In ÃÂÃÂÃÂÃÂÃÂÃÂABC, ÃÂÃÂÃÂâÃÂÃÂÃÂÃÂÃÂàBAC = 90ÃÂÃÂÃÂÃÂÃÂÃÂÃÂð and AD ÃÂÃÂÃÂâÃÂÃÂÃÂÃÂÃÂÃÂ¥ BC. If BD = 3 cm and CD = 4 cm, then length of AD is :
(A) $$2\sqrt 3 $$ cm
(B) 3.5 cm
(C) 6 cm
(D) 5 cm
Solution:
We know that, AD2 = CD × BD AD2 = 4 × 3 AD2 = 12 AD = $$2\sqrt 3 $$ cm
2.
If two angles of a triangle are 21ÃÂÃÂÃÂð and 38ÃÂÃÂÃÂð, then the triangle is :
(A) Right-angled triangle
(B) Acute-angled triangle
(C) Obtuse-angled triangle
(D) Isosceles triangle
Solution:
According to question, Given : ∠A = 21°, ∠C = 38° As we know that ∠A + ∠B + ∠C ∠B = 180° - 21° - 38° ∠B = 121° ∴ The triangle is obtuse-angled triangle.
3.
In ÃÂÃÂÃÂÃÂÃÂÃÂPQR, straight line parallel to the base QR cuts PQ at X and PR at Y. If PX : XQ = 5 : 6, then XY : QR will be
If ABC is an equilateral triangle and P, Q, R respectively denote the middle points of AB, BC, CA then
(A) PQR must be an equilateral triangle
(B) PQ + QR = PQR + AB
(C) PQ + QR = PR + 2AB
(D) PQR must be a right angled
Solution:
According to question, Given : P, Q and R are the mid points of AB, BC and AC PQ || AC and PQ = $$\frac{1}{2}$$ AC PR || BC and PR = $$\frac{1}{2}$$ BC RQ || AB and RQ = $$\frac{1}{2}$$ AB (mid point theorem) ∴ ΔPQR is an equilateral triangle
5.
In ÃÂÃÂÃÂÃÂÃÂÃÂABC, ÃÂÃÂÃÂâÃÂÃÂÃÂÃÂÃÂàA + ÃÂÃÂÃÂâÃÂÃÂÃÂÃÂÃÂàB = 65ÃÂÃÂÃÂÃÂÃÂÃÂÃÂð, ÃÂÃÂÃÂâÃÂÃÂÃÂÃÂÃÂàB + ÃÂÃÂÃÂâÃÂÃÂÃÂÃÂÃÂàC = 140ÃÂÃÂÃÂÃÂÃÂÃÂÃÂð, then find ÃÂÃÂÃÂâÃÂÃÂÃÂÃÂÃÂàB.
(A) 40°
(B) 25°
(C) 35°
(D) 20°
Solution:
According to question, Given : ∠A + ∠B = 65° ∠B + ∠C = 140° We know that ∠A + ∠B + ∠C = 180° ∠C = 180° - (∠A + ∠B) ∠C = 180° - 65° ∠C = 115° ∠B = 140° - 115° ∠B = 25°
6.
If the three angles of a triangle are: $${\left(x + 15 \right)^ \circ },$$ ÃÂÃÂ $${\left({\frac{{6x}}{5} + 6} \right)^ \circ }$$ ÃÂÃÂ and $${\left({\frac{{2x}}{3} + 30} \right)^ \circ }$$ ÃÂÃÂ then the triangle is:
ÃÂÃÂÃÂâÃÂÃÂÃÂÃÂÃÂàA + $$\frac{1}{2}$$ ÃÂÃÂÃÂâÃÂÃÂÃÂÃÂÃÂàB + ÃÂÃÂÃÂâÃÂÃÂÃÂÃÂÃÂàC = 140ÃÂÃÂÃÂÃÂÃÂÃÂÃÂð, then ÃÂÃÂÃÂâÃÂÃÂÃÂÃÂÃÂàB is :
(A) 50°
(B) 80°
(C) 40°
(D) 60°
Solution:
According to question, ∠A + $$\frac{1}{2}$$ ∠B + ∠C = 140° . . . . . . . . (i) As we know that ∠A + ∠B + ∠C = 180° . . . . . . . . . (ii) Compare equation (i) and (ii) $$\frac{1}{2}$$ ∠B = 40° ∴ ∠B = 80°
8.
If the measures of the sides of triangle are (x2 - 1), (x2 + 1) and 2x cm, then the triangle would be :
(A) Equilateral
(B) Acute-angled
(C) Right-Angled
(D) Isosceles
Solution:
According to question, Sides AB = x2 - 1 BC = 2x AC = x2 + 1 By using Pythagoras theorem AC2 = AB2 + BC2 (x2 + 1)2 = (x2 - 1)2 + (2x)2 x4 + 1 + 2x2 = x2 + 1 - 2x2 + 4x2 (x2 + 1)2 = (x2 + 1)2 ∴ The triangle is right angle Δ
9.
O is the incentre of ÃÂÃÂÃÂÃÂÃÂÃÂABC and ÃÂÃÂÃÂâÃÂÃÂÃÂÃÂÃÂàA = 30ÃÂÃÂÃÂÃÂÃÂÃÂÃÂð, then ÃÂÃÂÃÂâÃÂÃÂÃÂÃÂÃÂàBOC is
In ÃÂÃÂÃÂÃÂÃÂÃÂABC, ÃÂÃÂÃÂâÃÂÃÂÃÂÃÂÃÂàBAC = 90ÃÂÃÂÃÂÃÂÃÂÃÂÃÂð and AB = $$\frac{1}{2}$$ BC, Then the measure of ÃÂÃÂÃÂâÃÂÃÂÃÂÃÂÃÂàACB is :
(A) 60°
(B) 30°
(C) 45°
(D) 15°
Solution:
According to question, Given : BAC is right angle triangle AB = $$\frac{1}{2}$$BC $$\eqalign{ & \frac{{AB}}{{BC}} = \frac{1}{2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{P}{H} = \frac{1}{2} \cr & \sin \theta = \frac{P}{H} = \frac{1}{2} \cr & \sin \,{30^ \circ } = \frac{1}{2} \cr & \therefore \theta = \angle ACB = {30^ \circ } \cr} $$