If △ABC is an isosceles triangles with ∠C = 90° and AC = 5 cm, then AB is:
If △ABC is an isosceles triangles with ∠C = = 90° and AC = 5 cm, then AB is: [A] [B] [C] [D] Show Answer AC = BC = 5 cm ∴ AB Hence option [A] is the right answer.
If △ABC is an isosceles triangles with ∠C = = 90° and AC = 5 cm, then AB is: [A] [B] [C] [D] Show Answer AC = BC = 5 cm ∴ AB Hence option [A] is the right answer.
For an equilateral triangle, the ratio of the in-radius and the ex-radius is: [A] [B] [C] [D] Show Answer In-radius Circum-radius ∴ Required ratio Hence option [D] is the right answer.
The side BC of a triangle ABC is extended to D. If ∠ACD = 120° and ∠ABC ∠CAB, then the value of ∠ABC is: [A]20° [B]80° [C]40° [D]60° Show Answer 40° ∠CAB = 2 ∠ABC ∠ACB + ∠ACD = 180° = ∠ACB + 120° = 180° = ∠ACB = 180° – 120° = 60° ∴ […]
If in a triangle ABC as drawn in the figure, AB = AC and ∠ACD = 120°, then ∠A is equal to: [A]70° [B]60° [C]50° [D]80° Show Answer 60° ∠ACB = 180° – 120° = 60° AB = AC ∴ ∠ABC = ∠ACB = 60° ∴ ∠BAC = 60° Hence option [B] is the right […]
If the incentre of an equilateral triangle lies inside the triangle and its radius is 3 cm, then the side of the equilateral triangle is: [A] [B] [C] [D] Show Answer In radius = Hence option [C] is the right answer.
If ABC is an equilateral triangle and P, Q, R respectively denote the middle points of AB, BC, CA then [A]PQR must be a right angled triangle [B]PQR must be an equilateral triangle [C]PQ + QR + PR = AB [D]PQ + QR + PR = 2AB Show Answer PQR must be an equilateral triangle […]
The radius of the incircle of the equilateral triangle having each side 6 cm is [A] [B] [C] [D] Show Answer E = In-center, AD ⊥ BC AB = 6 cm, BD = 3 cm ∠ADB = 90° ∴ AD = ∴ In-radius = Hence option [B] is the right answer.
G is the centroid in the equilateral △ ABC. If AB = 10 cm then length of AG is : [A] [B] [C] [D] Show Answer AB = 10cm BD = 5 cm ∠ADB = 90° ∴ AD = Hence option [C] is the right answer.
ABC is an equilateral triangle and CD is the internal bisector of ∠C. If DC is produced to E such that AC = CE, then ∠CAE is equal to : [A]15° [B]45° [C]30° [D]60° Show Answer 15° ∠BCD = ∠DCA = 30° ∠DCE = 180° ∴ ∠ACE = 180° – 30° = 150° AC = […]
The side QR of an equilateral triangle PQR is produced to the point S in such a way that QR = RS and P is joined to S. Then the measure of ∠PSR is : [A]45° [B]30° [C]15° [D]60° Show Answer 30° ∠PRQ = 60° ∠PRS = 180° – 60° = 120° =>∠PSR + ∠RPS […]