Question 1
Q. 1 (A) Choose the correct alternative from given: (4 Marks)
(1) If \(\Delta ABC \sim \Delta DEF\), \(m\angle B=60^{\circ}\), then \(m\angle E=\)
Answer: (B) \(60^{\circ}\)
Explanation: In similar triangles, corresponding angles are congruent. Since \(\Delta ABC \sim \Delta DEF\), \(\angle B \cong \angle E\). Therefore, \(m\angle E = 60^{\circ}\).
Explanation: In similar triangles, corresponding angles are congruent. Since \(\Delta ABC \sim \Delta DEF\), \(\angle B \cong \angle E\). Therefore, \(m\angle E = 60^{\circ}\).
(2) Two circles of radii 5.5 cm and 4.2 cm touch each other externally then distance between their centres is
Answer: (A) 9.7 cm
Explanation: When circles touch externally, the distance between their centres is the sum of their radii. \(d = r_1 + r_2 = 5.5 + 4.2 = 9.7\) cm.
Explanation: When circles touch externally, the distance between their centres is the sum of their radii. \(d = r_1 + r_2 = 5.5 + 4.2 = 9.7\) cm.
(3) A line makes an angle of \(45^{\circ}\) with the positive direction of X-axis. So the slope of the line is
Answer: (C) 1
Explanation: Slope \(m = \tan \theta\). Here \(\theta = 45^{\circ}\). \(m = \tan 45^{\circ} = 1\).
Explanation: Slope \(m = \tan \theta\). Here \(\theta = 45^{\circ}\). \(m = \tan 45^{\circ} = 1\).
(4) The volume of a cube of side 2 cm is
Answer: (D) \(8~cm^{3}\)
Explanation: Volume of cube \(= (\text{side})^3 = 2^3 = 8~cm^3\).
Explanation: Volume of cube \(= (\text{side})^3 = 2^3 = 8~cm^3\).
Q. 1 (B) Solve the following subquestions: (4 Marks)
(1) Find the diagonal of square whose side is 10 cm.
Formula: Diagonal of square = \(\sqrt{2} \times \text{side}\)
\(\text{Diagonal} = \sqrt{2} \times 10 = 10\sqrt{2}\) cm.
\(\text{Diagonal} = \sqrt{2} \times 10 = 10\sqrt{2}\) cm.
(2) The ratio of corresponding sides of similar triangles is 3:5, then find the ratio of their areas.
By Theorem of Areas of Similar Triangles:
\(\frac{A_1}{A_2} = \frac{s_1^2}{s_2^2}\)
\(\frac{A_1}{A_2} = \frac{3^2}{5^2} = \frac{9}{25}\)
Ratio of areas is 9:25.
\(\frac{A_1}{A_2} = \frac{s_1^2}{s_2^2}\)
\(\frac{A_1}{A_2} = \frac{3^2}{5^2} = \frac{9}{25}\)
Ratio of areas is 9:25.
(3) Find the slope of the line passing through the points \(A(2, 3)\) and \(B(4,7)\).
Slope \(m = \frac{y_2 - y_1}{x_2 - x_1}\)
\(m = \frac{7 - 3}{4 - 2} = \frac{4}{2} = 2\).
\(m = \frac{7 - 3}{4 - 2} = \frac{4}{2} = 2\).
(4) If \(\sin \theta = \frac{7}{25}\), then find the value of \(\text{cosec } \theta\).
\(\text{cosec } \theta = \frac{1}{\sin \theta}\)
\(\text{cosec } \theta = \frac{1}{7/25} = \frac{25}{7}\).
\(\text{cosec } \theta = \frac{1}{7/25} = \frac{25}{7}\).
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Question 2
Q. 2 (A) Complete the following activities and rewrite it (any two): (4 Marks)
(1) In the given figure, \(AR \perp BC\), \(AR \perp PQ\), then complete the activity for finding \(\frac{A(\Delta ABC)}{A(\Delta APQ)}\).
[Figure shows two triangles ABC and APQ with a common height AR]
Activity:
\(\frac{A(\Delta ABC)}{A(\Delta APQ)} = \frac{\text{Base} \times \text{Height}}{\text{Base} \times \text{Height}}\)
\(\frac{A(\Delta ABC)}{A(\Delta APQ)} = \frac{\class{activity-box}{BC} \times AR}{PQ \times \class{activity-box}{AR}}\)
\(\frac{A(\Delta ABC)}{A(\Delta APQ)} = \frac{\class{activity-box}{BC}}{\class{activity-box}{PQ}}\)
\(\frac{A(\Delta ABC)}{A(\Delta APQ)} = \frac{\text{Base} \times \text{Height}}{\text{Base} \times \text{Height}}\)
\(\frac{A(\Delta ABC)}{A(\Delta APQ)} = \frac{\class{activity-box}{BC} \times AR}{PQ \times \class{activity-box}{AR}}\)
\(\frac{A(\Delta ABC)}{A(\Delta APQ)} = \frac{\class{activity-box}{BC}}{\class{activity-box}{PQ}}\)
(2) In the figure, seg PS is a tangent segment, line PR is a secant. If \(PQ=3.6\), \(QR=6.4\) then find PS by completing the following activity.
Activity:
\(PS^{2} = PQ \times \class{activity-box}{PR}\) ... (Tangent Secant Segments Theorem)
\(PS^{2} = PQ \times (PQ + \class{activity-box}{QR})\)
\(PS^{2} = 3.6 \times (3.6 + \class{activity-box}{6.4})\)
\(PS^{2} = 3.6 \times 10\)
\(PS^{2} = 36\)
\(PS = \class{activity-box}{6}\)
\(PS^{2} = PQ \times \class{activity-box}{PR}\) ... (Tangent Secant Segments Theorem)
\(PS^{2} = PQ \times (PQ + \class{activity-box}{QR})\)
\(PS^{2} = 3.6 \times (3.6 + \class{activity-box}{6.4})\)
\(PS^{2} = 3.6 \times 10\)
\(PS^{2} = 36\)
\(PS = \class{activity-box}{6}\)
(3) Measure of an arc of a circle is \(90^{\circ}\) and its radius is 14 cm. Complete the following activity to find the length of an arc.
Activity:
Length of an arc \( = \frac{\theta}{360} \times \class{activity-box}{2\pi r}\) ... (Formula)
\( = \frac{90}{360} \times 2 \times \frac{22}{7} \times \class{activity-box}{14}\)
\( = \frac{1}{4} \times \class{activity-box}{88}\)
Length of an arc \( = \class{activity-box}{22}\) cm
Length of an arc \( = \frac{\theta}{360} \times \class{activity-box}{2\pi r}\) ... (Formula)
\( = \frac{90}{360} \times 2 \times \frac{22}{7} \times \class{activity-box}{14}\)
\( = \frac{1}{4} \times \class{activity-box}{88}\)
Length of an arc \( = \class{activity-box}{22}\) cm
Q. 2 (B) Solve the following subquestions (any four): (8 Marks)
(1) In \(\Delta LMN\), ray MT bisects \(\angle LMN\). If \(LM=6\), \(MN=10\), \(TN=8\), then find LT.
By Theorem of Angle Bisector of a triangle:
\(\frac{LM}{MN} = \frac{LT}{TN}\)
\(\frac{6}{10} = \frac{LT}{8}\)
\(LT = \frac{6 \times 8}{10}\)
\(LT = \frac{48}{10}\)
\(LT = 4.8\)
\(\frac{LM}{MN} = \frac{LT}{TN}\)
\(\frac{6}{10} = \frac{LT}{8}\)
\(LT = \frac{6 \times 8}{10}\)
\(LT = \frac{48}{10}\)
\(LT = 4.8\)
(2) Find surface area of sphere of radius 7 cm.
Surface Area of Sphere \(= 4\pi r^2\)
\(= 4 \times \frac{22}{7} \times 7 \times 7\)
\(= 4 \times 22 \times 7\)
\(= 88 \times 7\)
Surface Area \(= 616\) sq. cm.
\(= 4 \times \frac{22}{7} \times 7 \times 7\)
\(= 4 \times 22 \times 7\)
\(= 88 \times 7\)
Surface Area \(= 616\) sq. cm.
(3) In the figure, \(m(\text{arc } NS)=125^{\circ}\), \(m(\text{arc } EF)=37^{\circ}\). Find \(m\angle NMS\).
Since chords intersect inside the circle:
\(m\angle NMS = \frac{1}{2} [m(\text{arc } NS) + m(\text{arc } EF)]\)
\(m\angle NMS = \frac{1}{2} [125^{\circ} + 37^{\circ}]\)
\(m\angle NMS = \frac{1}{2} [162^{\circ}]\)
\(m\angle NMS = 81^{\circ}\)
\(m\angle NMS = \frac{1}{2} [m(\text{arc } NS) + m(\text{arc } EF)]\)
\(m\angle NMS = \frac{1}{2} [125^{\circ} + 37^{\circ}]\)
\(m\angle NMS = \frac{1}{2} [162^{\circ}]\)
\(m\angle NMS = 81^{\circ}\)
(4) Find the co-ordinates of midpoint of the segment joining the points \(P(22,20)\) and \(Q(0,16)\).
By Midpoint Formula:
\(x = \frac{x_1 + x_2}{2}, y = \frac{y_1 + y_2}{2}\)
\(x = \frac{22 + 0}{2} = \frac{22}{2} = 11\)
\(y = \frac{20 + 16}{2} = \frac{36}{2} = 18\)
Co-ordinates of midpoint are (11, 18).
\(x = \frac{x_1 + x_2}{2}, y = \frac{y_1 + y_2}{2}\)
\(x = \frac{22 + 0}{2} = \frac{22}{2} = 11\)
\(y = \frac{20 + 16}{2} = \frac{36}{2} = 18\)
Co-ordinates of midpoint are (11, 18).
(5) Find the volume of a cone if radius of its base is 7 cm and its perpendicular height is 15 cm.
Volume of cone \(V = \frac{1}{3}\pi r^2 h\)
\(V = \frac{1}{3} \times \frac{22}{7} \times 7 \times 7 \times 15\)
\(V = 22 \times 7 \times 5\)
\(V = 154 \times 5\)
Volume \(= 770~cm^3\)
\(V = \frac{1}{3} \times \frac{22}{7} \times 7 \times 7 \times 15\)
\(V = 22 \times 7 \times 5\)
\(V = 154 \times 5\)
Volume \(= 770~cm^3\)
Question 3
Q. 3 (A) Complete the following activities and rewrite it (any one): (3 Marks)
(1) If \(\tan \theta=1\), then find the value of \(\frac{\sin \theta+\cos \theta}{\sec \theta+\text{cosec } \theta}\) by completing the activity.
Activity:
\(\tan \theta = 1\) (given)
but \(\tan \class{activity-box}{45^{\circ}} = 1\)
\(\therefore \theta = \class{activity-box}{45^{\circ}}\)
\(\frac{\sin \theta+\cos \theta}{\sec \theta+\text{cosec } \theta} = \frac{\sin 45^{\circ}+\cos 45^{\circ}}{\sec 45^{\circ}+\text{cosec } 45^{\circ}}\)
\( = \frac{\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}}{\sqrt{2} + \class{activity-box}{\sqrt{2}}}\)
\( = \frac{\frac{2}{\sqrt{2}}}{\class{activity-box}{2\sqrt{2}}}\)
Final Answer: \(\frac{\sin \theta+\cos \theta}{\sec \theta+\text{cosec } \theta} = \frac{1}{\class{activity-box}{2}}\)
\(\tan \theta = 1\) (given)
but \(\tan \class{activity-box}{45^{\circ}} = 1\)
\(\therefore \theta = \class{activity-box}{45^{\circ}}\)
\(\frac{\sin \theta+\cos \theta}{\sec \theta+\text{cosec } \theta} = \frac{\sin 45^{\circ}+\cos 45^{\circ}}{\sec 45^{\circ}+\text{cosec } 45^{\circ}}\)
\( = \frac{\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}}{\sqrt{2} + \class{activity-box}{\sqrt{2}}}\)
\( = \frac{\frac{2}{\sqrt{2}}}{\class{activity-box}{2\sqrt{2}}}\)
Final Answer: \(\frac{\sin \theta+\cos \theta}{\sec \theta+\text{cosec } \theta} = \frac{1}{\class{activity-box}{2}}\)
(2) In the figure, point O is the centre and chord AB is equal to the radius. Find measures of \(\angle AOB\), arc AB, and \(\angle ACB\).
Activity:
In \(\Delta AOB\),
\(AO = OB = AB\) (given and radii)
\(\therefore \Delta AOB\) is an \(\class{activity-box}{\text{equilateral}}\) triangle.
\(\therefore m\angle AOB = \class{activity-box}{60^{\circ}}\)
\(m\angle AOB = m(\text{arc } AB) = \class{activity-box}{60^{\circ}}\) (definition of measure of an arc)
\(m\angle ACB = \frac{1}{2} \times \class{activity-box}{m(\text{arc } AB)}\)
\(= \frac{1}{2} \times 60^{\circ}\)
\(m\angle ACB = \class{activity-box}{30^{\circ}}\)
In \(\Delta AOB\),
\(AO = OB = AB\) (given and radii)
\(\therefore \Delta AOB\) is an \(\class{activity-box}{\text{equilateral}}\) triangle.
\(\therefore m\angle AOB = \class{activity-box}{60^{\circ}}\)
\(m\angle AOB = m(\text{arc } AB) = \class{activity-box}{60^{\circ}}\) (definition of measure of an arc)
\(m\angle ACB = \frac{1}{2} \times \class{activity-box}{m(\text{arc } AB)}\)
\(= \frac{1}{2} \times 60^{\circ}\)
\(m\angle ACB = \class{activity-box}{30^{\circ}}\)
Q. 3 (B) Solve the following subquestions (any two): (6 Marks)
(1) Find co-ordinates of point P, if P divides the line segment joining the points \(A(-1,7)\) and \(B(4,-3)\) in the ratio 2:3.
Let \(A(x_1, y_1) = (-1, 7)\) and \(B(x_2, y_2) = (4, -3)\).
Ratio \(m:n = 2:3\).
By Section Formula:
\(x = \frac{mx_2 + nx_1}{m+n} = \frac{2(4) + 3(-1)}{2+3} = \frac{8 - 3}{5} = \frac{5}{5} = 1\)
\(y = \frac{my_2 + ny_1}{m+n} = \frac{2(-3) + 3(7)}{2+3} = \frac{-6 + 21}{5} = \frac{15}{5} = 3\)
Co-ordinates of P are (1, 3).
Ratio \(m:n = 2:3\).
By Section Formula:
\(x = \frac{mx_2 + nx_1}{m+n} = \frac{2(4) + 3(-1)}{2+3} = \frac{8 - 3}{5} = \frac{5}{5} = 1\)
\(y = \frac{my_2 + ny_1}{m+n} = \frac{2(-3) + 3(7)}{2+3} = \frac{-6 + 21}{5} = \frac{15}{5} = 3\)
Co-ordinates of P are (1, 3).
(2) Draw a circle with centre O of radius 3.4 cm. Draw a chord MN of length 5.7 cm in it. Construct tangents at point M and N to the circle.
Steps of Construction:
- Draw a circle with centre O and radius 3.4 cm.
- Take a point M on the circle. Using a ruler, measure 5.7 cm and mark point N on the circle to draw chord MN.
- Draw ray OM and ray ON extending outwards.
- Construct a perpendicular line to ray OM at point M. This is the tangent at M.
- Construct a perpendicular line to ray ON at point N. This is the tangent at N.
(3) A storm broke a tree and the treetop rested 20 m from the base of the tree, making an angle of \(60^{\circ}\) with the horizontal. Find the height of the tree.
Let AB be the remaining tree height and AC be the broken part (hypotenuse).
The top touches ground at C, 20m from base B. \(\angle C = 60^{\circ}\).
In \(\Delta ABC\) (Right angled at B):
\(\tan 60^{\circ} = \frac{AB}{BC} \Rightarrow \sqrt{3} = \frac{AB}{20} \Rightarrow AB = 20\sqrt{3}\) m.
\(\cos 60^{\circ} = \frac{BC}{AC} \Rightarrow \frac{1}{2} = \frac{20}{AC} \Rightarrow AC = 40\) m.
Total Height = Broken part + Remaining part = \(AC + AB\)
\(= 40 + 20\sqrt{3}\) m.
Height of tree is \(20(2 + \sqrt{3})\) m.
The top touches ground at C, 20m from base B. \(\angle C = 60^{\circ}\).
In \(\Delta ABC\) (Right angled at B):
\(\tan 60^{\circ} = \frac{AB}{BC} \Rightarrow \sqrt{3} = \frac{AB}{20} \Rightarrow AB = 20\sqrt{3}\) m.
\(\cos 60^{\circ} = \frac{BC}{AC} \Rightarrow \frac{1}{2} = \frac{20}{AC} \Rightarrow AC = 40\) m.
Total Height = Broken part + Remaining part = \(AC + AB\)
\(= 40 + 20\sqrt{3}\) m.
Height of tree is \(20(2 + \sqrt{3})\) m.
(4) Prove that, 'In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of remaining two sides'.
Proof (Pythagoras Theorem):
Given: In \(\Delta ABC, \angle ABC = 90^\circ\).
To Prove: \(AC^2 = AB^2 + BC^2\).
Construction: Draw \(BD \perp AC\), \(A-D-C\).
Proof:
In right angled \(\Delta ABC\), \(BD \perp AC\).
\(\therefore \Delta ABC \sim \Delta ADB\) (Similarity of right angled triangles)
\(\frac{AB}{AD} = \frac{AC}{AB} \Rightarrow AB^2 = AD \times AC\) ... (I)
Similarly, \(\Delta ABC \sim \Delta BDC\)
\(\frac{BC}{DC} = \frac{AC}{BC} \Rightarrow BC^2 = DC \times AC\) ... (II)
Adding (I) and (II):
\(AB^2 + BC^2 = AD \times AC + DC \times AC\)
\(AB^2 + BC^2 = AC (AD + DC)\)
\(AB^2 + BC^2 = AC \times AC\) (Since A-D-C)
\(AB^2 + BC^2 = AC^2\)
Hence Proved.
Given: In \(\Delta ABC, \angle ABC = 90^\circ\).
To Prove: \(AC^2 = AB^2 + BC^2\).
Construction: Draw \(BD \perp AC\), \(A-D-C\).
Proof:
In right angled \(\Delta ABC\), \(BD \perp AC\).
\(\therefore \Delta ABC \sim \Delta ADB\) (Similarity of right angled triangles)
\(\frac{AB}{AD} = \frac{AC}{AB} \Rightarrow AB^2 = AD \times AC\) ... (I)
Similarly, \(\Delta ABC \sim \Delta BDC\)
\(\frac{BC}{DC} = \frac{AC}{BC} \Rightarrow BC^2 = DC \times AC\) ... (II)
Adding (I) and (II):
\(AB^2 + BC^2 = AD \times AC + DC \times AC\)
\(AB^2 + BC^2 = AC (AD + DC)\)
\(AB^2 + BC^2 = AC \times AC\) (Since A-D-C)
\(AB^2 + BC^2 = AC^2\)
Hence Proved.
Question 4
Q. 4 Solve the following subquestions (any two): (8 Marks)
(1) \(\Delta ABC\) has sides of length 4 cm, 5 cm and 6 cm while \(\Delta PQR\) has perimeter of 90 cm. If \(\Delta ABC \sim \Delta PQR\), then find the length of corresponding sides of \(\Delta PQR\).
Perimeter of \(\Delta ABC = 4 + 5 + 6 = 15\) cm.
Since \(\Delta ABC \sim \Delta PQR\), ratio of perimeters equals ratio of corresponding sides.
\(\frac{\text{Perimeter}(\Delta PQR)}{\text{Perimeter}(\Delta ABC)} = \frac{90}{15} = 6\).
Therefore, scale factor \(k = 6\).
Sides of \(\Delta PQR\):
Side 1: \(4 \times 6 = 24\) cm.
Side 2: \(5 \times 6 = 30\) cm.
Side 3: \(6 \times 6 = 36\) cm.
Sides are 24 cm, 30 cm, and 36 cm.
Since \(\Delta ABC \sim \Delta PQR\), ratio of perimeters equals ratio of corresponding sides.
\(\frac{\text{Perimeter}(\Delta PQR)}{\text{Perimeter}(\Delta ABC)} = \frac{90}{15} = 6\).
Therefore, scale factor \(k = 6\).
Sides of \(\Delta PQR\):
Side 1: \(4 \times 6 = 24\) cm.
Side 2: \(5 \times 6 = 30\) cm.
Side 3: \(6 \times 6 = 36\) cm.
Sides are 24 cm, 30 cm, and 36 cm.
(2) \(\Delta ABC \sim \Delta PBR\), \(BC=8\) cm, \(AC=10\) cm, \(\angle B=90^{\circ}\), \(\frac{BC}{BR}=\frac{5}{4}\) then construct \(\Delta PBR\).
Construction Analysis:
Vertex B is common. \(\Delta ABC\) is given. We need to construct \(\Delta PBR\).
Since \(\frac{BC}{BR} = \frac{5}{4}\), side of PBR is smaller than ABC.
Steps:
1. Draw \(\Delta ABC\) with \(BC=8, \angle B=90^{\circ}\). Using Pythagoras, \(AB = \sqrt{10^2 - 8^2} = 6\) cm. Draw AB=6, BC=8, join AC.
2. Draw a ray from B making an acute angle with BC.
3. Mark 5 equal points on the ray (\(B_1\) to \(B_5\)).
4. Join \(B_5\) to C.
5. Draw a line parallel to \(B_5C\) from \(B_4\), intersecting BC at R.
6. From R, draw a line parallel to AC intersecting AB at P.
7. \(\Delta PBR\) is the required triangle.
Vertex B is common. \(\Delta ABC\) is given. We need to construct \(\Delta PBR\).
Since \(\frac{BC}{BR} = \frac{5}{4}\), side of PBR is smaller than ABC.
Steps:
1. Draw \(\Delta ABC\) with \(BC=8, \angle B=90^{\circ}\). Using Pythagoras, \(AB = \sqrt{10^2 - 8^2} = 6\) cm. Draw AB=6, BC=8, join AC.
2. Draw a ray from B making an acute angle with BC.
3. Mark 5 equal points on the ray (\(B_1\) to \(B_5\)).
4. Join \(B_5\) to C.
5. Draw a line parallel to \(B_5C\) from \(B_4\), intersecting BC at R.
6. From R, draw a line parallel to AC intersecting AB at P.
7. \(\Delta PBR\) is the required triangle.
(3) In the figure \(\Delta ABC\) is an isosceles triangle with perimeter 44 cm. The base BC is of length 12 cm. Side AB and AC are congruent. A circle touches the three sides of triangle as shown. Find the length of tangent segment from A to circle.
Perimeter = 44 cm, Base \(BC = 12\) cm.
\(AB + AC + BC = 44\)
\(2AB + 12 = 44\) (Since \(AB \cong AC\))
\(2AB = 32 \Rightarrow AB = 16\) cm.
Let points of contact be P on AB, Q on AC, R on BC.
By Tangent Segment Theorem: \(BP = BR\), \(CR = CQ\), \(AP = AQ\).
Since \(\Delta ABC\) is isosceles and circle touches base at R, R is the midpoint of BC (Altitude of isosceles triangle is median and angle bisector).
\(\therefore BR = \frac{12}{2} = 6\) cm.
Since \(BP = BR\), then \(BP = 6\) cm.
We need to find tangent from A, which is AP.
\(A - P - B\), so \(AP = AB - BP\)
\(AP = 16 - 6 = 10\) cm.
Length of tangent segment from A is 10 cm.
\(AB + AC + BC = 44\)
\(2AB + 12 = 44\) (Since \(AB \cong AC\))
\(2AB = 32 \Rightarrow AB = 16\) cm.
Let points of contact be P on AB, Q on AC, R on BC.
By Tangent Segment Theorem: \(BP = BR\), \(CR = CQ\), \(AP = AQ\).
Since \(\Delta ABC\) is isosceles and circle touches base at R, R is the midpoint of BC (Altitude of isosceles triangle is median and angle bisector).
\(\therefore BR = \frac{12}{2} = 6\) cm.
Since \(BP = BR\), then \(BP = 6\) cm.
We need to find tangent from A, which is AP.
\(A - P - B\), so \(AP = AB - BP\)
\(AP = 16 - 6 = 10\) cm.
Length of tangent segment from A is 10 cm.
Question 5
Q. 5 Solve the following subquestions (any one): (3 Marks)
(1) Draw right-angled \(\Delta ABC\) of lengths of sides 3 cm, 4 cm and 5 cm. Draw median on the hypotenuse of \(\Delta ABC\). Measure the length of median and write your observations.
Solution:
1. Construct \(\Delta ABC\) with sides 3, 4, and hypotenuse 5.
2. Find midpoint M of hypotenuse (5 cm).
3. Draw segment joining opposite vertex (Right angle) to M.
Measurement: The median length will be 2.5 cm.
Observation: In a right-angled triangle, the length of the median drawn to the hypotenuse is exactly half the length of the hypotenuse.
1. Construct \(\Delta ABC\) with sides 3, 4, and hypotenuse 5.
2. Find midpoint M of hypotenuse (5 cm).
3. Draw segment joining opposite vertex (Right angle) to M.
Measurement: The median length will be 2.5 cm.
Observation: In a right-angled triangle, the length of the median drawn to the hypotenuse is exactly half the length of the hypotenuse.
(2) Observe the given figure of a solid cone and answer:
- How many surfaces does a solid cone have?
- What are the names of slant height and perpendicular height in the figure?
- If slant height is 10 cm and perpendicular height is 8 cm, find diameter of base?
Answers:
(i) A solid cone has 2 surfaces (1 curved surface + 1 flat circular base).
(ii) In the figure (assuming standard notation P vertex, O center, A/B on rim):
- Slant height is seg PB or seg PA.
- Perpendicular height is seg PO.
(iii) Let slant height \(l = 10\), height \(h = 8\), radius \(r\).
\(r^2 + h^2 = l^2\)
\(r^2 + 8^2 = 10^2\)
\(r^2 + 64 = 100 \Rightarrow r^2 = 36 \Rightarrow r = 6\) cm.
Diameter \(= 2r = 2 \times 6 = 12\) cm.
Diameter of base is 12 cm.
(i) A solid cone has 2 surfaces (1 curved surface + 1 flat circular base).
(ii) In the figure (assuming standard notation P vertex, O center, A/B on rim):
- Slant height is seg PB or seg PA.
- Perpendicular height is seg PO.
(iii) Let slant height \(l = 10\), height \(h = 8\), radius \(r\).
\(r^2 + h^2 = l^2\)
\(r^2 + 8^2 = 10^2\)
\(r^2 + 64 = 100 \Rightarrow r^2 = 36 \Rightarrow r = 6\) cm.
Diameter \(= 2r = 2 \times 6 = 12\) cm.
Diameter of base is 12 cm.
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