Contents
GSEB Solutions for Class 10 mathematics – Similarity and The Theorem of Pythagoras (English Medium)
Exercise-7.1
Question 1:
∠B is a right angle in ∆ABC. \(\overline{BD}\bot \overline{AC}\) and D ∈ \(\overline{AC}\). If AD = 4DC, prove that BD = 2DC.
Solution :
Question 2:
5, 12, 13 are the lengths of the sides of a triangle. Show that the triangle is right angled. Find the length of altitude on the hypotenuse.
Solution :
Question 3:
In ∆PQR, \(\overline{QM}\) is the altitude to hypotenuse \(\overline{PR}\). If PM = 8, RM = 12, find PQ, QR and QM.
Solution :
Question 4:
In ∆ABC, m∠B = 90, \(\overline{BD}\bot \overline{AC}\), M ∈ \(\overline{AC}\). If AM – MC = 7, AB2 – BC2 = 175, find AC.
Solution :
Question 5:
∠A is right angle in ∆ABC. \(\overline{AD}\) is an altitude of the triangle. If AB = \(\sqrt{5}\), BD = 2, find the length of the hypotenuse of the triangle.
Solution :
Question 6:
m∠B = 90 in ∆ABC. \(\overline{BM}\) is altitude to \(\overline{AC}\).
- If AM = BM = 8, find AC.
- If BM = 15, AC = 34, find AB.
- If BM = 2\(\sqrt{30}\), MC = 6, find AC.
- If AB = \(\sqrt{10}\), AM = 2.5, find MC.
Question 6(1):
Solution :
Question 6(2):
Solution :
Question 6(3):
Solution :
Question 6(4):
Solution :
Question 7:
In APQR m∠Q = 90, PQ = x, QR = y and \(\overline{QD}\bot \overline{PR}\). D ∈ \(\overline{PR}\). Find PD, QD, RD in terms Of x and y.
Solution :
Question 8:
∠Q is a right angle in ∆PQR and \(\overline{QM}\bot \overline{PR}\), M ∈ \(\overline{PR}\). If PQ = 4QR, then prove that PM = 16RM.
Solution :
Question 9:
\(\square \) PQRS is a rectangle. If PQ + QR = 7 and PR + QS = 10, then find the area of \(\square \) PQRS.
Solution :
Question 10:
The diagonals of a convex \(\square \) ABCD intersect at right angles. Prove that AB2 + CD2 = AD2 + BC2.
Solution :
Given: The diagonals of a convex ⃞ABCD intersect at right angles at O.
To prove : AB2 + CD2 = AD2 + BC2
Proof : In ∆AOD, m∠O = 90°
∴ AD2 = OA2 + OD2 … … (1)
In ∆BOC, m∠O = 90°
∴ BC2 = OB2 + OC2 … … (2)
In ∆AOB, m∠O = 90°
∴ AB2 = OA2 + OB2 … … (3)
In ∆COD, m∠O = 90°
∴CD2 = OC2 + OD2 … … (4)
Adding results (3) and (4),
AB2 + CD2 = OA2 + OB2 + OC2 + OD2
∴AB2 + CD2 = (OA2 + OD2) + (OB2 + OC2)
∴AB2 + CD2 = AD2 + BC2 [from (1) and (2)]
Question 11:
In ∆PQR, m∠Q = 90, M ∈ \(\overline{QR}\) and N ∈ \(\overline{PQ}\). Prove that PM2 + RN2 = PR2 + MN2.
Solution :
Question 12:
The sides Of a triangle have lengths a2 + b2, 2ab, a2 – b2, where a > b and a, b ∈ R+. Prove that the angle opposite to the side having length a2 + b2 is a right angle.
Solution :
Question 13:
In ∆ABC, m∠B = 90 and \(\overline{BE}\) is a median. Prove that AB2 + BC2 + AC2 = 8BE2.
Solution :
Question 14:
AB = AC and ∠A is right angle in ∆ABC. If BC = \(\sqrt{2}\)a, then find the area of the triangle. (a ∈ R, a > 0)
Solution :
Exercise-7.2
Question 1:
In rectangle ∆BCD, AB + BC = 23, AC + BD = 34. Find the area of the rectangle.
Solution :
□ ABCD is a rectangle.
∴ AC = BD and AC + BD = 34
∴ 2AC = 34∴ AC = 17
Also, AB + BC = 23
Let, AB = x
∴ BC = 23 – x
In ∆ABC, m∠B = 90°
∴ AB2 + BC2 = AC2
∴ x2 + (23 – x)2 = 172
∴ x2 + 529 – 46x + x2 = 289
∴ 2x2 – 46x + 240 = 0
∴ x2 – 23x + 120 = 0
∴ (x – 15)(x – 8) = 0
∴ x = 15 or x = 8
∴ AB = 15 or AB = 8
If AB = 15, then BC = 23 – 15 = 8
If AB = 8,then BC = 23 – 8 = 15
Hence, under both conditions, the lengths of two adjacent sides of rectangle ABCD are 15 and 8.
Area of rectangle ABCD
= Product of two adjacent sides
= 15 × 8
= 120
∴ The area of rectangle ABCD is 120.
Question 2:
In ∆ABC m∠A = m∠B + m∠C, AB = 7, BC = 25. Find the perimeter of ∆ABC.
Solution :
In ∆ABC, m∠A = m∠B + m∠C and
m∠A + m∠B + m∠C = 180°
∴ m∠A + m∠A = 180°
∴ 2m∠A = 180°
∴ m∠A = 90°
In ∆ABC, m∠A = 90°
∴ AB2 + AC2 = BC2
∴ (7)2 + AC2 = (25)2
∴ AC2 = 625 – 49 = 576 = (24)2
∴ AC= 24
Perimeter of ∆ABC = AB + BC + AC
= 7 + 25 + 24 = 56
∴ The perimeter of ∆ABC is 56.
Question 3:
A staircase of length 6.5 meters touches a wall at height of 6 meter. Find the distance of base of the staircase from the wall.
Solution :
Question 4:
In ∆ABC AB = 7, AC = 5, AD = 5. Find BC, if the mid-point of \(\overline{BD}\) is D.
Solution :
Question 5:
In equilateral ∆ABC, D ∈ \(\overline{BC}\) such that BD : DC = 1 : 2. Prove that 3AD = \(\sqrt{7}\)AB.
Solution :
Question 6:
In ∆ABC, AB = 17, BC = 15, AC = 8, find the length of the median on the largest side.
Solution :
Question 7:
\(\overline{AD}\) is a median of ∆ABC. AB2 + AC2 = 148 and AD = 7. Find BC.
Solution :
Question 8:
In rectangle ABCD, AC = 25 and CD = 7. Find perimeter of the rectangle.
Solution :
□ABCD is a rectangle. ∴ m∠D = 90°
In ∆ADC, m∠D = 90°
∴ AC2 = AD2 + CD2
∴ 252 = AD2 + 72
∴ 625 = AD2 + 49
∴ AD2 = 576 = 242
∴ AD = 24
Perimeter of rectangle ABCD = 2(AD + CD)
= 2 (24 + 7)
= 2 × 31 = 62
∴ The perimeter of rectangle ABCD is 62.
Question 9:
In rhombus XYZW, XZ = 14 and YW = 48. Find XY.
Solution :
Question 10:
In ∆PQR, m∠Q : m∠R : m∠P = 1 : 2 : 1 If PQ = 2\(\sqrt{6}\), find PR.
Solution :
Exercise-7
Question 1:
\(\overline{AD}\), \(\overline{BE}\), \(\overline{CF}\) are the medians of ∆ ABC. If BE = 12, CF = 9 and AB2 + BC2 + AC2 = 600, BC = 10, find AD.
Solution :
Question 2:
\(\overline{AD}\) is the altitude of ∆ABC such that B – D – C. If AD2 = BD ∙ DC, prove that ∠BAC is right angle. [Hint : \(\overline{AD}\bot \overline{BC}\). So, B – D – C is given. So, ∆ADB and ∆ADC are right angled triangles to which Pythagoras’ theorem can be applied. Same method can be applied to solve Ex. 3, 4, 5.]
Solution :
Question 3:
In ∆ ABC, \(\overline{AD}\bot \overline{BC}\), B – D – C. If AB2 = BD ∙ BC, prove that ∠BAC is a right angle.
Solution :
Question 4:
In ∆ABC, \(\overline{AD}\bot \overline{BC}\), B – D – C. If AC2 = CD ∙ BC, prove that ∠BAC is a right angle.
Solution :
Question 5:
\(\overline{AD}\) is a median of ∆ABC. If BD = AD, prove that ∠A is a right angle in ∆ ABC.
Solution :
Question 6:
In figure 7.25, AC is the length of a pole standing vertical on the ground. The pole is bent at point B, so that the top of the pole touches the ground at a point 15 meters away from the base of the pole. If the length of the pole is 25, find the length of the upper part of the pole.
Solution :
When the pole is bent at point B, its top C touches the ground at point C’.
∴ BC = BC’
So, AC’ = 15 m and AC = 25 m.
AC = AB + BC
∴ 25 = AB + BC’
Let, AB = x m
∴ 25 = x + BC’
∴ BC’ = (25 – x) m
In ∆ABC, m∠A = 90°
∴ BC2 = AB2 + AC2
∴ (25 – x)2 = x2 + 152
∴ 625 – 50x + x2 = x2 +225
∴ 625 – 225 = 50x
∴ 50x = 400
∴ x = 8
∴ AB = 8 m
Length of the upper part of the pole = BC = BC’ = 25-x = 25 – 8 = 17 m
Question 7:
In ∆ABC, AB > AC, D is the mid-point of \(\overline{BC}\). \(\overline{AM}\bot \overleftrightarrow{BC}\) such that B – M – C. Prove that AB2 – AC2 = 2BC ∙ DM.
Solution :
Question 8:
In ∆ABC, \(\overline{BD}\bot \overline{AC}\), D ∈ \(\overline{AC}\) and ∠B is right angle. If AC = SCD, prove that BD = 2CD.
Solution :
Question 9:
Select a proper option (a), (b), (c) or (d) from given options :
Question 9(1):
In ∆PQR, if m∠P + m∠Q = m∠R. PR = 7, QR = 24, then PQ = …….
Solution :
b. 25
In ∆PQR,
m∠P + m∠Q = m∠R
∴m∠R = 90°
∴PQ2 = PR2 + QR2
= 72 + 24
= 49 + 576
= 625
= 252
∴PQ = 25
Question 9(2):
In ∆ABC, \(\overline{AD}\) is an altitude and ∠A is right angle. If AB = \(\sqrt{20}\) BD = 4, CD = ………
Solution :
Question 9(3):
In ∆ABC, AB2 + AC2 = 50. The length of the median AD = 3. So, BC = ……..
Solution :
Question 9(4):
In ∆ABC, m∠B = 90, AB = BC. Then AB : AC = ………
Solution :
Question 9(5):
In ∆ABC, m∠B = 90, AC = 10. The length of the median BM = …….
Solution :
Question 9(6):
In ∆ABC, AB = BC = \(\frac{AC}{\sqrt{2}}\). m∠B ………
Solution :
Question 9(7):
In ∆ABC, if \(\frac{AB}{1}=\frac{AC}{2}=\frac{BC}{\sqrt{2}}\), then m∠B = ……….
Solution :
Question 9(8):
In ∆XYZ, m∠X : m∠Y : m∠Z = 1 : 2 : 3. If XY = 15, YZ = ……
Solution :
Question 9(9):
In ∆ABC, ∠B is a right angle and \(\overline{BD}\) is an altitude. If AD = BD = 5, then DC = …….
Solution :
Question 9(10):
In ∆ABC, \(\overline{AD}\) is median. If AB2 + AC2 = 130 and AD = 7, then BD = ……..
Solution :
Question 9(11):
The diagonal of a square is 5\(\sqrt{2}\). The length of the side of the square is …….
Solution :
Question 9(12):
The length of a diagonal of a rectangle is 13. If one of the side of the rectangle is 5, the perimeter of the rectangle is ……….
Solution :
b. 34
In any rectangle,
(Diagonal)2 = (Length)2 + (Breadth)2
∴ 132 = (Length)2 + 52
∴ 169 – 25 = (Length)2
∴ (Length)2 = 144
∴ (Length)2 = 12
Perimeter of rectangle = 2(Length + Breadth)
= 2(12 + 5) = 2(17) = 34
Question 9(13):
The length of a median of an equilateral triangle is \(\sqrt{3}\). Length of the side of the triangle is .
Solution :
Question 9(14):
The perimeter of an equilateral triangle is 6. The length of the altitude of the triangle is ………
Solution :
Question 9(15):
In ∆ABC, m∠A = 90. \(\overline{AD}\) is a median. If AD = 6, AB = 10, then AC = ……..
Solution :
Question 9(16):
In ∆PQR, m∠Q = 90 and PQ = QR. \(\overline{QM}\bot \overline{PR}\), M ∈ \(\overline{PR}\). If QM = 2, PQ = ………
Solution :
Question 9(17):
In ∆ABC, m∠A = 90, \(\overline{AD}\) is an altitude. So AB2 =
Solution :
Question 9(18):
In ∆ABC, m∠A = 90, \(\overline{AD}\) is an altitude. Therefore BD ∙ DC = …….
Solution :