Mathematics (71) Geometry - Part II
SSC 10th Board Exam Solution | 2025 Revised Course
Marks: 40 | Time: 2 Hours
Q. 1 (A) Choose the correct alternative from given : (4 Marks)
(1)
Out of the following which is a Pythagorean triplet?
Explanation:
In (3, 4, 5), \(3^2 + 4^2 = 9 + 16 = 25\) and \(5^2 = 25\). Since \(a^2 + b^2 = c^2\), it is a Pythagorean triplet.
Answer: (B)
Answer: (B)
(2)
\(\angle ACB\) is inscribed angle in a circle with centre O. If \(\angle ACB = 65^\circ\), then what is measure of its intercepted arc AXB?
Solution:
Measure of intercepted arc = \(2 \times\) Measure of inscribed angle.
\(m(\text{arc } AXB) = 2 \times 65^\circ = 130^\circ\).
Answer: (D)
\(m(\text{arc } AXB) = 2 \times 65^\circ = 130^\circ\).
Answer: (D)
(3)
Distance of point (3, 4) from the origin is .................... .
Solution:
Distance = \(\sqrt{x^2 + y^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5\).
Answer: (C)
Answer: (C)
(4)
If radius of cone is 5 cm and its perpendicular height is 12 cm, then the slant height is ..................... .
Solution:
Slant height \(l = \sqrt{r^2 + h^2} = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13\) cm.
Answer: (C)
Answer: (C)
Q. 1 (B) Solve the following sub-questions : (4 Marks)
(1)
In the following figure \(\Delta ABC\), \(B-D-C\) and \(BD = 7, BC = 20\), then find \(\frac{A(\Delta ABD)}{A(\Delta ABC)}\).
Solution:
Triangles \(\Delta ABD\) and \(\Delta ABC\) have a common vertex A and their bases lie on the same line BC. Thus, they have equal heights.
Areas of triangles with equal heights are proportional to their bases.
\(\therefore \frac{A(\Delta ABD)}{A(\Delta ABC)} = \frac{BD}{BC}\)
\(= \frac{7}{20}\)
Ans: 7/20
Areas of triangles with equal heights are proportional to their bases.
\(\therefore \frac{A(\Delta ABD)}{A(\Delta ABC)} = \frac{BD}{BC}\)
\(= \frac{7}{20}\)
Ans: 7/20
(2)
In the following figure \(\angle MNP = 90^\circ\), seg \(NQ \perp\) seg \(MP\), \(MQ = 9\), \(QP = 4\), find NQ.
Solution:
By Theorem of Geometric Mean:
\(NQ^2 = MQ \times QP\)
\(NQ^2 = 9 \times 4\)
\(NQ^2 = 36\)
\(NQ = 6\) units.
\(NQ^2 = MQ \times QP\)
\(NQ^2 = 9 \times 4\)
\(NQ^2 = 36\)
\(NQ = 6\) units.
(3)
Angle made by a line with the positive direction of X-axis is \(30^\circ\). Find slope of that line.
Solution:
Here, inclination \(\theta = 30^\circ\).
Slope \(m = \tan \theta\)
\(m = \tan 30^\circ\)
\(m = \frac{1}{\sqrt{3}}\)
Slope \(m = \tan \theta\)
\(m = \tan 30^\circ\)
\(m = \frac{1}{\sqrt{3}}\)
(4)
In cyclic quadrilateral ABCD, \(m\angle A = 100^\circ\), then find \(m\angle C\).
Solution:
Opposite angles of a cyclic quadrilateral are supplementary.
\(\therefore \angle A + \angle C = 180^\circ\)
\(100^\circ + \angle C = 180^\circ\)
\(\angle C = 180^\circ - 100^\circ = 80^\circ\).
\(\therefore \angle A + \angle C = 180^\circ\)
\(100^\circ + \angle C = 180^\circ\)
\(\angle C = 180^\circ - 100^\circ = 80^\circ\).
Q. 2 (A) Complete the following activities and rewrite it (any two) : (4 Marks)
(1)
The radius of a circle with centre ‘P’ is 10 cm. If chord AB of the circle subtends a right angle at P, find area of minor sector by using the following activity. (\(\pi = 3.14\))
Solution:
\(r = 10\) cm, \(\theta = 90^\circ\), \(\pi = 3.14\).
\(A(P-AXB) = \frac{\theta}{360} \times\) \(\pi r^2\)
\(= \frac{\text{\class{input-box}{90}}}{360} \times 3.14 \times 10^2\)
\(= \frac{1}{4} \times\) 314
\(A(P-AXB) =\) 78.5 sq. cm.
\(A(P-AXB) = \frac{\theta}{360} \times\) \(\pi r^2\)
\(= \frac{\text{\class{input-box}{90}}}{360} \times 3.14 \times 10^2\)
\(= \frac{1}{4} \times\) 314
\(A(P-AXB) =\) 78.5 sq. cm.
(2)
In the following figure chord MN and chord RS intersect at point D. If RD = 15, DS = 4, MD = 8, find DN by completing the activity.
Solution:
Activity :
\(MD \times DN =\) RD \(\times DS\) (Theorem of internal division of chords)
8 \(\times DN = 15 \times 4\)
\(DN = \frac{\text{\class{input-box}{60}}}{8}\)
\(DN =\) 7.5
\(MD \times DN =\) RD \(\times DS\) (Theorem of internal division of chords)
8 \(\times DN = 15 \times 4\)
\(DN = \frac{\text{\class{input-box}{60}}}{8}\)
\(DN =\) 7.5
(3)
An observer at a distance of 10 m from tree looks at the top of the tree, the angle of elevation is \(60^\circ\). To find the height of tree complete the activity. (\(\sqrt{3} = 1.73\))
Solution:
\(\tan \theta = \frac{\text{\class{input-box}{AB}}}{BC}\) ... (I)
\(\tan 60^\circ =\) \(\sqrt{3}\) ... (II)
\(\frac{AB}{BC} = \sqrt{3}\)
\(AB = BC \times \sqrt{3} = 10\sqrt{3}\)
\(AB = 10 \times 1.73 =\) 17.3
\(\therefore\) height of the tree is 17.3 m.
\(\tan 60^\circ =\) \(\sqrt{3}\) ... (II)
\(\frac{AB}{BC} = \sqrt{3}\)
\(AB = BC \times \sqrt{3} = 10\sqrt{3}\)
\(AB = 10 \times 1.73 =\) 17.3
\(\therefore\) height of the tree is 17.3 m.
Q. 2 (B) Solve the following sub-questions (any four) : (8 Marks)
(1)
In \(\Delta ABC\), \(DE \parallel BC\). If \(DB = 5.4\) cm, \(AD = 1.8\) cm, \(EC = 7.2\) cm, then find AE.
Solution:
In \(\Delta ABC\), \(DE \parallel BC\). By Basic Proportionality Theorem (BPT):
\(\frac{AD}{DB} = \frac{AE}{EC}\)
\(\frac{1.8}{5.4} = \frac{AE}{7.2}\)
\(\frac{1}{3} = \frac{AE}{7.2}\)
\(AE = \frac{7.2}{3}\)
\(AE = 2.4\) cm.
\(\frac{AD}{DB} = \frac{AE}{EC}\)
\(\frac{1.8}{5.4} = \frac{AE}{7.2}\)
\(\frac{1}{3} = \frac{AE}{7.2}\)
\(AE = \frac{7.2}{3}\)
\(AE = 2.4\) cm.
(2)
In the figure given, find RS and PS using the information given in \(\Delta PSR\). (Given: \(\angle S = 90^\circ, \angle P = 30^\circ, PR = 12\))
Solution:
In \(\Delta PSR\), \(\angle S = 90^\circ, \angle P = 30^\circ\), so \(\angle R = 60^\circ\).
This is a \(30^\circ-60^\circ-90^\circ\) triangle.
Side opposite to \(30^\circ\) (RS) is half the hypotenuse (PR).
\(RS = \frac{1}{2} PR = \frac{1}{2} \times 12 = 6\) units.
Side opposite to \(60^\circ\) (PS) is \(\frac{\sqrt{3}}{2}\) times the hypotenuse.
\(PS = \frac{\sqrt{3}}{2} PR = \frac{\sqrt{3}}{2} \times 12 = 6\sqrt{3}\) units.
This is a \(30^\circ-60^\circ-90^\circ\) triangle.
Side opposite to \(30^\circ\) (RS) is half the hypotenuse (PR).
\(RS = \frac{1}{2} PR = \frac{1}{2} \times 12 = 6\) units.
Side opposite to \(60^\circ\) (PS) is \(\frac{\sqrt{3}}{2}\) times the hypotenuse.
\(PS = \frac{\sqrt{3}}{2} PR = \frac{\sqrt{3}}{2} \times 12 = 6\sqrt{3}\) units.
(3)
In the figure, circle with centre D touches the sides of \(\angle ACB\) at A and B. If \(\angle ACB = 52^\circ\), find measure of \(\angle ADB\).
Solution:
Radius is perpendicular to the tangent at the point of contact.
\(\therefore \angle CAD = 90^\circ\) and \(\angle CBD = 90^\circ\).
In quadrilateral ACBD, sum of all angles is \(360^\circ\).
\(\angle ACB + \angle CAD + \angle ADB + \angle CBD = 360^\circ\)
\(52^\circ + 90^\circ + \angle ADB + 90^\circ = 360^\circ\)
\(232^\circ + \angle ADB = 360^\circ\)
\(\angle ADB = 360^\circ - 232^\circ = 128^\circ\).
\(\therefore \angle CAD = 90^\circ\) and \(\angle CBD = 90^\circ\).
In quadrilateral ACBD, sum of all angles is \(360^\circ\).
\(\angle ACB + \angle CAD + \angle ADB + \angle CBD = 360^\circ\)
\(52^\circ + 90^\circ + \angle ADB + 90^\circ = 360^\circ\)
\(232^\circ + \angle ADB = 360^\circ\)
\(\angle ADB = 360^\circ - 232^\circ = 128^\circ\).
(4)
Verify, whether points A(1, –3), B(2, –5) and C(–4, 7) are collinear or not.
Solution:
Slope of line AB = \(\frac{y_2 - y_1}{x_2 - x_1} = \frac{-5 - (-3)}{2 - 1} = \frac{-2}{1} = -2\).
Slope of line BC = \(\frac{y_3 - y_2}{x_3 - x_2} = \frac{7 - (-5)}{-4 - 2} = \frac{12}{-6} = -2\).
Since Slope of AB = Slope of BC, and point B is common, the points A, B, and C are collinear.
Slope of line BC = \(\frac{y_3 - y_2}{x_3 - x_2} = \frac{7 - (-5)}{-4 - 2} = \frac{12}{-6} = -2\).
Since Slope of AB = Slope of BC, and point B is common, the points A, B, and C are collinear.
(5)
If \(\sin \theta = \frac{11}{61}\), find the values of \(\cos \theta\) using trigonometric identity.
Solution:
Identity: \(\sin^2 \theta + \cos^2 \theta = 1\)
\((\frac{11}{61})^2 + \cos^2 \theta = 1\)
\(\frac{121}{3721} + \cos^2 \theta = 1\)
\(\cos^2 \theta = 1 - \frac{121}{3721} = \frac{3721 - 121}{3721} = \frac{3600}{3721}\)
Taking square roots,
\(\cos \theta = \frac{60}{61}\).
\((\frac{11}{61})^2 + \cos^2 \theta = 1\)
\(\frac{121}{3721} + \cos^2 \theta = 1\)
\(\cos^2 \theta = 1 - \frac{121}{3721} = \frac{3721 - 121}{3721} = \frac{3600}{3721}\)
Taking square roots,
\(\cos \theta = \frac{60}{61}\).
Q. 3 (A) Complete the following activities and rewrite it (any one) : (3 Marks)
(1)
In figure, XY \(\parallel\) seg AC. If \(2AX = 3BX\) and \(XY = 9\). Complete the activity to find value of AC.
Solution:
\(2AX = 3BX \implies \frac{AX}{BX} = \) \(\frac{3}{2}\)
\(\frac{AX + BX}{BX} = \frac{3+2}{2}\) ... (by componendo)
\(\frac{\text{\class{input-box}{AB}}}{BX} = \frac{5}{2}\) ... (I)
Now \(\Delta BCA \sim \Delta BYX\) ... (AA test of similarity)
\(\therefore \frac{BA}{BX} = \frac{AC}{XY}\)
\(\frac{\text{\class{input-box}{5}}}{\text{\class{input-box}{2}}} = \frac{AC}{9}\) ... from (I)
\(\therefore AC =\) 22.5
\(\frac{AX + BX}{BX} = \frac{3+2}{2}\) ... (by componendo)
\(\frac{\text{\class{input-box}{AB}}}{BX} = \frac{5}{2}\) ... (I)
Now \(\Delta BCA \sim \Delta BYX\) ... (AA test of similarity)
\(\therefore \frac{BA}{BX} = \frac{AC}{XY}\)
\(\frac{\text{\class{input-box}{5}}}{\text{\class{input-box}{2}}} = \frac{AC}{9}\) ... from (I)
\(\therefore AC =\) 22.5
(2)
Complete the activity to prove that sum of squares of diagonals of a rhombus is equal to sum of squares of the sides. (Using Apollonius theorem)
Solution:
PQRS is a rhombus. Diagonals bisect at T.
In \(\Delta PQS\), PT is median. By Apollonius:
\(PQ^2 + PS^2 =\) \(2PT^2\) \(+ 2QT^2\) ... (I)
Similarly in \(\Delta QRS\), RT is median:
\(QR^2 + SR^2 =\) \(2RT^2\) \(+ 2QT^2\) ... (II)
Adding (I) and (II):
\(PQ^2+PS^2+QR^2+SR^2 = 2(PT^2 +\) \(RT^2\) \() + 4QT^2\)
Since \(PT=RT\), \(2(PT^2+PT^2) = 4PT^2\).
\(= 4PT^2 + 4QT^2\)
\(= (\) \(2PT\) \()^2 + (2QT)^2\) (Note: \(2PT = PR\))
\(\therefore PQ^2+PS^2+QR^2+SR^2 = PR^2 +\) \(QS^2\)
In \(\Delta PQS\), PT is median. By Apollonius:
\(PQ^2 + PS^2 =\) \(2PT^2\) \(+ 2QT^2\) ... (I)
Similarly in \(\Delta QRS\), RT is median:
\(QR^2 + SR^2 =\) \(2RT^2\) \(+ 2QT^2\) ... (II)
Adding (I) and (II):
\(PQ^2+PS^2+QR^2+SR^2 = 2(PT^2 +\) \(RT^2\) \() + 4QT^2\)
Since \(PT=RT\), \(2(PT^2+PT^2) = 4PT^2\).
\(= 4PT^2 + 4QT^2\)
\(= (\) \(2PT\) \()^2 + (2QT)^2\) (Note: \(2PT = PR\))
\(\therefore PQ^2+PS^2+QR^2+SR^2 = PR^2 +\) \(QS^2\)
Q. 3 (B) Solve the following sub-questions (any two) : (6 Marks)
(1)
Show that points P(1, –2), Q(5, 2), R(3, –1), S(–1, –5) are the vertices of a parallelogram.
Solution:
A quadrilateral is a parallelogram if its diagonals bisect each other.
Midpoint of diagonal PR = \(\left(\frac{1+3}{2}, \frac{-2+(-1)}{2}\right) = (2, -1.5)\).
Midpoint of diagonal QS = \(\left(\frac{5+(-1)}{2}, \frac{2+(-5)}{2}\right) = \left(\frac{4}{2}, \frac{-3}{2}\right) = (2, -1.5)\).
Since the midpoints are the same, the diagonals bisect each other.
\(\therefore \square PQRS\) is a parallelogram.
Midpoint of diagonal PR = \(\left(\frac{1+3}{2}, \frac{-2+(-1)}{2}\right) = (2, -1.5)\).
Midpoint of diagonal QS = \(\left(\frac{5+(-1)}{2}, \frac{2+(-5)}{2}\right) = \left(\frac{4}{2}, \frac{-3}{2}\right) = (2, -1.5)\).
Since the midpoints are the same, the diagonals bisect each other.
\(\therefore \square PQRS\) is a parallelogram.
(2)
Prove that tangent segments drawn from an external point to a circle are congruent.
Proof:
Given: A circle with centre O. Point P is outside. PA and PB are tangent segments at A and B.
To prove: Seg PA \(\cong\) Seg PB.
Construction: Draw radius OA and OB. Draw seg OP.
Proof: In \(\Delta OAP\) and \(\Delta OBP\),
\(\angle OAP = \angle OBP = 90^\circ\) (Tangent-Radius theorem)
Hypotenuse OP \(\cong\) Hypotenuse OP (Common side)
Side OA \(\cong\) Side OB (Radii of same circle)
\(\therefore \Delta OAP \cong \Delta OBP\) (Hypotenuse-Side test)
\(\therefore\) Seg PA \(\cong\) Seg PB (c.s.c.t)
Hence Proved.
To prove: Seg PA \(\cong\) Seg PB.
Construction: Draw radius OA and OB. Draw seg OP.
Proof: In \(\Delta OAP\) and \(\Delta OBP\),
\(\angle OAP = \angle OBP = 90^\circ\) (Tangent-Radius theorem)
Hypotenuse OP \(\cong\) Hypotenuse OP (Common side)
Side OA \(\cong\) Side OB (Radii of same circle)
\(\therefore \Delta OAP \cong \Delta OBP\) (Hypotenuse-Side test)
\(\therefore\) Seg PA \(\cong\) Seg PB (c.s.c.t)
Hence Proved.
(3)
Draw a circle with radius 4.1 cm. Construct tangents to the circle from a point at a distance 7.3 cm from the centre.
Steps of Construction:
1. Draw a circle with centre O and radius 4.1 cm.
2. Take a point P such that OP = 7.3 cm.
3. Draw the perpendicular bisector of segment OP. Let M be the midpoint.
4. With M as centre and radius OM, draw arcs intersecting the circle at points A and B.
5. Draw lines PA and PB. These are the required tangents.
2. Take a point P such that OP = 7.3 cm.
3. Draw the perpendicular bisector of segment OP. Let M be the midpoint.
4. With M as centre and radius OM, draw arcs intersecting the circle at points A and B.
5. Draw lines PA and PB. These are the required tangents.
(4)
How many solid cylinders of radius 10 cm and height 6 cm can be made by melting a solid sphere of radius 30 cm?
Solution:
Let \(n\) be the number of cylinders.
Volume of Sphere = \(n \times\) Volume of one Cylinder
Sphere: \(R = 30\). Cylinder: \(r = 10, h = 6\).
\(\frac{4}{3} \pi R^3 = n \times \pi r^2 h\)
\(\frac{4}{3} (30)^3 = n (10)^2 (6)\)
\(\frac{4}{3} \times 27000 = n \times 100 \times 6\)
\(36000 = 600n\)
\(n = \frac{36000}{600} = 60\).
Ans: 60 cylinders.
Volume of Sphere = \(n \times\) Volume of one Cylinder
Sphere: \(R = 30\). Cylinder: \(r = 10, h = 6\).
\(\frac{4}{3} \pi R^3 = n \times \pi r^2 h\)
\(\frac{4}{3} (30)^3 = n (10)^2 (6)\)
\(\frac{4}{3} \times 27000 = n \times 100 \times 6\)
\(36000 = 600n\)
\(n = \frac{36000}{600} = 60\).
Ans: 60 cylinders.
Q. 4 Solve the following sub-questions (any two) : (8 Marks)
(1)
In figure \(DE \parallel BC\). (i) If DE=4, BC=8, A(\(\Delta\)ADE)=25, find A(\(\Delta\)ABC). (ii) If DE:BC = 3:5, find A(\(\Delta\)ADE) : A(\(\square\)DBCE).
Solution:
Since \(DE \parallel BC\), \(\Delta ADE \sim \Delta ABC\) (AA Test).
(i) By theorem of areas of similar triangles:
\(\frac{A(\Delta ADE)}{A(\Delta ABC)} = \frac{DE^2}{BC^2} = (\frac{4}{8})^2 = (\frac{1}{2})^2 = \frac{1}{4}\)
\(\frac{25}{A(\Delta ABC)} = \frac{1}{4} \implies A(\Delta ABC) = 100\) cm\(^2\).
(ii) If \(DE:BC = 3:5\), then Area Ratio = \(3^2:5^2 = 9:25\).
Let \(A(\Delta ADE) = 9k\) and \(A(\Delta ABC) = 25k\).
Area of quadrilateral DBCE = \(A(\Delta ABC) - A(\Delta ADE) = 25k - 9k = 16k\).
Ratio \(A(\Delta ADE) : A(\square DBCE) = 9k : 16k = 9:16\).
(i) By theorem of areas of similar triangles:
\(\frac{A(\Delta ADE)}{A(\Delta ABC)} = \frac{DE^2}{BC^2} = (\frac{4}{8})^2 = (\frac{1}{2})^2 = \frac{1}{4}\)
\(\frac{25}{A(\Delta ABC)} = \frac{1}{4} \implies A(\Delta ABC) = 100\) cm\(^2\).
(ii) If \(DE:BC = 3:5\), then Area Ratio = \(3^2:5^2 = 9:25\).
Let \(A(\Delta ADE) = 9k\) and \(A(\Delta ABC) = 25k\).
Area of quadrilateral DBCE = \(A(\Delta ABC) - A(\Delta ADE) = 25k - 9k = 16k\).
Ratio \(A(\Delta ADE) : A(\square DBCE) = 9k : 16k = 9:16\).
(2)
\(\Delta ABC \sim \Delta PQR\). In \(\Delta ABC\), \(AB=3.6, BC=4, AC=4.2\). Sides are in ratio 2:3. Construct both triangles.
Solution:
Given \(\frac{AB}{PQ} = \frac{BC}{QR} = \frac{AC}{PR} = \frac{2}{3}\).
Calculation for \(\Delta PQR\):
\(PQ = \frac{3}{2} AB = 1.5 \times 3.6 = 5.4\) cm.
\(QR = \frac{3}{2} BC = 1.5 \times 4 = 6.0\) cm.
\(PR = \frac{3}{2} AC = 1.5 \times 4.2 = 6.3\) cm.
Construction:
1. Draw \(\Delta ABC\) with sides 3.6, 4, 4.2.
2. Draw \(\Delta PQR\) with sides 5.4, 6, 6.3 separately.
Calculation for \(\Delta PQR\):
\(PQ = \frac{3}{2} AB = 1.5 \times 3.6 = 5.4\) cm.
\(QR = \frac{3}{2} BC = 1.5 \times 4 = 6.0\) cm.
\(PR = \frac{3}{2} AC = 1.5 \times 4.2 = 6.3\) cm.
Construction:
1. Draw \(\Delta ABC\) with sides 3.6, 4, 4.2.
2. Draw \(\Delta PQR\) with sides 5.4, 6, 6.3 separately.
(3)
Frustum of cone: Radii 14 cm and 8 cm. Height 8 cm. Find (i) CSA, (ii) TSA, (iii) Volume. (\(\pi = 3.14\))
Solution:
\(r_1 = 14, r_2 = 8, h = 8\).
Slant height \(l = \sqrt{h^2 + (r_1-r_2)^2} = \sqrt{8^2 + 6^2} = \sqrt{64+36} = 10\) cm.
(i) CSA = \(\pi(r_1+r_2)l = 3.14(14+8)10 = 3.14(220) = 690.8\) cm\(^2\).
(ii) TSA = CSA + \(\pi r_1^2 + \pi r_2^2\)
\(= 690.8 + 3.14(14^2) + 3.14(8^2)\)
\(= 690.8 + 3.14(196) + 3.14(64) = 690.8 + 615.44 + 200.96 = 1507.2\) cm\(^2\).
(iii) Volume = \(\frac{1}{3} \pi h (r_1^2 + r_2^2 + r_1r_2)\)
\(= \frac{1}{3} (3.14)(8) (196 + 64 + 112)\)
\(= \frac{25.12}{3} (372) = 25.12 \times 124 = 3114.88\) cm\(^3\).
Slant height \(l = \sqrt{h^2 + (r_1-r_2)^2} = \sqrt{8^2 + 6^2} = \sqrt{64+36} = 10\) cm.
(i) CSA = \(\pi(r_1+r_2)l = 3.14(14+8)10 = 3.14(220) = 690.8\) cm\(^2\).
(ii) TSA = CSA + \(\pi r_1^2 + \pi r_2^2\)
\(= 690.8 + 3.14(14^2) + 3.14(8^2)\)
\(= 690.8 + 3.14(196) + 3.14(64) = 690.8 + 615.44 + 200.96 = 1507.2\) cm\(^2\).
(iii) Volume = \(\frac{1}{3} \pi h (r_1^2 + r_2^2 + r_1r_2)\)
\(= \frac{1}{3} (3.14)(8) (196 + 64 + 112)\)
\(= \frac{25.12}{3} (372) = 25.12 \times 124 = 3114.88\) cm\(^3\).
Q. 5 Solve the following sub-questions (any one) : (3 Marks)
(1)
ABCD is a rectangle. AD is diameter of a semicircle AXD intersecting diagonal BD at X. AB = 12, AD = 9. Find BD and BX.
Solution:
In \(\Delta DAB\), \(\angle A = 90^\circ\).
\(BD = \sqrt{12^2 + 9^2} = \sqrt{144+81} = \sqrt{225} = 15\) cm.
Since AD is diameter, \(\angle AXD = 90^\circ\) (angle in semicircle).
So, \(AX \perp BD\).
In right \(\Delta DAB\), seg AX is altitude to hypotenuse.
By Geometric Mean (Leg Rule): \(AB^2 = BX \times BD\).
\(12^2 = BX \times 15\)
\(144 = 15 BX\)
\(BX = \frac{144}{15} = 9.6\) cm.
Ans: BD = 15 cm, BX = 9.6 cm.
\(BD = \sqrt{12^2 + 9^2} = \sqrt{144+81} = \sqrt{225} = 15\) cm.
Since AD is diameter, \(\angle AXD = 90^\circ\) (angle in semicircle).
So, \(AX \perp BD\).
In right \(\Delta DAB\), seg AX is altitude to hypotenuse.
By Geometric Mean (Leg Rule): \(AB^2 = BX \times BD\).
\(12^2 = BX \times 15\)
\(144 = 15 BX\)
\(BX = \frac{144}{15} = 9.6\) cm.
Ans: BD = 15 cm, BX = 9.6 cm.
(2)
Taking \(\theta = 30^\circ\) verify the following Trigonometric identities : (i) \(\sin^2 \theta + \cos^2 \theta = 1\), (ii) \(1 + \tan^2 \theta = \sec^2 \theta\), (iii) \(1 + \cot^2 \theta = \text{cosec}^2 \theta\).
Solution:
(i) \(\sin^2 30^\circ + \cos^2 30^\circ = (\frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2 = \frac{1}{4} + \frac{3}{4} = \frac{4}{4} = 1\). Verified.
(ii) \(1 + \tan^2 30^\circ = 1 + (\frac{1}{\sqrt{3}})^2 = 1 + \frac{1}{3} = \frac{4}{3}\).
\(\sec^2 30^\circ = (\frac{2}{\sqrt{3}})^2 = \frac{4}{3}\). LHS = RHS. Verified.
(iii) \(1 + \cot^2 30^\circ = 1 + (\sqrt{3})^2 = 1 + 3 = 4\).
\(\text{cosec}^2 30^\circ = (2)^2 = 4\). LHS = RHS. Verified.
(ii) \(1 + \tan^2 30^\circ = 1 + (\frac{1}{\sqrt{3}})^2 = 1 + \frac{1}{3} = \frac{4}{3}\).
\(\sec^2 30^\circ = (\frac{2}{\sqrt{3}})^2 = \frac{4}{3}\). LHS = RHS. Verified.
(iii) \(1 + \cot^2 30^\circ = 1 + (\sqrt{3})^2 = 1 + 3 = 4\).
\(\text{cosec}^2 30^\circ = (2)^2 = 4\). LHS = RHS. Verified.