Triangles - Study Mode
[#11] ABC is a triangle, PQ is line segment intersecting AB is P and AC in Q and PQ || BC. The ratio of AP : BP = 3 : 5 and length of PQ is 18 cm. The length of BC is:
Correct Answer
(B) 48 cm
Explanation
Solution: ∵ PQ || BC So ∠AQP = ∠ACB = α and ∠APQ = ∠ABC = β So, ΔABC and ΔAPQ $$eqalign{
& frac{{AP}}{{AB}} = frac{{PQ}}{{BC}} cr
& frac{3}{8} = frac{{PQ}}{{BC}} cr
& frac{3}{8} = frac{{18}}{{BC}} cr
& BC = 48,{ ext{cm}} cr} $$
[#12] Which of the following is a true statement
Correct Answer
(C) Two triangles are similar if their corresponding sides are proportional
Explanation
Solution: Two triangles are similar if their corresponding sides are proportional
[#13] ΔABC is similar to ΔDEF is area of ΔABC is 9 sq. cm. and area of ΔDEF is 16 sq. cm. and BC = 21 cm. Then the length of EF will be:
Correct Answer
(B) 2.8 cm
Explanation
Solution: ∵ ΔABC ≅ ΔDEF $$eqalign{
& herefore frac{{AB}}{{DE}} = frac{{BC}}{{EF}} = frac{{sqrt 9 }}{{sqrt {16} }} cr
& = frac{{2.1}}{{EF}} = frac{3}{4} cr
& EF = 2.8,{ ext{cm}} cr} $$
[#14] In a triangle ABC, AB + BC = 12 cm, BC + CA = 14 cm and CA + AB = 18 cm. Find the radius of the circle (in cm) which has the same perimeter as the triangle
Correct Answer
(B) $$frac{7}{2}$$
Explanation
Solution: According to question, AB + BC = 12 cm BC + CA = 14 cm CA + AB = 18 cm 2(AB + BC + CA) = 44 cm AB + BC + CA = $$frac{{44}}{2}$$ cm AB + BC + CA = 22 cm Perimeter of triangle = 22 cm Perimeter of triangle = Perimeter of circle 22 = 2πr 2 × $$frac{{22}}{7}$$ × r = 22 r = $$frac{7}{2}$$ cm
[#15] ABC is an isosceles triangle such that AB = AC and ∠B = 35°, AD is the median to the base BC. Then ∠BAD is
Correct Answer
(D) 55°
Explanation
Solution: According to question, AB = AC, xa0 xa0 xa0 xa0 ∠B = ∠C ∠A + ∠B + ∠C = 180° ∠A + 2∠B = 180° ∠A = 180° - 70° ∠A = 110° Note : In isosceles triangle median bisects the opposite side and make angle 90° on opposite side. It also bisects the vertex angle. ∠BAD = $$frac{{angle A}}{2}$$ ∠BAD = $$frac{{{{110}^ circ }}}{2}$$ ∠BAD = 55°