Geometry - Study Mode
[#66] AB is a diameter of a circle with centre O, CB is a tangent to the circle at B. AC intersects the circle at G. If the radius of the circle is 6 cm and AG = 8 cm, then the length of BC is:
Correct Answer
(D) 6√5 cm
Explanation
Solution: $$eqalign{
& Delta AGB sim Delta BGC cr
& frac{{AB}}{{BC}} = frac{{BG}}{{GC}} = frac{{AG}}{{BG}} cr
& frac{{12}}{{BC}} = frac{{4sqrt 5 }}{{GC}} = frac{8}{{4sqrt 5 }} cr
& GC = 10 cr
& frac{{12}}{{BC}} = frac{{4sqrt 5 }}{{10}} cr
& BC = 6sqrt 5 cr} $$
[#67] A circle is inscribed in a triangle ABC. It touches side AB, BC and AC at points R, P and Q, respectively. If AQ = 2.6 cm, PC = 2.7 cm and BR = 3 cm, then the perimeter (in cm) of the triangle ΔABC is:
Correct Answer
(C) 16.6
Explanation
Solution: Because tangent drawn from a single point are equal. So perimeter of ABC = (2.6 + 2.7 + 3) × 2 = 16.6
[#68] PQR is a triangle. S and T are the midpoints of the sides PQ and PR respectively. which of the following is TRUE? I. Triangle PST is similar to triangle PQR. II. ST = $$frac{1}{2}$$ (QR) III. ST is parallel to QR.
Correct Answer
(D) All I, II and III
Explanation
Solution: Triangle PST is similar to triangle PQR [By AAA] By mid point theorem. ST = $$frac{1}{2}$$(QR) ST is parallel to QR. All I, II and III are true.
[#69] B 1 is a point on the side AC of ΔABC and B 1 B is joined. A line is drawn through A parallel to B 1 B meeting BC at A 1 and another line is drawn through C parallel to B 1 B meeting AB produced at C 1 . Then
Correct Answer
(B) $$frac{1}{{{ ext{C}}{{ ext{C}}_1}}} + frac{1}{{{ ext{A}}{{ ext{A}}_1}}} = frac{1}{{{ ext{B}}{{ ext{B}}_1}}}$$
[#70] In a circle with centre O, ABCD is a cyclic quadrilateral and AC is the diameter. Chords AB and DC are produced to meet at E. If ∠CAE = 34° and ∠E = 30°, then ∠CBD is equal to:
Correct Answer
(C) 26°
Explanation
Solution: In, ΔBCE ∠B + ∠C + ∠E = 180° 90° + ∠C + 30° = 180° ∠C = 60° In, ΔABC ∠A + ∠B + ∠C = 180° 34° + 90° + ∠C = 180° ∠C = 56° ∠ACD = 64° = ∠ABD ∠CBD = 90° - 64° = 26°