Solution: (i) The point O where the angle bisectors meet is called the incenter of the triangle. (ii) The perpendicular drawn from point O to AB and CA are equal. i.e., OR and OQ. (iii) ∠ACO = ∠BCO....
Using ruler and compasses only, draw an equilateral triangle of side 5 cm and draw its inscribed circle. Measure the radius of the circle.
Solution: Steps to construct: Step 1: Draw a line segment BC = 5cm. Step 2: With Center as B and radius 5cm, with center as C and radius 5cm draw two arcs which intersect each other at point A. Step...
Construct a tangent to a circle of radius 4cm from a point on the concentric circle of radius 6 cm and measure its length. Also, verify the measurement by actual calculation.
Solution: Steps to construct: Step 1: Mark a point O. Step 2: With center O and radius 4cm and 6cm, draw two concentric circles. Step 3: Join OA and mark its mid-point as M. Step 4: With center M...
a) In the figure (i) given below, O is the centre of the circle. Prove that ∠AOC = 2 (∠ACB + ∠BAC). (b) In the figure (ii) given below, O is the centre of the circle. Prove that x + y = z
Solution : (a) Given: O is the center of the circle. To Prove : ∠AOC = 2 (∠ACB + ∠BAC). Proof: In ∆ABC, ∠ACB + ∠BAC + ∠ABC = 180° (Angles of a triangle) ∠ABC = 180o – (∠ACB + ∠BAC)….(i) In the...
(a) In the figure (i) given below, triangle ABC is equilateral. Find ∠BDC and ∠BEC. (b) In the figure (ii) given below, AB is a diameter of a circle with center O. OD is perpendicular to AB and C is a point on the arc DB. Find ∠BAD and ∠ACD
Solution: (a) triangle ABC is an equilateral triangle Each angle = 60o ∠A = 60o But ∠A = ∠D (Angles in the same segment) ∠D = 600 Now ABEC is a cyclic quadrilateral, ∠A = ∠E = 180o 60o + ∠E = 180o...
Three circles of radii 2 cm, 3 cm and 4 cm touch each other externally. Find the perimeter of the triangle obtained on joining the centers of these circles.
Solution: Three circles with centers A, B and C touch each other externally at P, Q and R respectively and the radii of these circles are 2 cm, 3 cm and 4 cm. By joining the centers of triangle ABC...
(a) If a, b, c are the sides of a right triangle where c is the hypotenuse, prove that the radius r of the circle which touches the sides of the triangle is given by r = /frac (a + b – c) – (2) (b) In the given figure, PB is a tangent to a circle with center O at B. AB is a chord of length 24 cm at a distance of 5 cm from the center. If the length of the tangent is 20 cm, find the length of OP.
Solution: (a) Let the circle touch the sides BC, CA and AB of the right triangle ABC at points D, E and F respectively, where BC = a, CA = b and AB = c (as showing in the given figure). As the...
(a) In the figure (i) given below, O is the center of the circle and AB is a tangent at B. If AB = 15 cm and AC = 7.5 cm, find the radius of the circle. (b) In the figure (ii) given below, from an external point P, tangents PA and PB are drawn to a circle. CE is a tangent to the circle at D. If AP = 15 cm, find the perimeter of the triangle PEC.
Solution: (i) Join OB ∠OBA = 90° (Radius through the point of contact is perpendicular to the tangent) OB2 = OA2 – AB2 r2 = (r + 7.5)2 – 152 r2 = r2 + 56.25 + 15r – 225 15r = 168.75 r = 11.25 Hence,...
(a) In figure (i) given below, quadrilateral ABCD is circumscribed; find the perimeter of quadrilateral ABCD. (b) In figure (ii) given below, quadrilateral ABCD is circumscribed and AD ⊥ DC ; find x if radius of incircle is 10 cm.
Solution: (a) From A, AP and AS are the tangents to the circle ∴AS = AP = 6 From B, BP and BQ are the tangents ∴BQ = BP = 5 From C, CQ and CR are the tangents CR = CQ From D, DS and DR are the...
Two circles of radii 5 cm and 2-8 cm touch each other. Find the distance between their centers if they touch : (i) externally (ii) internally.
Solution: Radii of the circles are 5 cm and 2.8 cm. i.e. OP = 5 cm and CP = 2.8 cm. (i) When the circles touch externally, then the distance between their centers = OC = 5 + 2.8 = 7.8 cm. (ii) When...
A point P is at a distance 13 cm from the center C of a circle and PT is a tangent to the given circle. If PT = 12 cm, find the radius of the circle.
Solution: CT is the radius CP = 13 cm and tangent PT = 12 cm CT is the radius and TP is the tangent CT is perpendicular TP Now in right angled triangle CPT, CP2 = CT2 + PT2 [using Pythagoras axiom]...
(a) In the figure (i) given below, AB is a diameter of the circle. If ∠ADC = 120°, find ∠CAB. (b) In the figure (ii) given below, sides AB and DC of a cyclic quadrilateral ABCD are produced to meet at E, the sides AD and BC are produced to meet at F. If x : y : z = 3 : 4 : 5, find the values of x, y and z.
Solution: (a) Construction: Join BC, and AC then ABCD is a cyclic quadrilateral. Now in ∆DCF Ext. ∠2 = x + z and in ∆CBE Ext. ∠1 = x + y Adding (i) and (ii) x + y + x + z = ∠1 + ∠2 2 x + y + z =...
(a) In the figure given below, PQ is a diameter. Chord SR is parallel to PQ.Given ∠PQR = 58°, calculate (i) ∠RPQ (ii) ∠STP (T is a point on the minor arc SP)
(b) In the figure given below, if ∠ACE = 43° and ∠CAF = 62°, find the values of a, b and c (2007) Solution: (a) In ∆PQR, ∠PRQ = 90° (Angle in a semi-circle) and ∠PQR = 58° ∠RPQ = 90° – ∠PQR = 90° –...
(a) In the figure (i) given below, ABCD is a parallelogram. A circle passes through A and D and cuts AB at E and DC at F. Given that ∠BEF = 80°, find ∠ABC. (b) In the figure (ii) given below, ABCD is a cyclic trapezium in which AD is parallel to BC and ∠B = 70°, find: (i)∠BAD (ii) DBCD.
Solution: (a) ADFE is a cyclic quadrilateral Ext. ∠FEB = ∠ADF ⇒ ∠ADF = 80° ABCD is a parallelogram ∠B = ∠D = ∠ADF = 80° or ∠ABC = 80° (b)In trapezium ABCD, AD || BC (i) ∠B + ∠A = 180° ⇒ 70° + ∠A =...
In a ∆ABC, D and E are points on the sides AB and AC respectively such that AD = 5.7cm, BD = 9.5cm, AE = 3.3cm and AC = 8.8cm. Is DE || BC? Justify your answer.
Solution:- From the question it is given that, In a ∆ABC, D and E are points on the sides AB and AC respectively. AD = 5.7cm, BD = 9.5cm, AE = 3.3cm and AC = 8.8cm Consider the ∆ABC, EC = AC – AE =...
In the adjoining figure, AB = AC. If PM ⊥ AB and PN ⊥ AC, show that PM x PC = PN x PB.
Solution:- From the given figure, AB = AC. If PM ⊥ AB and PN ⊥ AC We have to show that, PM x PC = PN x PB Consider the ∆ABC, AB = AC … [given] ∠B = ∠C Then, consider ∆CPN and ∆BPM ∠N = ∠M … [both...
Given that ∆s ABC and PQR are similar. Find: (i) The ratio of the area of ∆ABC to the area of ∆PQR if their corresponding sides are in the ratio 1 : 3. (ii) the ratio of their corresponding sides if area of ∆ABC : area of ∆PQR = 25 : 36.
Solution:- From the question it is given that, (i) The area of ∆ABC to the area of ∆PQR if their corresponding sides are in the ratio 1 : 3 Then, ∆ABC ~ ∆PQR area of ∆ABC/area of ∆PQR = BC2/QR2 So,...
In the figure (ii) given below, PQ || AC, AP = 4 cm, PB = 6 cm and BC = 8 cm. Find CQ and BQ.
Solution:- From the figure, PQ || AC, AP = 4 cm, PB = 6 cm and BC = 8 cm ∠BQP = ∠BCA … [because alternate angles are equal] Also, ∠B = ∠B … [common for both the triangles] Therefore, ∆ABC ~ ∆BPQ...
In the given figure, CM and RN are respectively the medians of ∆ABC and ∆PQR. If ∆ABC ~ ∆PQR, prove that:
(i) ∆AMC ~ ∆PQR
(ii) CM/RN = AB/PQ
(iii) ∆CMB ~ ∆RNQ
Solution:- From the given figure it is given that, CM and RN are respectively the medians of ∆ABC and ∆PQR. (i) We have to prove that, ∆AMC ~ ∆PQR Consider the ∆ABC and ∆PQR As ∆ABC ~ ∆PQR ∠A = ∠P,...
In the given figure, DB ⊥ BC, DE ⊥ AB and AC ⊥ BC. Prove that BE/DE = AC/BC
Solution:- From the figure, DB ⊥ BC, DE ⊥ AB and AC ⊥ BC We have to prove that, BE/DE = AC/BC Consider the ∆ABC and ∆DEB, ∠C = 90o ∠A + ∠ABC = 90o [from the figure equation (i)] Now in ∆DEB ∠DBE +...
(a) In the figure (1) given below, AP = 2PB and CP = 2PD.
(i) Prove that ∆ACP is similar to ∆BDP and AC || BD.
(ii) If AC = 4.5 cm, calculate the length of BD.
Solution:- From the question it is give that, AP = 2PB, CP = 2PD (i) We have to prove that, ∆ACP is similar to ∆BDP and AC || BD AP = 2PB AP/PB = 2/1 Then, CP = 2PD CP/PD = 2/1 ∠APC = ∠BPD [from...
A card is drawn from a well-shuffled pack of 52 cards. Find the probability of getting:
(v) a king or a queen
(vi) a non-face card
(v) Let E be the event of getting a king or a queen. There will be 4 cards of king and 4 cards of queen. Number of favourable outcomes = 4+4 = 8 P(E) = 8/52 = 2/13 Hence the probability of getting a...
How many spherical lead shots of diameter 4 cm can be made out of a solid cube of lead whose edge measures 44 cm?
Solution: Edge of the cube, a = 44 cm Volume of cube = a3 = 443 = 85184 cm3 Diameter of shot = 4 cm So radius of shot, r = 4/2 = 2 cm Volume of a shot = (4/3)r3 = (4/3)×(22/7)×23 = 704/21 cm3 Number...
Two spheres of the same metal weigh 1 kg and 7 kg. The radius of the smaller sphere is 3 cm. The two spheres are melted to form a single big sphere. Find the diameter of the big sphere.
Solution: For same material, density will be same. Density = mass/Volume Mass of the smaller sphere, m1 = 1 kg Mass of the bigger sphere, m2 = 7 kg The spheres are melted to form a new sphere. So...
A solid metallic hemisphere of radius 8 cm is melted and recasted into right circular cone of base radius 6 cm. Determine the height of the cone.
Solution: Given radius of the hemisphere, r = 8 cm Volume of the hemisphere, V = (2/3)r3 = (2/3)×83 = (1024/3) cm3 Radius of cone, R = 6 cm Since hemisphere is melted and recasted into a cone, the...
An exhibition tent is in the form of a cylinder surmounted by a cone. The height of the tent above the ground is 85 m and the height of the cylindrical part is 50 m. If the diameter of the base is 168 m, find the quantity of canvas required to make the tent. Allow 20% extra for folds and stitching. Give your answer to the nearest m².
solution; Given height of the tent above the ground = 85 m Height of the cylindrical part, H = 50 m height of the cone, h = 85-50 h = 35 m Diameter of the base, d = 168 m Radius of the base of...
From a cube of edge 14 cm, a cone of maximum size is carved out. Find the volume of the remaining material.
Given edge of the cube, a = 14 cm Radius of the cone, r = 14/2 = 7 cm Height of the cone, h = 14 cm Volume of the cube = a3 = 143 = 14×14×14 = 2744 cm3 Volume of the cone = (1/3)r2h =...
Find the surface area of a sphere of radius : (i) 14 cm (ii) 10.5 cm
Solution: (i) Given radius of the sphere, r = 14 cm Surface area of the sphere = 4r2 = 4×(22/7)×142 = 4×22×14×2 = 2464 cm3 Hence the surface area of the sphere is 2464 cm2. (ii) Given radius of the...
The volume of a right circular cone is 9856 cm3 and the area of its base is 616 cm2 . Find
(i) the slant height of the cone.
(ii) total surface area of the cone.
Solution: Given base area of the cone = 616 cm2 r2 = 616 (22/7)×r2 = 616 r2 = 616×7/22 r2 = 196 r = 14 Given volume of the cone = 9856 cm3 (1/3)r2h = 9856 (1/3)×(22/7)×142 ×h = 9856 h =...
A conical tent is 10 m high and the radius of its base is 24 m. Find : (i) slant height of the tent. (ii) cost of the canvas required to make the tent, if the cost of 1 m2 canvas is Rs 70.
Solution: (i)Given height of the tent, h = 10 m Radius, r = 24 m We know that l2 = h2+r2 l2 = 102+242 l2 = 100+576 l2 = 676 l = √676 l = 26 (ii) Curved surface area = rl = (22/7)×24×26 = 13728/7 m2...
If the volume of a right circular cone of height 9 cm is 48π cm3 , find the diameter of its base.
Solution: Given height of a cone, h = 9 cm Volume of the cone = 48 (1/3)r2h = 48 (1/3)r2×9 = 48 3r2 = 48 r2 = 48/3 = 16 r = 4 So diameter = 2×radius = 2×4 = 8 cm Hence the diameter of the cone is 8...
Find the curved surface area of a right circular cone whose slant height is 10 cm and base radius is 7 cm.
Solution: Given slant height of the cone, l = 10 cm Base radius, r = 7 cm Curved surface area of the cone = rl = (22/7)×7×10 = 220 cm2 Hence the curved surface area of the cone is 220...
Find the equation of the line which is parallel to 3x – 2y = – 4 and passes through the point (0, 3).
Solution: Given line: 3x – 2y = -4 Slope (m1) is given by 2y = 3x + 4 y = (3/2) x + 2 So, m1 = 3/2 Now, the slope of the line parallel to the given line will have the same slope as 3/2 = m And the...
The co-ordinates of two points A and B are ( – 3, 3) and (12, – 7) respectively. P is a point on the line segment AB such that AP : PB = 2 : 3. Find the co-ordinates of P.
Solution: Let the co-ordinates of P(x, y) divides AB in the ratio m:n. A(-3,3) and B(12,-7) are the given points. Given m:n = 2:3 x1 = -3 , y1 = 3 , x2 = 12 , y2 = -7 , m = 2 and n = 3 By Section...