In ÃÂÃÂÃÂÃÂÃÂÃÂABC, AD ÃÂÃÂÃÂâÃÂÃÂÃÂÃÂÃÂÃÂ¥ BC and AD = BD ÃÂÃÂÃÂÃÂÃÂàDC. The measure of ÃÂÃÂÃÂâÃÂÃÂÃÂÃÂÃÂàBAC is :
2
(A) 75°
(B) 90°
(C) 45°
(D) 60°
Solution:
AD2 = BD.DC ΔADC ∼ ΔCAB (Property of a right angle Δ) ∠BAC = ∠ADC = 90°
92.
In a triangle ABC, ÃÂÃÂÃÂâÃÂÃÂÃÂÃÂÃÂàA = 90ÃÂÃÂÃÂÃÂÃÂÃÂÃÂð, ÃÂÃÂÃÂâÃÂÃÂÃÂÃÂÃÂàC = 55ÃÂÃÂÃÂÃÂÃÂÃÂÃÂð, $${AD}$$ ÃÂÃÂÃÂâÃÂÃÂÃÂÃÂÃÂÃÂ¥ $${BC}$$. What is the value of ÃÂÃÂÃÂâÃÂÃÂÃÂÃÂÃÂàBAD ?
(A) 35°
(B) 60°
(C) 45°
(D) 55°
Solution:
According to question, In right angle ΔBAC ∠A + ∠B + ∠C = 180° ∠B = 180° - 55° - 90° ∠B = 35° In right angle ΔADB ∠ADB + ∠ABD + ∠BAD = 180° ∠BAD = 180° - 35° - 90° ∠BAD = 55° Alternate ΔBAC ∼ ΔBDA ∴ ∠BCA = ∠BAD = 55°
93.
For a triangle ABC, D and E are two points on AB and AC such that AD = $$\frac{1}{4}$$ AB, AE = $$\frac{1}{4}$$ AC. If BC = 12 cm, then DE is :
(A) 5 cm
(B) 4 cm
(C) 3 cm
(D) 6 cm
Solution:
According to question, By using B.P.T $$\eqalign{ & \frac{{AD}}{{AB}} = \frac{{AE}}{{AC}} = \frac{{DE}}{{BC}} \cr & \frac{{AD}}{{AB}} = \frac{{DE}}{{BC}} \cr & \Rightarrow \,\frac{1}{4} = \frac{{DE}}{{12}} \cr & \Rightarrow DE = 3\,{\text{cm}} \cr} $$
94.
In a ÃÂÃÂÃÂÃÂÃÂÃÂABC, AB = AC and BA is produced to D such that AC = AD. Then the ÃÂÃÂÃÂâÃÂÃÂÃÂÃÂÃÂàBCD is :
Solution:
If triangle's side are a, b, c then must be:- a + b > c or a - b c only option (B) satisfy 3 + 4 > 5 7 > 5
96.
If the measure of the angles of a triangle are in the ratio 1 : 2 : 3 and if the length of the smallest side of the triangle is 10 cm, then the length of the longest side is:
(A) 20 cm
(B) 25 cm
(C) 30 cm
(D) 35 cm
Solution:
Let angle = x, 2x, 3x x + 2x + 3x = 180 (∵ Sum of internal angle of a Δ) 6x = 180° x = 30° So, angle = 30, 60, 90 Smallest side of Δ = 1 unit = 10 cm Largest side of Δ = 2 units = 20 cm
97.
ABC is a right angled triangled, right angled at C and P is the length of the perpendicular from C on AB. If a, b and c are the length of the sides BC, CA and AB respectively, then
Solution:
According to question, ACB is a right angle triangle ∴ area of ΔACB $$\frac{1}{2}$$ × AC × BC = $$\frac{1}{2}$$ × AB × PC $$\frac{1}{2}$$ × b × a = $$\frac{1}{2}$$ × c × p c = $$\frac{{ab}}{p}$$ . . . . . . . (i) By using pythagoras theorem AB2 = AC2 + BC2 c2 = b2 + a2 . . . . . . . . . . (ii) Put the value of C in equation (ii) $$\eqalign{ & {\left( {\frac{{ab}}{p}} \right)^2} = {a^2} + {b^2} \cr & \Rightarrow \frac{{{a^2}{b^2}}}{{{p^2}}} = {a^2} + {b^2} \cr & \Rightarrow \frac{1}{{{p^2}}} = \frac{{{a^2}}}{{{a^2}{b^2}}} + \frac{{{b^2}}}{{{a^2}{b^2}}} \cr & \Rightarrow \frac{1}{{{p^2}}} = \frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} \cr} $$ Alternate : From figure, $$\eqalign{ & P = \frac{{ab}}{c} \cr & P = \frac{{ab}}{{\sqrt {{a^2} + {b^2}} }}\,\left( {\because {a^2} + {b^2} = {c^2}} \right) \cr & {P^2} = \frac{{{a^2}{b^2}}}{{{a^2} + {b^2}}} \cr & \frac{1}{{{p^2}}} = \frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} \cr} $$
98.
In a ÃÂÃÂÃÂÃÂÃÂÃÂABC, If 2ÃÂÃÂÃÂâÃÂÃÂÃÂÃÂÃÂàA = 3ÃÂÃÂÃÂâÃÂÃÂÃÂÃÂÃÂàB = 6ÃÂÃÂÃÂâÃÂÃÂÃÂÃÂÃÂàC, then the value of ÃÂÃÂÃÂâÃÂÃÂÃÂÃÂÃÂàB is:
(A) 60°
(B) 30°
(C) 45°
(D) 90°
Solution:
According to question, Given : $$\eqalign{ & 2\angle A = 3\angle B \cr & \frac{{\angle A}}{{\angle B}} = \frac{3}{2}\,\,\,\,\,\,\,\,\,\,\,\,\,3\angle B = 6\angle C \cr & \frac{{\angle B}}{{\angle C}} = \frac{6}{3} = \frac{2}{1} \cr} $$ To make angle ∠B same ∴ ∠A : ∠B : ∠C 3 : 2 : 1 As we know that ∠A + ∠B + ∠C = 180° 3x + 2x + x = 180° x = 30° ∠B = 2x ∠B = 60°
99.
BL and CM are medians of ÃÂÃÂÃÂÃÂÃÂÃÂABC right-angled at A and BC = 5 cm. If BL = $$\frac{{3\sqrt 5 }}{2}$$ cm, then the length of CM is
(A) $$2\sqrt 5 $$ cm
(B) $$5\sqrt 2 $$ cm
(C) $$10\sqrt 2 $$ cm
(D) $$4\sqrt 5 $$ cm
Solution:
According to question, According to figure, when two medians intersect each other in a right angled triangle then we use this equation. ⇒ 4 (BL2 + CM2) = 5BC2 ⇒ 4 × $${\left( {\frac{{3\sqrt 5 }}{2}} \right)^2}$$ + 4CM2 = 5BC2 ⇒ 45 + 4CM2 = 125 ⇒ CM2 = $$\frac{{125 - 45}}{4}$$ ⇒ CM2 = 20 ⇒ CM = $$2\sqrt 5 $$ cm
100.
In a ÃÂÃÂÃÂÃÂÃÂÃÂABC, ÃÂÃÂÃÂâÃÂÃÂÃÂÃÂÃÂàA + ÃÂÃÂÃÂâÃÂÃÂÃÂÃÂÃÂàB = 75ÃÂÃÂÃÂÃÂÃÂÃÂÃÂð and ÃÂÃÂÃÂâÃÂÃÂÃÂÃÂÃÂàB + ÃÂÃÂÃÂâÃÂÃÂÃÂÃÂÃÂàC = 140ÃÂÃÂÃÂÃÂÃÂÃÂÃÂð, then ÃÂÃÂÃÂâÃÂÃÂÃÂÃÂÃÂàB is: