝑆 = 𝑄𝑇/𝑄𝑃               [using (ii)]
∠𝑄 = ∠𝑄
∴ 𝛥𝑃𝑄𝑆 ~ 𝛥𝑇𝑄𝑅              [SAS similarity criterion]

Q.5) 𝑆 and 𝑇 are point on sides PR and QR of 𝛥𝑃𝑄𝑅 such that ∠𝑃 = ∠𝑅𝑇𝑆. Show that 𝛥𝑅𝑃𝑄 ~ 𝛥𝑅𝑇𝑆.
Sol.5) In 𝛥𝑅𝑃𝑄 and 𝛥𝑅𝑆𝑇,
∠𝑅𝑇𝑆 = ∠𝑄𝑃𝑆 (Given)
∠𝑅 = ∠𝑅                        (Common angle)
∴ 𝛥𝑅𝑃𝑄 ~ 𝛥𝑅𝑇𝑆              (By AA similarity criterion)

Q.6) In the fig., if 𝛥𝐴𝐵𝐸 ≅ 𝛥𝐴𝐶𝐷, show that 𝛥𝐴𝐷𝐸 ~ 𝛥𝐴𝐵𝐶.
Sol.6) It is given that 𝛥𝐴𝐵𝐸 ≅ 𝛥𝐴𝐶𝐷.
∴ 𝐴𝐵 = 𝐴𝐶          [By cpct]       ...(i)
And, 𝐴𝐷 = 𝐴𝐸     [By cpct]      ...(ii)
In 𝛥𝐴𝐷𝐸 and 𝛥𝐴𝐵𝐶,
𝐴𝐷/𝐴𝐵 = 𝐴𝐸/𝐴𝐶
[Dividing equation (ii) by (i)]
∠𝐴 = ∠𝐴 [Common angle]
∴ 𝛥𝐴𝐷𝐸 ~ 𝛥𝐴𝐵𝐶 [By SAS similarity criterion]

Q.7) In the fig., altitudes AD and CE of 𝛥𝐴𝐵𝐶 intersect each other at the point P. Show that:
(i) 𝛥𝐴𝐸𝑃 ~ 𝛥𝐶𝐷𝑃 (ii) 𝛥𝐴𝐵𝐷 ~ 𝛥𝐶𝐵𝐸
(iii) 𝛥𝐴𝐸𝑃 ~ 𝛥𝐴𝐷𝐵 (iv) 𝛥𝑃𝐷𝐶 ~ 𝛥𝐵𝐸𝐶
Sol.7) (i) In 𝛥𝐴𝐸𝑃 and 𝛥𝐶𝐷𝑃,
∠𝐴𝐸𝑃 = ∠𝐶𝐷𝑃          (Each 90°)
∠𝐴𝑃𝐸 = ∠𝐶𝑃𝐷          (Vertically opposite angles)
Hence, by using AA similarity criterion,
𝛥𝐴𝐸𝑃 ~ 𝛥𝐶𝐷𝑃

(ii) In 𝛥𝐴𝐵𝐷 and 𝛥𝐶𝐵𝐸,
∠𝐴𝐷𝐵 = ∠𝐶𝐸𝐵          (Each 90°)
∠𝐴𝐵𝐷 = ∠𝐶𝐵𝐸          (Common)
Hence, by using AA similarity criterion,
𝛥𝐴𝐵𝐷 ~ 𝛥𝐶𝐵𝐸

(iii) In 𝛥𝐴𝐸𝑃 and 𝛥𝐴𝐷𝐵,
∠𝐴𝐸𝑃 = ∠𝐴𝐷𝐵          (Each 90°)
∠𝑃𝐴𝐸 = ∠𝐷𝐴𝐵          (Common)
Hence, by using AA similarity criterion,
𝛥𝐴𝐸𝑃 ~ 𝛥𝐴𝐷𝐵

(iv) In 𝛥𝑃𝐷𝐶 and 𝛥𝐵𝐸𝐶,
∠𝑃𝐷𝐶 = ∠𝐵𝐸𝐶          (Each 90°)
∠𝑃𝐶𝐷 = ∠𝐵𝐶𝐸          (Common angle)
Hence, by using AA similarity criterion,
𝛥𝑃𝐷𝐶 ~ 𝛥𝐵𝐸𝐶

Q.8) E is a point on the side AD produced of a parallelogram 𝐴𝐵𝐶𝐷 and BE intersects CD at F. Show that 𝛥𝐴𝐵𝐸 ~ 𝛥𝐶𝐹𝐵
Sol.8) In 𝛥𝐴𝐵𝐸 and 𝛥𝐶𝐹𝐵,
∠𝐴 = ∠𝐶                    (Opposite angles of a parallelogram)
∠𝐴𝐸𝐵 = ∠𝐶𝐵𝐹             (Alternate interior angles as 𝐴𝐸 || 𝐵𝐶)
∴ 𝛥𝐴𝐵𝐸 ~ 𝛥𝐶𝐹𝐵          (By AA similarity criterion)

Q.9) In the fig., 𝐴𝐵𝐶 and 𝐴𝑀𝑃 are two right triangles, right angled at B and M respectively, prove that: (i) 𝛥𝐴𝐵𝐶 ~ 𝛥𝐴𝑀𝑃 (ii)
𝐶𝐴/𝑃𝐴 = 𝐵𝐶/𝑀𝑃
Sol.9) (i) In 𝛥𝐴𝐵𝐶 and 𝛥𝐴𝑀𝑃, we have
∠𝐴 = ∠𝐴          (common angle)
∠𝐴𝐵𝐶 = ∠𝐴𝑀𝑃 = 90°         (each 90°)
∴ 𝛥𝐴𝐵𝐶 ~ 𝛥𝐴𝑀𝑃                (By AA similarity criterion)
(ii) As, 𝛥𝐴𝐵𝐶 ~ 𝛥𝐴𝑀𝑃       (By AA similarity criterion)
If two triangles are similar then the corresponding sides are equal,
Hence,
𝐶𝐴/𝑃𝐴 = 𝐵𝐶/𝑀𝑃

Q.10) CD and GH are respectively the bisectors of ∠𝐴𝐶𝐵 and ∠𝐸𝐺𝐹 such that D and H lie on sides AB and FE of 𝛥𝐴𝐵𝐶 and 𝛥𝐸𝐹𝐺 respectively. If 𝛥𝐴𝐵𝐶 ~ 𝛥𝐹𝐸𝐺, Show that:
(i) 𝐶𝐷/𝐺𝐻 = 𝐴𝐶/𝐹𝐺
(ii) 𝛥𝐷𝐶𝐵 ~ 𝛥𝐻𝐺𝐸 (iii) 𝛥𝐷𝐶𝐴 ~ 𝛥𝐻𝐺𝐹
Sol.10) (i) It is given that 𝛥𝐴𝐵𝐶 ~ 𝛥𝐹𝐸𝐺.
∴ ∠𝐴 = ∠𝐹, ∠𝐵 = ∠𝐸, and
∠𝐴𝐶𝐵 = ∠𝐹𝐺𝐸
∴ ∠𝐴𝐶𝐷 = ∠𝐹𝐺𝐻             (Angle bisector)
And,
∠𝐷𝐶𝐵 = ∠𝐻𝐺𝐸               (Angle bisector)
In 𝛥𝐴𝐶𝐷 and 𝛥𝐹𝐺𝐻,
∠𝐴 = ∠𝐹                       (Proved above)
∠𝐴𝐶𝐷 = ∠𝐹𝐺𝐻               (Proved above)
∴ 𝛥𝐴𝐶𝐷 ~ 𝛥𝐹𝐺𝐻             (By AA similarity criterion)
⇒ 𝐶𝐷/𝐺𝐻 = 𝐴𝐶/𝐹𝐺
ii) In 𝛥𝐷𝐶𝐵 and 𝛥𝐻𝐺𝐸,
∠𝐷𝐶𝐵 = ∠𝐻𝐺𝐸               (Proved above)
∠𝐵 = ∠𝐸                      (Proved above)
∴ 𝛥𝐷𝐶𝐵 ~ 𝛥𝐻𝐺𝐸            (By AA similarity criterion)
(iii) In 𝛥𝐷𝐶𝐴 and 𝛥𝐻𝐺𝐹,
∠𝐴𝐶𝐷 = ∠𝐹𝐺𝐻               (Proved above)
∠𝐴 = ∠𝐹                      (Proved above)
∴ 𝛥𝐷𝐶𝐴 ~ 𝛥𝐻𝐺𝐹            (By AA similarity criterion)

Q.11) In the following figure, E is a point on side CB produced of an isosceles triangle ABC with 𝐴𝐵 = 𝐴𝐶. If 𝐴𝐷 ⊥ 𝐵𝐶 and 𝐸𝐹 ⊥ 𝐴𝐶, prove that 𝛥𝐴𝐵𝐷 ~ 𝛥𝐸𝐶𝐹.
Sol.11) It is given that ABC is an isosceles triangle.
∴ 𝐴𝐵 = 𝐴𝐶
⇒ ∠𝐴𝐵𝐷 = ∠𝐸𝐶𝐹
In 𝛥𝐴𝐵𝐷 and 𝛥𝐸𝐶𝐹,
∠𝐴𝐷𝐵 = ∠𝐸𝐹𝐶               (Each 90°)
∠𝐵𝐴𝐷 = ∠𝐶𝐸𝐹               (Proved above)
∴ 𝛥𝐴𝐵𝐷 ~ 𝛥𝐸𝐶𝐹            (By using AA similarity criterion)

Q.12) Sides 𝐴𝐵 and 𝐵𝐶 and median AD of a triangle ABC are respectively proportional to sides 𝑃𝑄 and 𝑄𝑅 and median 𝑃𝑀 of 𝛥𝑃𝑄𝑅 (see Fig.). Show that 𝛥𝐴𝐵𝐶 ~ 𝛥𝑃𝑄𝑅.
Sol.12) Given:
𝛥𝐴𝐵𝐶 and 𝛥𝑃𝑄𝑅, AB, BC and
median AD of 𝛥𝐴𝐵𝐶 are proportional to sides PQ, QR and

Q.13) D is a point on the side BC of a triangle ABC such that ∠𝐴𝐷𝐶 = ∠𝐵𝐴𝐶. Show that 𝐶𝐴² = 𝐶𝐵. 𝐶𝐷

Sol.13) In 𝛥𝐴𝐷𝐶 and 𝛥𝐵𝐴𝐶,
∠𝐴𝐷𝐶 = ∠𝐵𝐴𝐶 (Given)
∠𝐴𝐶𝐷 = ∠𝐵𝐶𝐴 (Common angle)
∴ 𝛥𝐴𝐷𝐶 ~ 𝛥𝐵𝐴𝐶 (By AA similarity criterion)
We know that corresponding sides of similar triangles are in proportion.
∴ 𝐶𝐴/𝐶𝐵 = 𝐶𝐷/𝐶𝐴
⇒ 𝐶𝐴² = 𝐶𝐵. 𝐶𝐷

Q.14) Sides AB and 𝐴𝐶 and median 𝐴𝐷 of a triangle 𝐴𝐵𝐶 are respectively proportional to sides 𝑃𝑄 and 𝑃𝑅 and median 𝑃𝑀 of another triangle 𝑃𝑄𝑅. Show that 𝛥𝐴𝐵𝐶 ~ 𝛥𝑃𝑄𝑅.
Sol.14) Given:
Two triangles 𝛥𝐴𝐵𝐶 and 𝛥𝑃𝑄𝑅 in which AD and PM are medians such that AB/PQ = 𝐴𝐶/𝑃𝑅 = 𝐴𝐷/𝑃𝑀
To Prove: 𝛥𝐴𝐵𝐶 ~ 𝛥𝑃𝑄𝑅
Construction:
Produce 𝐴𝐷 to E so that 𝐴𝐷 = 𝐷𝐸.
Join CE, Similarly produce PM to N such that 𝑃𝑀 = 𝑀𝑁, also Join RN.
Proof:
In 𝛥𝐴𝐵𝐷 and 𝛥𝐶𝐷𝐸, we have
𝐴𝐷 = 𝐷𝐸             [By Construction]
𝐵𝐷 = 𝐷𝐶             [∴ 𝐴𝑃 is the median] and,
∠𝐴𝐷𝐵 = ∠𝐶𝐷𝐸      [Vertically opp. angles]

∴ 𝛥𝐴𝐵𝐷 ≅ 𝛥𝐶𝐷𝐸           [By SAS criterion of congruence]
⇒ 𝐴𝐵 = 𝐶𝐸                    [CPCT]             ...(i)
Also, in 𝛥𝑃𝑄𝑀 and 𝛥𝑀𝑁𝑅, we have
𝑃𝑀 = 𝑀𝑁                      [By Construction]
𝑄𝑀 = 𝑀𝑅                      [∴ 𝑃𝑀 is the median] and,
∠𝑃𝑀𝑄 = ∠𝑁𝑀𝑅            [Vertically opposite angles]
∴ 𝛥𝑃𝑄𝑀 = 𝛥𝑀𝑁𝑅         [By SAS criterion of congruence]
⇒ 𝑃𝑄 = 𝑅𝑁                   [CPCT]       ...(ii)

Therefore, ∠2 = ∠4
Similarly, ∠1 = ∠3
∴ ∠1 + ∠2 = ∠3 + ∠4
⇒ ∠𝐴 = ∠𝑃 ...(iii)
Now, In 𝛥𝐴𝐵𝐶 and 𝛥𝑃𝑄𝑅, we have
𝐴𝐵/𝑃𝑄 = 𝐴𝐶/𝑃𝑅
(Given)
∠𝐴 = ∠𝑃 [From (iii)]
∴ 𝛥𝐴𝐵𝐶 ~ 𝛥𝑃𝑄𝑅 [By SAS similarity criterion]

Q.15) A vertical pole of a length 6 𝑚 casts a shadow 4𝑚 long on the ground and at the same time a tower casts a shadow 28 𝑚 long. Find the height of the tower.
Sol.15) Length of the vertical pole = 6𝑚       (Given)
Shadow of the pole = 4 𝑚 (Given)
Let Height of tower = ℎ 𝑚
Length of shadow of the tower = 28 𝑚           (Given)/
∴ 𝐴𝐵/𝐷𝐹 = 𝐵𝐶/𝐸𝐹
(If two triangles are similar corresponding sides are proportional)
∴ 6/ℎ = 4/28
⇒ ℎ = 6 × 28/4
⇒ ℎ = 6 × 7
⇒ ℎ = 42 𝑚
Hence, the height of the tower is 42 𝑚.

Q.16) If 𝐴𝐷 and 𝑃𝑀 are medians of triangles 𝐴𝐵𝐶 and 𝑃𝑄𝑅, respectively where 𝛥𝐴𝐵𝐶 ~ 𝛥𝑃𝑄𝑅 prove that 𝐴𝐵/𝑃𝑄 = 𝐴𝐷/𝑃𝑀.
Sol.16) It is given that 𝛥𝐴𝐵𝐶 ~ 𝛥𝑃𝑄𝑅
We know that the corresponding sides of similar triangles are in proportion.

Exercise 6.4

Q.1) Let 𝛥𝐴𝐵𝐶 ~ 𝛥𝐷𝐸𝐹 and their areas be, respectively, 64 𝑐𝑚² and 121 𝑐𝑚² . If 𝐸𝐹 = 15.4 𝑐𝑚, find 𝐵𝐶.
Sol.1) It is given that,
Area of 𝛥𝐴𝐵𝐶 = 64 𝑐𝑚²
Area of 𝛥𝐷𝐸𝐹 = 121 𝑐𝑚²,
𝐸𝐹 = 15.4 𝑐𝑚 and,
𝛥𝐴𝐵𝐶 ~ 𝛥𝐷𝐸𝐹
∴ Area of 𝛥𝐴𝐵𝐶/Area of 𝛥𝐷𝐸𝐹 = 𝐴𝐵2/𝐷𝐸2 = 𝐴𝐶²/𝐷𝐹² = 𝐵𝐶²/𝐸𝐹²         ...(i)
[If two triangles are similar, ratio of their areas are equal to the square of the ratio of their corresponding sides]
∴ 64/121 = 𝐵𝐶²/𝐸𝐹²
⇒ (8/11)² = (𝐵𝐶/15.4)²
⇒ 8/11 = 𝐵𝐶/15.4
⇒ 𝐵𝐶 = 8 × 15.4/11
⇒ 𝐵𝐶 = 8 × 1.4
⇒ 𝐵𝐶 = 11.2 𝑐𝑚

Q.2) Diagonals of a trapezium 𝐴𝐵𝐶𝐷 with 𝐴𝐵 || 𝐷𝐶 intersect each other at the point 𝑂. If 𝐴𝐵 = 2𝐶𝐷, find the ratio of the areas of triangles 𝐴𝑂𝐵 and 𝐶𝑂𝐷.
Sol.2) 𝐴𝐵𝐶𝐷 is a trapezium with 𝐴𝐵 || 𝐷𝐶.
Diagonals AC and BD intersect each other at point O.
In 𝛥𝐴𝑂𝐵 and 𝛥𝐶𝑂𝐷, we have
∠1 = ∠2      (Alternate angles)
∠3 = ∠4      (Alternate angles)
∠5 = ∠6       (Vertically opposite angle)
∴ 𝛥𝐴𝑂𝐵 ~ 𝛥𝐶𝑂𝐷          [By AAA similarity criterion]
Now, Area of (𝛥𝐴𝑂𝐵)/Area of (𝛥𝐶𝑂𝐷) = 𝐴𝐵2/𝐶𝐷2 [If two triangles are similar then the ratio of their areas are equal to the square of the ratio of their corresponding sides]

Hence, the required ratio of the area of 𝛥𝐴𝑂𝐵 and 𝛥𝐶𝑂𝐷 = 4: 1

Q.3) In the fig., ABC and DBC are two triangles on the same base BC. If 𝐴𝐷 intersects 𝐵𝐶 at O, show that area (𝛥𝐴𝐵𝐶)/area (𝛥𝐷𝐵𝐶) = 𝐴𝑂/𝐷𝑂.

Sol.3) Given:
𝐴𝐵𝐶 and 𝐷𝐵𝐶 are triangles on the same base 𝐵𝐶. 𝐴𝐷 intersects 𝐵𝐶 at O.
To Prove:
area (𝛥𝐴𝐵𝐶)/area (𝛥𝐷𝐵𝐶) = 𝐴𝑂/𝐷𝑂.
Construction:
Let us draw two perpendiculars 𝐴𝑃 and 𝐷𝑀 on line BC.
Proof:
We know that area of a triangle = 1/2 × 𝐵𝑎𝑠𝑒 × 𝐻𝑒𝑖𝑔ℎ𝑡

Q.4) If the areas of two similar triangles are equal, prove that they are congruent.
Sol.4) Given:
𝛥𝐴𝐵𝐶 and 𝛥𝑃𝑄𝑅 are similar and equal in area.
To Prove: 𝛥𝐴𝐵𝐶 ≅ 𝛥𝑃𝑄𝑅
Proof:
Since, 𝛥𝐴𝐵𝐶 ~ 𝛥𝑃𝑄𝑅
∴ Area of (𝛥𝐴𝐵𝐶)/Area of (𝛥𝑃𝑄𝑅) = 𝐵𝐶²/𝑄𝑅²
⇒ 𝐵𝐶²/𝑄𝑅² = 1
Since, Area(𝛥𝐴𝐵𝐶) = (𝛥𝑃𝑄𝑅)
⇒ 𝐵𝐶²/𝑄𝑅² ⇒ 𝐵𝐶 = 𝑄𝑅
Similarly, we can prove that 𝐴𝐵 = 𝑃𝑄 and 𝐴𝐶 = 𝑃𝑅
Thus, 𝛥𝐴𝐵𝐶 ≅ 𝛥𝑃𝑄𝑅                      [BY SSS criterion of congruence]

Q.5) D, E and F are respectively the mid-points of sides AB, BC and CA of 𝛥𝐴𝐵𝐶. Find the ratio of the area of 𝛥𝐷𝐸𝐹 and 𝛥𝐴𝐵𝐶
Sol.5) Given:
D, E and F are the mid-points of the sides AB, BC and CA respectively of the 𝛥𝐴𝐵𝐶.
To Find: area(𝛥𝐷𝐸𝐹) and area(𝛥𝐴𝐵𝐶)
Solution:
In 𝛥𝐴𝐵𝐶, we have F is the mid-point of AB               (Given)
E is the mid-point of 𝐴𝐶               (Given)
So, by the mid-point theorem, we have
𝐹𝐸 || 𝐵𝐶 and 𝐹𝐸 = ½𝐵𝐶
⇒ 𝐹𝐸 || 𝐵𝐶 and 𝐹𝐸 || 𝐵𝐷               [𝐵𝐷 = 1/2 𝐵𝐶]
∴ 𝐵𝐷𝐸𝐹 is parallelogram [Opposite sides of parallelogram are equal and parallel]
Similarly in 𝛥𝐹𝐵𝐷 and 𝛥𝐷𝐸𝐹, we have
𝐹𝐵 = 𝐷𝐸               (Opposite sides of parallelogram 𝐵𝐷𝐸𝐹)
𝐹𝐷 = 𝐹𝐷               (Common)
𝐵𝐷 = 𝐹𝐸               (Opposite sides of parallelogram 𝐵𝐷𝐸𝐹)
∴ 𝛥𝐹𝐵𝐷 ≅ 𝛥𝐷𝐸𝐹
Similarly, we can prove that :
𝛥𝐴𝐹𝐸 ≅ 𝛥𝐷𝐸𝐹
𝛥𝐸𝐷𝐶 ≅ 𝛥𝐷𝐸𝐹
If triangles are congruent, then they are equal in area.
So, 𝑎𝑟𝑒𝑎(𝛥𝐹𝐵𝐷) = 𝑎𝑟𝑒𝑎(𝛥𝐷𝐸𝐹) ...(i)
𝑎𝑟𝑒𝑎(𝛥𝐴𝐹𝐸) = 𝑎𝑟𝑒𝑎(𝛥𝐷𝐸𝐹) ...(ii) and,
𝑎𝑟𝑒𝑎(𝛥𝐸𝐷𝐶) = 𝑎𝑟𝑒𝑎(𝛥𝐷𝐸𝐹) ...(iii)
Now,
𝑎𝑟𝑒𝑎(𝛥𝐴𝐵𝐶) = 𝑎𝑟𝑒𝑎(𝛥𝐹𝐵𝐷) + 𝑎𝑟𝑒𝑎(𝛥𝐷𝐸𝐹) + 𝑎𝑟𝑒𝑎(𝛥𝐴𝐹𝐸) + 𝑎𝑟𝑒𝑎(𝛥𝐸𝐷𝐶) ...(iv)
𝑎𝑟𝑒𝑎(𝛥𝐴𝐵𝐶) = 𝑎𝑟𝑒𝑎(𝛥𝐷𝐸𝐹) + 𝑎𝑟𝑒𝑎(𝛥𝐷𝐸𝐹) + 𝑎𝑟𝑒𝑎(𝛥𝐷𝐸𝐹) + 𝑎𝑟𝑒𝑎(𝛥𝐷𝐸𝐹)
⇒ 𝑎𝑟𝑒𝑎(𝛥𝐷𝐸𝐹) = 1/4
𝑎𝑟𝑒𝑎(𝛥𝐴𝐵𝐶)                [From (i), (ii) and (iii)]
⇒ 𝑎𝑟𝑒𝑎(𝛥𝐷𝐸𝐹)/𝑎𝑟𝑒𝑎(𝛥𝐴𝐵𝐶) = 1/4
Hence, 𝑎𝑟𝑒𝑎(𝛥𝐷𝐸𝐹): 𝑎𝑟𝑒𝑎(𝛥𝐴𝐵𝐶) = 1: 4

Q.6) Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.
Sol.6) Given: 𝐴𝑀 and 𝐷𝑁 are the medians of triangles 𝐴𝐵𝐶 and 𝐷𝐸𝐹 respectively and 𝛥𝐴𝐵𝐶 ~ 𝛥𝐷𝐸𝐹.

Q.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.
Sol.7) Given:
ABCD is a square whose one diagonal is AC.
𝛥𝐴𝑃𝐶 and 𝛥𝐵𝑄𝐶 are two equilateral triangles described on the diagonals AC and side BC of the square ABCD.
To Prove:
𝑎𝑟𝑒𝑎(𝛥𝐵𝑄𝐶) = 1/2 𝑎𝑟𝑒𝑎(𝛥𝐴𝑃𝐶)
Proof:
𝛥𝐴𝑃𝐶 and 𝛥𝐵𝑄𝐶 are both equilateral triangles (Given)
∴ 𝛥𝐴𝑃𝐶 ~ 𝛥𝐵𝑄𝐶 [AAA similarity criterion]

Q.8) Tick the correct answer and justify:
𝐴𝐵𝐶 and 𝐵𝐷𝐸 are two equilateral triangles such that D is the mid-point of BC. Ratio of the area of triangles 𝐴𝐵𝐶 and 𝐵𝐷𝐸 is
(A) 2 : 1 (B) 1 : 2 (C) 4 : 1 (D) 1 : 4
Sol.8) 𝛥𝐴𝐵𝐶 and 𝛥𝐵𝐷𝐸 are two equilateral triangle.
D is the mid-point of BC.

Q.9) Sides of two similar triangles are in the ratio 4 : 9. Areas of these triangles are in the ratio
(A) 2 : 3 (B) 4 : 9 (C) 81 : 16 (D) 16 : 81
Sol.9) Let 𝐴𝐵𝐶 and 𝐷𝐸𝐹 are two similarity triangles 𝛥𝐴𝐵𝐶 ~ 𝛥𝐷𝐸𝐹 (Given) and,

Exercise 6.5

Q.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.
(i) 7 cm, 24 cm, 25 cm (ii) 3 cm, 8 cm, 6 cm
(iii) 50 cm, 80 cm, 100 cm (iv) 13 cm, 12 cm, 5 cm
Sol.1) (i) Given that the sides of the triangle are 7 cm, 24 cm, and 25 cm.
Squaring the lengths of these sides, we will get 49, 576, and 625.
49 + 576 = 625
(7)² + (24)² = (25)²
The sides of the given triangle are satisfying Pythagoras theorem.
Hence, it is right angled triangle.
Length of Hypotenuse = 25 𝑐𝑚
(ii) Given that the sides of the triangle are 3 cm, 8 cm, and 6 cm.
Squaring the lengths of these sides, we will get 9, 64, and 36.
However, 9 + 36 ≠ 64 Or, 32 + 62 ≠ 82
Clearly, the sum of the squares of the lengths of two sides is not equal to the square of the length of the third side.
Therefore, the given triangle is not satisfying Pythagoras theorem.
(iii) Given that sides are 50 cm, 80 cm, and 100 cm.
Squaring the lengths of these sides, we will get 2500, 6400, and 10000.
However, 2500 + 6400 ≠ 10000 Or, 502 + 802 ≠ 1002
Clearly, the sum of the squares of the lengths of two sides is not equal to the square of
the length of the third side.
Therefore, the given triangle is not satisfying Pythagoras theorem.
Hence, it is not a right triangle.
(iv) Given that sides are 13 cm, 12 cm, and 5 cm.
Squaring the lengths of these sides, we will get 169, 144, and 25.
Clearly, 144 + 25 = 169 Or, 122 + 52 = 132
The sides of the given triangle are satisfying Pythagoras theorem.
Therefore, it is a right triangle.
Length of the hypotenuse of this triangle is 13 cm.

Q.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.
Sol.2) Given:
𝛥𝑃𝑄𝑅 is right angled at P is a point on QR such that 𝑃𝑀 ⊥ 𝑄𝑅.
To prove: 𝑃𝑀² = 𝑄𝑀 × 𝑀𝑅
Proof: In 𝛥𝑃𝑄𝑀, we have 𝑃𝑄² = 𝑃𝑀² + 𝑄𝑀² [By Pythagoras theorem] Or,
𝑃𝑀² = 𝑃𝑄² − 𝑄𝑀² ...(i)
In 𝛥𝑃𝑀𝑅, we have 𝑃𝑅² = 𝑃𝑀² + 𝑀𝑅² [By Pythagoras theorem] Or,
𝑃𝑀² = 𝑃𝑅² − 𝑀𝑅² ...(ii)
Adding (i) and (ii), we get
2𝑃𝑀² = (𝑃𝑄² + 𝑃𝑀²) − (𝑄𝑀² + 𝑀𝑅²)
= 𝑄𝑅² − 𝑄𝑀² − 𝑀𝑅² [∴ 𝑄𝑅² = 𝑃𝑄² + 𝑃𝑅² ]
= (𝑄𝑀 + 𝑀𝑅)² − 𝑄𝑀² − 𝑀𝑅² = 2𝑄𝑀 × 𝑀𝑅
∴ 𝑃𝑀² = 𝑄𝑀 × 𝑀𝑅

Q.3) In Fig., 𝐴𝐵𝐷 is a triangle right angled at A and 𝐴𝐶 ⊥ 𝐵𝐷. Show that
(i) 𝐴𝐵² = 𝐵𝐶 × 𝐵𝐷 (ii) 𝐴𝐶² = 𝐵𝐶 × 𝐷𝐶 (iii) 𝐴𝐷² = 𝐵𝐷 × 𝐶𝐷
Sol.3) (i) In 𝛥𝐴𝐷𝐵 and 𝛥𝐶𝐴𝐵, we have
∠𝐷𝐴𝐵 = ∠𝐴𝐶𝐵 (Each equals to 90°)
∠𝐴𝐵𝐷 = ∠𝐶𝐵𝐴 (Common angle)
∴ 𝛥𝐴𝐷𝐵 ~ 𝛥𝐶𝐴𝐵 [AA similarity criterion]
⇒ 𝐴𝐵/𝐶𝐵 = 𝐵𝐷/𝐴𝐵
⇒ 𝐴𝐵² = 𝐶𝐵 × 𝐵𝐷
(ii) Let ∠𝐶𝐴𝐵 = 𝑥
In 𝛥𝐶𝐵𝐴,
∠𝐶𝐵𝐴 = 180° − 90° − 𝑥
∠𝐶𝐵𝐴 = 90° − 𝑥
Similarly, in 𝛥𝐶𝐴𝐷
∠𝐶𝐴𝐷 = 90° − ∠𝐶𝐵𝐴
= 90° − 𝑥
∠𝐶𝐷𝐴 = 180° − 90° − (90° − 𝑥)
∠𝐶𝐷𝐴 = 𝑥
In 𝛥𝐶𝐵𝐴 and 𝛥𝐶𝐴𝐷, we have
∠𝐶𝐵𝐴 = ∠𝐶𝐴𝐷
∠𝐶𝐴𝐵 = ∠𝐶𝐷𝐴
∠𝐴𝐶𝐵 = ∠𝐷𝐶𝐴 (Each equals to 90°)
∴ 𝛥𝐶𝐵𝐴 ~ 𝛥𝐶𝐴𝐷 [By AAA similarity criterion]
⇒ 𝐴𝐶/𝐷𝐶 = 𝐵𝐶/𝐴𝐶
⇒ 𝐴𝐶² = 𝐷𝐶 × 𝐵𝐶
(iii) In 𝛥𝐷𝐶𝐴 and 𝛥𝐷𝐴𝐵, we have
∠DCA = ∠DAB (Each equals to 90°)
∠𝐶𝐷𝐴 = ∠𝐴𝐷𝐵 (common angle)
∴ 𝛥𝐷𝐶𝐴 ~ 𝛥𝐷𝐴𝐵 [By AA similarity criterion]
⇒ 𝐷𝐶/𝐷𝐴
= 𝐷𝐴/𝐷𝐴
⇒ 𝐴𝐷² = 𝐵𝐷 × 𝐶𝐷

Q.4) 𝐴𝐵𝐶 is an isosceles triangle right angled at C. Prove that 𝐴𝐵² = 2𝐴𝐶²
Sol.4) Given that 𝛥𝐴𝐵𝐶 is an isosceles triangle right angled at C.
In 𝛥𝐴𝐶𝐵,
∠𝐶 = 90°
𝐴𝐶 = 𝐵𝐶 (Given)
𝐴𝐵² = 𝐴𝐶² + 𝐵𝐶² [By using Pythagoras theorem]
= 𝐴𝐶² + 𝐴𝐶² [Since, AC = BC]
𝐴𝐵² = 2𝐴𝐶²

Q.5) ABC is an isosceles triangle with 𝐴𝐶 = 𝐵𝐶. If 𝐴𝐵² = 2𝐴𝐶² , prove that 𝐴𝐵𝐶 is a right triangle.
Sol.5) Given that 𝛥𝐴𝐵𝐶 is an isosceles triangle having
𝐴𝐶 = 𝐵𝐶 and 𝐴𝐵² = 2𝐴𝐶²
In 𝛥𝐴𝐶𝐵,
𝐴𝐶 = 𝐵𝐶 (Given)
𝐴𝐵² = 2𝐴𝐶² (Given)
𝐴𝐵² = 𝐴𝐶² + 𝐴𝐶²
= 𝐴𝐶² + 𝐵𝐶² [Since, AC = BC]
Hence, By Pythagoras theorem 𝛥𝐴𝐵𝐶 is right angle triangle.

Q.6) 𝐴𝐵𝐶 is an equilateral triangle of side 2𝑎. Find each of its altitudes.
Sol.6) ABC is an equilateral triangle of side 2𝑎.
Draw, 𝐴𝐷 ⊥ 𝐵𝐶
In 𝛥𝐴𝐷𝐵 and 𝛥𝐴𝐷𝐶, we have
𝐴𝐵 = 𝐴𝐶 [Given]
𝐴𝐷 = 𝐴𝐷 [Given]
∠𝐴𝐷𝐵 = ∠𝐴𝐷𝐶 [equal to 90°]
Therefore, 𝛥𝐴𝐷𝐵 ≅ 𝛥𝐴𝐷𝐶 by RHS congruence.
Hence, 𝐵𝐷 = 𝐷𝐶 [by CPCT]
In right angled 𝛥𝐴𝐷𝐵,
𝐴𝐵² = 𝐴𝐷² + 𝐵𝐷²
(2𝑎)² = 𝐴𝐷² + 𝑎²
⇒ 𝐴𝐷² = 4𝑎² − 𝑎²
⇒ 𝐴𝐷² = 3𝑎² ⇒ 𝐴𝐷 = √3𝑎

Q.7) Prove that the sum of the squares of the sides of rhombus is equal to the sum of the squares of its diagonals.
Sol.7) 𝐴𝐵𝐶𝐷 is a rhombus whose diagonals AC and BD intersect at O. [Given]
We have to prove that, 𝐴𝐵² + 𝐵𝐶² + 𝐶𝐷² + 𝐴𝐷²
= 𝐴𝐶² + 𝐵𝐷²
Since, the diagonals of a rhombus bisect each other at right angles.
Therefore, 𝐴𝑂 = 𝐶𝑂 and 𝐵𝑂 = 𝐷𝑂
In 𝛥𝐴𝑂𝐵,
∠𝐴𝑂𝐵 = 90°
𝐴𝐵² = 𝐴𝑂² + 𝐵𝑂² ... (i) [By Pythagoras]
Similarly, 𝐴𝐷² = 𝐴𝑂² + 𝐷𝑂² ... (ii)
𝐷𝐶² = 𝐷𝑂² + 𝐶𝑂² ... (iii)
𝐵𝐶² = 𝐶𝑂² + 𝐵𝑂² ... (iv)
Adding equations (i) + (ii) + (iii) + (iv) we get,
𝐴𝐵² + 𝐴𝐷² + 𝐷𝐶² + 𝐵𝐶² = 2(𝐴𝑂² + 𝐵𝑂² + 𝐷𝑂² + 𝐶𝑂²)
= 4𝐴𝑂² + 4𝐵𝑂² [Since, 𝐴𝑂 = 𝐶𝑂 and 𝐵𝑂 = 𝐷𝑂]
= (2𝐴𝑂)² + (2𝐵𝑂)² = 𝐴𝐶² + 𝐵𝐷²

Q.8) In Fig., O is a point in the interior of a triangle. 𝐴𝐵𝐶, 𝑂𝐷 ⊥ 𝐵𝐶, 𝑂𝐸 ⊥ 𝐴𝐶 and 𝑂𝐹 ⊥ 𝐴𝐵. Show that
(i) 𝑂𝑎² + 𝑂𝐵² + 𝑂𝐶² – 𝑂𝐷² – 𝑂𝐸²– 𝑂𝐹² = 𝐴𝐹² + 𝐵𝐷² + 𝐶𝐸²
(ii) 𝐴𝐹² + 𝐵𝐷² + 𝐶𝐸² = 𝐴𝐸² + 𝐶𝐷² + 𝐵𝐹
Sol.8) Join 𝑂𝐴, 𝑂𝐵 and 𝑂𝐶
(i) Applying Pythagoras theorem in 𝛥𝐴𝑂𝐹, we have
𝑂𝐴² = 𝑂𝐹² + 𝐴𝐹²
Similarly,
in 𝛥𝐵𝑂𝐷, 𝑂𝐵² = 𝑂𝐷² + 𝐵𝐷²
Similarly,
in 𝛥𝐶𝑂𝐸, 𝑂𝐶² = 𝑂𝐸² + 𝐸𝐶²
Adding these equations,
𝑂𝐴² + 𝑂𝐵² + 𝑂𝐶² = 𝑂𝐹² + 𝐴² + 𝑂𝐷² + 𝐵𝐷² + 𝑂𝐸² + 𝐸𝐶²
𝑂𝐴² + 𝑂𝐵² + 𝑂𝐶²– 𝑂𝐷² – 𝑂𝐸²– 𝑂𝐹² = 𝐴𝐹² + 𝐵𝐷² + 𝐶𝐸² .
(ii) 𝐴𝐹² + 𝐵𝐷² + 𝐸𝐶² = (𝑂𝐴² − 𝑂𝐸²) + (𝑂𝐶² − 𝑂𝐷²) + (𝑂𝐵² − 𝑂𝐹²)
∴ 𝐴𝐹² + 𝐵𝐷² + 𝐶𝐸² = 𝐴𝐸² + 𝐶𝐷² + 𝐵𝐹² .

Q.9) A ladder 10 m long reaches a window 8 m above the ground. Find the distance of the foot of the ladder from base of the wall.
Sol.9) Let BA be the wall and Ac be the ladder,
Therefore, by Pythagoras theorem, we have
𝐴𝐶² = 𝐴𝐵² + 𝐵𝐶²
102 = 82 + 𝐵𝐶²
𝐵𝐶² = 100 – 64
𝐵𝐶² = 36
𝐵𝐶 = 6𝑚
Therefore, the distance of the foot of the ladder from the base of the wall is 6 𝑚.

Q.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 ?
Sol.10) Let AB be the pole and 𝐴𝐶 be the wire.
By Pythagoras theorem,
𝐴𝐶² = 𝐴𝐵² + 𝐵𝐶²
242 = 182 + 𝐵𝐶²
𝐵𝐶² = 576 − 324
𝐵𝐶² = 252
𝐵𝐶 = 6√7𝑚
Therefore, the distance from the base is 6√7𝑚

Q.11) An aeroplane leaves an airport and flies due north at a speed of 1,000 km per hour. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1,200 km per hour. How far apart will be the two planes after 1(1/2) hours?
Sol.11) Speed of first aeroplane = 1000 𝑘𝑚/ℎ𝑟

⇒ 𝐴𝐵 = 300√61 𝑘𝑚
Hence, the distance between two aeroplanes will be 300√61 𝑘𝑚

Q.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.
Sol.12) Let CD and AB be the poles of height 11 𝑚 and 6 𝑚.
Therefore, 𝐶𝑃 = 11 − 6 = 5 𝑚
From the figure, it can be observed that 𝐴𝑃 = 12𝑚
Applying Pythagoras theorem for 𝛥𝐴𝑃𝐶, we get
𝐴𝑃2 = 𝑃𝐶² + 𝐴𝐶²
(12𝑚)² + (5𝑚)² = (𝐴𝐶)²
𝐴𝐶² = (144 + 25)𝑚² = 169 𝑚²
𝐴𝐶 = 13𝑚
Therefore, the distance between their tops is 13 𝑚.

Q.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
Sol.13) Applying Pythagoras theorem in 𝛥𝐴𝐶𝐸, we get
𝐴𝐶² + 𝐶𝐸² = 𝐴𝐸² ....(i)
Applying Pythagoras theorem in 𝛥𝐵𝐶𝐷, we get
𝐵𝐶² + 𝐶𝐷² = 𝐵𝐷²....(ii)
Using equations (i) and (ii), we get
𝐴𝐶² + 𝐶𝐸² + 𝐵𝐶² + 𝐶𝐷² = 𝐴𝐸² + 𝐵𝐷² ...(iii)
Applying Pythagoras theorem in 𝛥𝐶𝐷𝐸, we get
𝐷𝐸² = 𝐶𝐷² + 𝐶𝐸²
Applying Pythagoras theorem in 𝛥𝐴𝐵𝐶, we get
𝐴𝐵² = 𝐴𝐶² + 𝐶𝐵²
Putting these values in equation (iii), we get
𝐷𝐸² + 𝐴𝐵² = 𝐴𝐸² + 𝐵𝐷²

Q.14) The perpendicular from A on side BC of a 𝛥𝐴𝐵𝐶 intersects 𝐵𝐶 at D such that 𝐷𝐵 = 3𝐶𝐷 (see Fig.). Prove that 2𝐴𝐵² = 2𝐴𝐶² + 𝐵𝐶² .

Sol.14) Given that in 𝛥𝐴𝐵𝐶, we have
𝐴𝐷 ⊥ 𝐵𝐶 and 𝐵𝐷 = 3𝐶𝐷
In right angle triangles 𝐴𝐷𝐵 and 𝐴𝐷𝐶, we have
𝐴𝐵² = 𝐴𝐷²  + 𝐵𝐷²              ...(i)
𝐴𝐶² = 𝐴𝐷²  + 𝐷𝐶²             ...(ii) [By Pythagoras theorem]
Subtracting equation (ii) from equation (i), we get
𝐴𝐵² − 𝐴𝐶² = 𝐵𝐷²  − 𝐷𝐶² = 9𝐶𝐷²  − 𝐶𝐷²  [∴ 𝐵𝐷 = 3𝐶𝐷]
= 9𝐶𝐷²  = 8 (𝐵𝐶/4)²
[Since, 𝐵𝐶 = 𝐷𝐵 + 𝐶𝐷 = 3𝐶𝐷 + 𝐶𝐷 = 4𝐶𝐷]
Therefore, 𝐴𝐵² − 𝐴𝐶2 = 𝐵𝐶²/2
⇒ 2(𝐴𝐵² − 𝐴𝐶² ) = 𝐵𝐶²
⇒ 2𝐴𝐵² − 2𝐴𝐶² = 𝐵𝐶²
∴ 2𝐴𝐵² = 2𝐴𝐶² + 𝐵𝐶²

Q.15) In an equilateral triangle 𝐴𝐵𝐶, 𝐷 is a point on side 𝐵𝐶 such that 𝐵𝐷 = 1/3 𝐵𝐶. Prove that 9𝐴𝐷2 = 7𝐴𝐵2
Sol.15) Let the side of the equilateral triangle be 𝑎, and 𝐴𝐸 be the altitude of 𝛥𝐴𝐵𝐶.

Q.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.
Sol.16) Let the side of the equilateral triangle be 𝑎, and
𝐴𝐸 be the altitude of 𝛥𝐴𝐵𝐶.

4𝐴𝐸² = 3𝑎²
⇒ 4 × (𝑆𝑞𝑢𝑎𝑟𝑒 𝑜𝑓 𝑎𝑙𝑡𝑖𝑡𝑢𝑑𝑒) = 3 × (𝑆𝑞𝑢𝑎𝑟𝑒 𝑜𝑓 𝑜𝑛𝑒 𝑠𝑖𝑑𝑒)

Q.17) Tick the correct answer and justify: In 𝛥𝐴𝐵𝐶, 𝐴𝐵 = 6√3 𝑐𝑚, 𝐴𝐶 = 12 𝑐𝑚 and 𝐵𝐶 = 6 𝑐𝑚. The angle B is:
(A) 120° (B) 60° (C) 90° (D) 45°
Sol.17) Given that, 𝐴𝐵 = 6√3 𝑐𝑚, 𝐴𝐶 = 12 𝑐𝑚, and 𝐵𝐶 = 6 𝑐𝑚
We can observe that
𝐴𝐵² = 108
𝐴𝐶² = 144 And, 𝐵𝐶² = 36
𝐴𝐵² + 𝐵𝐶² = 𝐴𝐶²
The given triangle, 𝛥𝐴𝐵𝐶, is satisfying Pythagoras theorem.
Therefore, the triangle is a right triangle, right-angled at B.
∴ ∠𝐵 = 90°
Hence, the correct option is (C).