Triangles - Study Mode

[#106] If in a triangle ABC, D and E are on the sides AB and AC, such that, DE is parallel to BC and $$frac{{AD}}{{BD}}$$ = $$frac{3}{5}$$. If AC = 4 cm, then AE is
Correct Answer

(A) 1.5 cm

Explanation

Solution: According to question, Given : AD = 3 BD = 5 AB = 8 AC = 4 AE = ? By applying B. P. T $$eqalign{
& frac{{AD}}{{AB}} = frac{{AE}}{{AC}} = frac{{DE}}{{BC}} cr
& frac{3}{8} = frac{{AE}}{4} cr
& AE = frac{3}{2} = 1.5,{ ext{cm}} cr} $$

[#107] In a ΔABC, ∠A + ∠B = 118°, ∠A + ∠C = 96°. Find the value of ∠A.
Correct Answer

(D) 34°

Explanation

Solution: According to question, ∠A + ∠B = 118° ∠A + ∠C = 96° ∠A = ? As we know that ∠A + ∠B + ∠C = 180° ∠C = 180° - (∠A + ∠B) ∠C = 180° - 118° ∴ ∠C = 62° ∠A = 96° - 62° ∠A = 34°

[#108] 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 :
Correct Answer

(C) 3 cm

Explanation

Solution: According to question, By using B.P.T $$eqalign{
& frac{{AD}}{{AB}} = frac{{AE}}{{AC}} = frac{{DE}}{{BC}} cr
& frac{{AD}}{{AB}} = frac{{DE}}{{BC}} cr
& Rightarrow ,frac{1}{4} = frac{{DE}}{{12}} cr
& Rightarrow DE = 3,{ ext{cm}} cr} $$

[#109] In triangle ABC a straight line parallel to BC intersects AB and AC at D and E respectively. If AB = 2AD, then DE : BC is
Correct Answer

(C) 1 : 2

Explanation

Solution: According to question, Given : AB = 2AD $$frac{{AB}}{{AD}} = frac{2}{1}$$ By applying B. P. T $$eqalign{
& frac{{AD}}{{AB}} = frac{{DE}}{{BC}} = frac{{AE}}{{AC}} cr
& frac{{DE}}{{BC}} = frac{1}{2} cr
& herefore DE:BC = 1:2 cr} $$

[#110] ABC is a triangle and the sides AB, BC and CA are produced to E, F and G respectively. If ∠CBE = ∠ACF = 130°, then the value of ∠GAB is :
Correct Answer

(A) 100°

Explanation

Solution: We know that ⇒ Add of total exterior angle of a triangle (polygon) = 360° ⇒ So, 130° + 130° + x° = 360° x° = 100°