Practice MCQ Questions and Answer on Triangles

If in ÃÂÃÂÃÂÃÂÃÂÃÂABC, DE || BC, AB = 7.5 cm BD = 6 cm and DE = 2 cm then the length of BC in cm is:

(A) 6 cm
(B) 8 cm
(C) 10 cm
(D) 10.5 cm

Solution: $\begin{aligned} △ A D E ≃ △ A B C \\ \frac{A D}{A B} = \frac{D E}{B C} \\ \frac{1.5}{7.5} = \frac{2}{B C} \\ B C = 10 cm \end{aligned}$

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 ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ PSR is

(A) 30°
(B) 15°
(C) 60°
(D) 45°

Solution: According to question, Given : PQR is an equilateral triangle QR = RS PR = RS ∠SRP = 180° - 60° (Exterior ∠) ∠SRP = 120° ∴ ∠RPS = ∠RSP ∴ ∠RPS + ∠PRS + ∠RSP = 180° 2∠PSR = 180° - 120° ∠PSR = $\frac{60^{\circ}}{2}$ ∠PSR = 30°

If ÃÂÃÂÃÂÃÂÃÂÃÂPQR and ÃÂÃÂÃÂÃÂÃÂÃÂLMN are similar and 3PQ = LM and MN = 9 cm, then QR is equal to:

(A) 12 cm
(B) 6 cm
(C) 9 cm
(D) 3 cm

Solution: $\begin{aligned} \frac{P Q}{L M} = \frac{Q R}{M N} = \frac{P R}{L N} \\ (△ P Q R and △ L M N are similar) \\ \frac{P Q}{L M} = \frac{Q R}{M N} \\ \frac{1}{3} = \frac{Q R}{9} \\ Q R = 3 cm \end{aligned}$

ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ A + $\frac{1}{2}$ ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ B + ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ C = 140ÃÂÃÂÃÂÃÂÃÂÃÂÃÂÃÂ°, then ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ B 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°

If in a triangle, the orthocentre lies on vertex, then the triangle is

(A) Acute angled
(B) Isosceles
(C) Right angled
(D) Equilateral

If in a triangle ABC, BE and CF are two medians perpendicular to each other and if AB = 19 cm and AC = 22 cm then the length of BC is :

(A) 20.5 cm
(B) 19.5 cm
(C) 26 cm
(D) 13 cm

The circumcentre of a triangle ABC is O. If ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ BAC = 85ÃÂÃÂÃÂÃÂÃÂÃÂÃÂÃÂ° and ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ BCA = 75ÃÂÃÂÃÂÃÂÃÂÃÂÃÂÃÂ°, then the value of ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ OAC is

(A) 40°
(B) 60°
(C) 70°
(D) 90°

Angle between the internal bisectors of two angles of a triangle ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ B and ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ C is 120ÃÂÃÂÃÂÃÂÃÂÃÂÃÂÃÂ°, then ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ A is :

(A) 20°
(B) 30°
(C) 60°
(D) 90°

Solution: According to question, Given : ∠BIC = 120° ∠BIC = 90° + $\frac{1}{2}$ ∠A $\frac{∠ A}{2}$ = (120° - 90°) $\frac{∠ A}{2}$ = 30° ∠A = 60°

In triangle ABC, ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ BAC = 75ÃÂÃÂÃÂÃÂÃÂÃÂÃÂÃÂ°, ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ ABC = 45ÃÂÃÂÃÂÃÂÃÂÃÂÃÂÃÂ°, $\overset{―}{B C}$ is produced to D. If ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ ACD = xÃÂÃÂÃÂÃÂÃÂÃÂÃÂÃÂ°, then $\frac{x}{3}$ % of 60ÃÂÃÂÃÂÃÂÃÂÃÂÃÂÃÂ° is

(A) 30°
(B) 48°
(C) 15°
(D) 24°

Solution: According to question, Given : ∠A = 75°, ∠B = 45° ∴ ∠ACD = ∠A + ∠B x° = ∠ACD = 120° Now, $\frac{x}{3}$ % of 60° is = $\frac{120}{3}$ % of 60° = 40% of 60° = $\frac{40}{100}$ × 60° = 24°

10.

In a right-angle ÃÂÃÂÃÂÃÂÃÂÃÂABC, ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ ABC = 90ÃÂÃÂÃÂÃÂÃÂÃÂÃÂÃÂ°, AB = 5 cm and BC = 12 cm. The radius of the circumcircle of the triangle ABC is

(A) 7.5 cm
(B) 6 cm
(C) 6.5 cm
(D) 7 cm

Solution: According to question, ABC is a right angled triangle ∴ By using Pythagoras theorem AC2 = AB2 + BC2 AC2 = (5)2 + (12)2 AC2 = 25 + 144 AC2 = 169 AC = $\sqrt{169}$ AC = 13 cm BD = IR = Circumradius = $\frac{A C}{2}$ ∴ IR = $\frac{13}{2}$ IR = 6.5 cm

«
1
2
3
4
5
6
7
8
9
10
...
13
14
Next ›
Last »

Indian Polity Mcq IAS GK Question India GK World GK GK questions GK Quiz Biology GK Science GK Physics GK Chemistry GK General Awareness Computer GK Political GK Indian Political GK Economics GK History GK Sports GK Geography GK Teaching Aptitude GK CTET/TET GK B Ed Entrance GK Banking GK SSC GK UPSC GK Indian Army GK Hindi Grammar GK Hindi Grammar MCQ Reasoning questions Bihar GK Rajasthan GK Jharkhand GK MP GK UP GK Chhattisgarh GK Haryana GK HP GK Maharashtra GK Electronics GK Agriculture GK States/Capitals GK Books Name & Author Arthimetic Verbal Reasoning Non Verbal Reasoning English

Practice MCQ Questions and Answer on Triangles

General Knowledge