Students can access the CBSE Sample Papers for Class 9 Maths with Solutions and marking scheme Set 4 will help students understand the difficulty level of the exam.
CBSE Sample Papers for Class 9 Maths Set 4 with Solutions
Time Allowed: 3 Hours
Maximum Marks: 80
General Instructions:
- This question paper contains 38 questions. All questions are compulsory.
- The Question paper is divided into Five sections – Sections A, B, C, D, and E.
- In Section A, questions number 1 to 18 are multiple-choice questions (MCQs) and questions number 19 and 20 are Assertion-Reason based questions of 1 mark each.
- In Section B, questions number 21 to 25 are very short answer (VSA) type questions of 2 marks each.
- In Section C, questions number 26 to 31 are short answer (SA) type questions carrying 3 marks each.
- In Section D, questions number 32 to 35 are long answer (LA) type questions carrying 5 marks each.
- In Section E, questions number 36 to 38 are case-based integrated units of assessment questions carrying 4 marks each. Internal choice is provided in 2 marks questions in each case study.
- There is no overall choice. However, an internal choice has been provided in 2 questions in Section B, 2 questions in Section C, and 2 questions in Section D.
- Draw neat figures wherever required. Take π = \(\frac{22}{7}\) wherever required if not stated.
- Use of the calculator is not allowed.
Section-A
Consists of Multiple Choice Type questions of 1 mark each.
Question 1.
The value of \(\frac{\sqrt{32}+\sqrt{48}}{\sqrt{8}+\sqrt{12}}\) is equal to
(A) \(\sqrt{2}\)
(B) 2
(C) 4
(D) 8
Answer:
(B) 2
Explanation:
Question 2.
A soap manufacturer makes fragrant and non-fragrant liquid soaps. The liquid soaps are filled in plastic bottles and packed in equal size cartons for transportation. Each carton contains 50 bottles. The mass of a full bottle of soap is 220 gm and that of a half-filled bottles is 120 gm. What will be the mass (gm) of the empty bottle?
(A) 10
(B) 20
(C) 100
(D) 110
Answer:
(B) 20
Explanation:
Since,
Mass of full bottle – Mass of half-filled bottle = Mass of half-filled – Mass of empty bottle
⇒ 220 gm – 120 gm = 120 gm – Mass of empty bottle
⇒ 100 gm = 120 gm – Mass of empty bottle
⇒ Mass of empty bottle = 120 gm – 100 gm = 20 gm
Question 3.
The map shows three cities Conlen (C), Stratford (S), and Texhoma (T) on a straight highway.
Which of the following is true for the length of the highway between them?
(A) The length of the highway between C and S is equal to the length of the highway between S and T.
(B) The length of the highway between C and S is three-fourth of the length of the highway between S and T.
(C) The length of the highway between S and T is the sum of the lengths of the highway between CT and CS.
(D) The length of the highway between C and T is the sum of the lengths of the highway between CS and ST.
Answer:
(D) The length of the highway between C and T is the sum of the lengths of the highway between CS and ST.
Question 4.
The figure below shows the side view of a shopping trolley. The metal plate is fixed on the side by the store keeper for advertisement.
Three angles of the basket are obtuse. Which type of angle is the fourth?
(A) Acute
(B) Obtuse
(C) Right
(D) Reflex
Answer:
(A) Acute
Explanation:
The fourth angle of a quadrilateral should be acute.
Question 5.
In parallelogram ABCD, ∠A = (4x + 3)° and ∠D = (5x – 3)°, then the value of ∠B is
(A) 87°
(B) 97°
(C) 65°
(D) 77°
Answer:
(B) 97°
Explanation:
In ||gm ABCD,
∠A + ∠D = 180° (Adjacent angles are supplementary)
⇒ 4x + 3° + 5x – 3° = 180°
⇒ 9x = 180°
Hence, x = 20°
∠B = ∠D = 5x – 3° (Opposite angles of ||gm are equal)
So, ∠B = 5 × 20° – 3°
= 100° – 3°
= 97°
Question 6.
Given below is the map giving the position of four housing societies in a township connected by a circular Road A. Society 2 and 3 are connected by straight Road B, society 4 and 2 are connected by straight Road C and society 4 and 3 are connected by Road D. Point P denotes the position of a park. The park is equidistant to all four societies. Two new roads, Road E and Road F were constructed between society 4 and 1 and society 1 and 2.
Road G, perpendicular to Road F was constructed to connect the park and Road F. Which of the following is true for Road G and Road F?
(A) Road G and Road F are of same length.
(B) Road F divides Road G into two equal parts.
(C) Road G divides Road F into two equal parts.
(D) The length of Road G is one-fourth of the length of Road F.
Answer:
(C) Road G divides Road F into two equal parts.
Explanation:
Road G ⊥ Road F (given)
∴ Road G divides Road F in two equal parts.
(∵ Perpendicular drawn from the centre of a circle to the chord bisects the chord).
Question 7.
A charity surveys the people of a village for their haemoglobin counts. 25 out of 100 adult females in the village were tested. The result is given in this table.
Haemoglobin (mg/dl) Counts | No. of Females |
5 | 3 |
6 | 3 |
7 | 2 |
8 | 5 |
9 | 1 |
10 | 1 |
11 | 3 |
12 | 4 |
13 | 2 |
14 | 1 |
A haemoglobin count below 12 is considered deficient. What proportion of females in the survey can be considered deficient?
(A) \(\frac{3}{25}\)
(B) \(\frac{4}{25}\)
(C) \(\frac{18}{25}\)
(D) \(\frac{22}{25}\)
Answer:
(C) \(\frac{18}{25}\)
Explanation:
Total number of females = 25
Number of females considered deficient = 18
Proportion of deficient females = \(\frac{18}{25}\)
Question 8.
Points (1, -1), (2, -2), (4, -5), (-3, -4)
(A) lie in II quadrant
(B) lie in III quadrant
(C) lie in IV quadrant
(D) Do not lie in the same quadrant
Answer:
(D) Do not lie in the same quadrant
Explanation:
In points (1, -1), (2, -2) and (4, -5) x-coordinate is positive and y-coordinate is negative. So, they all lie in IVth quadrant.
In point (-3, -4), x-coordinate and y-coordinate both are negative, so it lies in III quadrant. So, all the points do not lie in same quadrant.
Question 9.
x = 5, y = 2 is a solution of the linear equation
(A) x + 2y = 7
(B) 5x + 2y = 7
(C) x + y = 7
(D) 5x + y = 7
Answer:
(C) x + y = 7
Explanation:
Given x = 5, y = 2
(A) x + 2y = 5 + 2(2) = 5 + 4 = 9
(B) 5x + 2y = 5(5) + 2(2) = 25 + 4 = 29
(C) x + y = 5 + 2 = 7
(D) 5x + y = 5(5) + 2 = 25 + 2 = 27
From above we can see only (C) x + y = 7.
Question 10.
The sum of 2√5 and 3√7 is ______
(A) 5\(\sqrt{12}\)
(B) 2√5 + 3√7
(C) 2 + 3√5 + √7
(D) 5√5 + √7
Answer:
(B) 2√5 + 3√7
Explanation:
2√5 is an irrational number.
3√7 is an irrational number.
Sum of two irrational number is also an irrational number.
∴ 2√5 + 3√7 is an answer.
Question 11.
(x + a)(x + b) = x2 + ______x + ab
(A) a + b
(B) ab
(C) a – b
(D) \(\frac{a}{b}\)
Answer:
(A) a + b
Question 12.
If the coordinates of the two points are P(-2, 3) and Q(-3, 5), then (abscissa of P) – (abscissa of Q) is
(A) -5
(B) 1
(C) -1
(D) -2
Answer:
(B) 1
Explanation:
Abscissa of P = -2
Abscissa of Q = -3
(Abscissa of P) – (Abscissa of Q) = (-2) – (-3) = -2 + 3 = 1
Question 13.
Two sides of a triangle are 13 cm and 14 cm and its semi-perimeter is 18 cm, then what will be the third side of the triangle?
(A) 9 cm
(B) 15 cm
(C) 6 cm
(D) 12 cm
Answer:
(A) 9 cm
Explanation:
Let a = 13 cm, b = 14 cm, third side = c cm
Semi perimeter, s = \(\frac{a+b+c}{2}\)
⇒ 18 = \(\frac{13+14+c}{2}\)
⇒ c = 36 – 27
⇒ c = 9 cm
Question 14.
Balan says, ‘The measure of all right angles cannot be equal as their arms can be of different lengths.’ Why is Balan’s statement not true?
(A) The measure of an angle depends upon its orientation.
(B) The measure of an angle depends upon the instrument used to measure it.
(C) The measure of an angle depends on the length of its angle arms.
(D) The measure of an angle depends upon the rotation of one arm on another.
Answer:
(D) The measure of an angle depends upon the rotation of one arm on another.
Question 15.
In the given figure, quadrilateral PQRS is cyclic. If ∠P = 80°, then ∠R is equal to ______
(A) 120°
(B) 100°
(C) 110°
(D) 90°
Answer:
(B) 100°
Explanation:
Since quadrilateral PQRS is cyclic.
∴ ∠P + ∠R = 180° (opposite angles of a cyclic quadrilateral are supplementary)
⇒ ∠80° + ∠R = 180°
⇒ ∠R = 100°
Question 16.
How can a parallelogram be formed by using paper folding?
(A) By joining any two vertices
(B) By joining one pair of opposite vertices
(C) By joining mid-points of sides of a quadrilateral
(D) None of the above
Answer:
(C) By joining mid-points of sides of a quadrilateral
Question 17.
D, E, F are the mid-points of sides BC, CA and AB of ΔABC. If perimeter of ΔABC is 12.8 cm, then perimeter of ΔDEF is
(A) 4.6 cm
(B) 6.4 cm
(C) 6.5 cm
(D) 4.2 cm
Answer:
(B) 6.4 cm
Explanation:
Given, perimeter of ΔABC = 12.8 cm
∴ Perimeter of ΔDEF = \(\frac{1}{2}\)(perimeter of ΔABC)
= \(\frac{12.8}{2}\) cm
= 6.4 cm
Question 18.
(x, y) = (y, x), if
(A) x > y
(B) x < y
(C) \(\frac{x}{y}\) = 1
(D) x = \(\frac{1}{y}\)
Answer:
(C) \(\frac{x}{y}\) = 1
Explanation:
Since, \(\frac{x}{y}\) = 1
⇒ x = y
Also, (x, y) = (y, x), if x = y
Directions: In the following questions, a statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as:
(A) Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false, but Reason (R) is true.
Question 19.
Assertion (A): In the given figure, if OA = 5 cm, AB = 8 cm and OD is perpendicular to AB, then CD is equal to 2 cm.
Reason (R): Perpendiculars bisectors of two chords of a circle intersect each other at the centre of the circle.
Answer:
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
Explanation:
In case of Assertion (A): The perpendicular drawn from the centre to a chord bisects the chord,
∴ AC = \(\frac{1}{2}\)AB
= \(\frac{1}{2}\) × 8
= 4 cm
In right-angle triangle AOC
∴ OC2 = OA2 – AC2
⇒ OC2 = 52 – 42
⇒ OC2 = 25 – 16
⇒ OC = √9
⇒ OC = 3 cm
Now, CD = OD – OC
= 5 – 3
= 2 cm
∴ The Assertion is true.
In case of Reason (R): Reason is true but not the correct explanation of Assertion.
Question 20.
Assertion (A): The point P(-6, -4) lies in quadrant III.
Reason (R): The signs of points in quadrants I, II, III, IV are respectively (+, +), (-, +), (-, -), (+, -).
Answer:
(A) Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).
Explanation:
In case of assertion (A): P(-6, -4) lies in III quadrant as the points having negative signs lies in quadrant III.
In case of reason (R): This statement is fact which is always true. Therefore, both A and R are true and R is the correct explanation of A.
Section-B
Consists of 5 questions of 2 marks each.
Question 21.
A soap manufacturer makes fragrant and non-fragrant liquid soaps. The liquid soaps are filled in plastic bottles and packed in equal size cartons for transportation. Each carton contains 50 bottles. Write an equation representing the number of fragrant and non-fragrant bottles in the carton.
Answer:
Let number of fragrant bottles = x
Number of non-fragrant bottles = y
Total number of bottles in cartoon = 50
∴ Equation is x + y = 50
Question 22.
If in ΔABC, ∠A = ∠B + ∠C, then write the shape of the given triangle.
Answer:
Given in ΔABC,
∠A = ∠B + ∠C ……(i)
By angle sum property of a triangle, we have
∠A + ∠B + ∠C = 180°
⇒ ∠A + ∠A = 180° [from equation (i)]
⇒ 2∠A = 180°
⇒ ∠A = 90°
Hence, the given triangle is a right angled triangle.
Question 23.
Five friends Anchal, Amisha, Mahi, Vaishu and Sahar are living in a hostel. At the end of every month, they calculate the expenses on food and shopping. The table given below shows their monthly expenses for the month of November.
Name | Anchal | Amisha | Mahi | Vishu | Sahar |
Expenditure (in ₹) | 3000 | 5000 | 6000 | 4500 | 7000 |
What is the average expense of the friends for the month of November?
Answer:
Total expenditure for the month of November = 3000 + 5000 + 6000 + 4500 + 7000 = ₹ 25500
Average expense = \(\frac{25500}{5}\) = ₹ 5100
Question 24.
If -2y + 3x = 14, then express y in terms of x.
OR
Express \(\frac{x}{4}\) – 3y = 7 in the form of ax + by + c = 0.
Answer:
-2y + 3x = 14
⇒ 3x – 14 = 2y
⇒ y = \(\frac{3 x-14}{2}\)
OR
\(\frac{x}{4}\) – 3y = 7
⇒ \(\frac{x-12 y}{4}\) = 7
⇒ x – 12y = 28
⇒ x – 12y – 28 = 0
Question 25.
Find the value of ‘a’ for which (x – 1) is a factor of the polynomial a2x3 – 4ax + 4a – 1.
OR
For what value of k, is the polynomial p(x) = 2x3 – kx2 + 3x + 10 exactly divisible by (x + 2)?
Answer:
Let f(x) = a2x3 – 4ax + 4a – 1
Since, (x – 1) is a factor of f(x)
Then, f(1) = 0
⇒ a2(1)3 – 4a(1) + 4a – 1 = 0
⇒ a2 – 4a + 4a – 1 = 0
⇒ a2 – 1 = 0
⇒ a = ±1
OR
Since (x + 2) is a factor of p(x).
Thus, p(-2) = 0
⇒ 2(-2)3 – k(-2)2 + 3(-2) + 10 = 0
⇒ -16 – 4k – 6 + 10 = 0
⇒ k = -3
Section-C
Consists of 6 questions of 3 marks each.
Question 26.
The area of a rectangle is (3x2 + x – 2) square units. Its width is (1 + x) units. What is the length of the rectangle?
Answer:
Area of rectangle = 3x2 + x – 2 square units.
Width of rectangle = (1 + x) units.
Question 27.
The figure below consists of a square and an equilateral triangle connected together with a common side.
In the design, DF and IG are two iron rods perpendicular to BC. The measure of ∠BAC = 120°. The length of IG is half of the length of GC. Write a proof for the statement.
Answer:
IG ⊥ BC (given)
∴ ∆IGC is a right angled triangle.
Also, ∆IGH is an equilateral triangle
∴ ∠IGH = 60°
Therefore, ∠HGC = ∠IGC – ∠IGH = 90° – 60° = 30°
Also, GH = HC
∠HGC = ∠HCG = 30° (angles opposite to equal sides are equal)
∴ ∠HCG = ∠ICG = 30°
In ∆IGC,
∠CIG + ∠ICG + ∠IGC = 180°
So, ∠CIG = 180° – 120°
∴ ∠CIG = 60°
Hence, the sides of the triangle IGC are in the ratio 2 : 1.
Question 28.
Draw a frequency polygon for the data given below, without drawing a histogram.
Classes | Frequency |
150 – 160 | 5 |
160 – 170 | 15 |
170 – 180 | 20 |
180 – 190 | 25 |
190 – 200 | 15 |
200 – 210 | 5 |
OR
The following data on the number of girls (to the nearest ten) per thousand boys in different sections of Indian Society is given below:
Sections of Society | Number of Girls Per Thousand Boys |
Schedule Caste (SC) | 940 |
Schedule Tribe (ST) | 970 |
Non-SC/ST | 920 |
Backward Districts | 950 |
Non-backward Districts | 920 |
Rural | 930 |
Urban | 910 |
Represent the information above by a bar graph.
Answer:
Frequency polygon with class marks.
Classes Marks | Frequency |
145 | 0 |
155 | 5 |
165 | 15 |
175 | 20 |
185 | 25 |
195 | 15 |
205 | 5 |
215 | 0 |
Frequency polygon showing various frequencies for the given class intervals.
OR
Question 29.
In ∆ABC, if AD is a median, then show that AB + BC > ∠AD.
OR
There is a slide in a park. One of its side walls has been painted in some colour with a message “KEEP THE PARK GREEN AND CLEAN”. If the sides of the wall are 15 m, 11 m and 6 m, find the area painted in colour.
Answer:
Construction: Produce AD to E, such that AD = DE
Now, in ∆ADB and ∆EDC, we have
BD = CD (Given AD is a median)
AD = DE (By construction)
∠1 = ∠2 (Vertically opposite angles)
∴ ∆ADB ≅ ∆EDC (By SAS rule)
Therefore, AB = CE (By CPCT) …..(i)
Now, in ∆AEC,
AC + CE > AE
AC + AB > AD + DE [from equation (i)]
AB + AC > 2AD [∵ AD = DE by construction]
Hence proved.
OR
Let a = 11 m, b = 6 m, c = 15 m
Question 30.
\(\sqrt{2}\) is an irrational number. Prove that 3 – \(\sqrt{2}\) is also an irrational number.
Answer:
\(\sqrt{2}\) is an irrational number (given)
Let us assume 3 – \(\sqrt{2}\) be a rational number ‘r’.
So, 3 – \(\sqrt{2}\) = r
3 – r = \(\sqrt{2}\)
We know that r is rational.
So, 3 – r is rational, so \(\sqrt{2}\) is also rational which contradicts the statement that \(\sqrt{2}\) is an irrational number.
∴ 3 – \(\sqrt{2}\) is irrational number.
Hence Proved.
Question 31.
If p(x) = x2 – 3x + 2, then what is the value of p(0) + p(2)?
Answer:
Putting, x = 0
p(0) = 02 – 3 × 0 + 2 = 2
Putting, x = 2
p(2) = 22 – 3 × 2 + 2 = 4 – 6 + 2 = 0
Thus, p(0) + p(2) = 2 + 0 = 2
Section-D
Consists of 4 questions of 5 marks each.
Question 32.
Simplify \(2 \sqrt[4]{81}-8 \sqrt[3]{216}+15 \sqrt[5]{32}+\sqrt{225}-\sqrt[4]{16}\)
OR
Evaluate \(\left(\frac{81}{16}\right)^{\frac{-3}{4}} \times\left[\left(\frac{9}{25}\right)^{\frac{3}{2}} \div\left(\frac{5}{2}\right)^{-3}\right]\)
Answer:
\(2 \sqrt[4]{81}-8 \sqrt[3]{216}+15 \sqrt[5]{32}+\sqrt{225}-\sqrt[4]{16}\)
= \(2\left(3^4\right)^{\frac{1}{4}}-8\left(6^3\right)^{\frac{1}{3}}+15\left(2^5\right)^{\frac{1}{5}}+15-\left(2^4\right)^{\frac{1}{4}}\)
= 2 × 3 – 8 × 6 + 15 × 2 + 15 – 2
= 6 – 48 + 30 + 15 – 2
= 51 – 50
= 1
OR
\(\left(\frac{81}{16}\right)^{\frac{-3}{4}} \times\left[\left(\frac{9}{25}\right)^{\frac{3}{2}} \div\left(\frac{5}{2}\right)^{-3}\right]\)
Question 33.
The triangular side walls of a flyover have been used for advertisements. The sides of the walls are 122 m, 22 m and 120 m. The advertisements yield an earning of ₹ 5,000 per m2 per year. A company hired one of its walls for 3 months.
(a) What is the area of the wall?
(b) How much rent did it pay?
Answer:
(i) Let a = 122 m, b = 22 m, c = 120 m
Thus, area of wall is 1320 m2.
(ii) Rent of 1 m2 area per year = ₹ 5,000
Rent of 1 m2 area per month = \(\frac{5000}{12}\)
Rent of 1320 m2 area for 3 months = ₹(\(\frac{1}{2}\) × 3 × 1320) = ₹ 1,650,000
Hence, the company had to pay ₹ 1,650,000 for three months.
Question 34.
The figure below shows an equilateral triangle bounded by two straight lines.
What is the sum of the four marked angles?
OR
If the given figure, the side QR of ΔPQR is produced to a point S. If the bisects of ∠PQR and ∠PRS meet at point T, then prove that ∠QTR = \(\frac{1}{2}\)∠QPR.
Answer:
∠1 + ∠2 + ∠3 = 180° (straight line angle)
∠2 = 60° (Angle of equilateral triangle)
∠1 + ∠3 = 180° – 60°
∴ ∠1 + ∠3 = 120° ……(i)
Similarly ∠4 + ∠5 + ∠6 = 180° (straight line angle)
∠5 = 60° (Angle of equilateral triangle)
∴ ∠4 + ∠6 = 120° …….(ii)
Thus on adding (i) and (ii) we get
∠1 + ∠3 + ∠4 + ∠6 = 120° + 120° = 240°
Hence, the sum of four marked angles = 240°
OR
In ΔPQR, QR is produced to point S,
∴ ∠PRS = ∠P + ∠PQR (exterior angle property)
So, \(\frac{1}{2}\)∠PRS = \(\frac{1}{2}\)∠P + \(\frac{1}{2}\)∠PQR
∠TRS = \(\frac{1}{2}\)∠P + ∠TQR ……(i)
(∵ QT and RT are bisectors of ∠PQR and ∠PRS respectively)
Now in ΔQTR,
∠TRS = ∠T + ∠TQR …..(ii)
(exterior angle property of a triangle)
From (i) and (ii) we have
\(\frac{1}{2}\)∠P + ∠TQR = ∠TQR + ∠T
⇒ \(\frac{1}{2}\)∠P = ∠T
Hence, \(\frac{1}{2}\)∠QPR = ∠QTR
⇒ ∠QTR = \(\frac{1}{2}\)∠QPR
Hence Proved.
Question 35.
If x – \(\frac{1}{x}\) = 2, find \(x^4+\frac{1}{x^4}\).
Answer:
On squaring both sides
Section-E
Cased-Based Subjective Questions.
Question 36.
‘Nikita has to make her project on ‘Monument in India’. She decided to make her project on Gol Gumbaz monument. She already knows following things about it:
It is located in a small town in Northern Karnataka.
It reaches up to 51 meters in height while the giant dome has an external diameter of 44 meters, making it one of the largest domes ever built.
At each of the four corners of the cube is a dome-shaped octagonal tower seven stories high with a staircase inside.
Help her in making project by answering the following questions:
(i) Which mathematical concept is be used here?
(ii) What is the curved surface area of hemispherical dome?
OR
What is the circumference of the base of the dome?
(iii) Find the cost of painting the dome, given the cost of painting is ₹ 100 per cm2.
Answer:
(i) Mathematical concept used here is volume and surface area of hemisphere.
(ii) Diameter = 44 m
Radius = 22 m
Curved surface area of hemispherical dome = 2πr2
= 2π(22)2
= 968π m2
OR
Circumference of the base of the dome = 2πr
= 2π(22)
= 44π m
(iii) Curved surface area of hemispherical dome = 968π m2
Cost of painting 100 cm2 = ₹ 10
Cost of painting 1 m2 = ₹ 1000
Thus, cost of painting the dome = ₹ 1000 × 968π cm2 = ₹ 968000π cm2
Question 37.
Three STD booth are placed at A, B and C in the figure and these are operated by handicapped persons. These three booth are equidistant from each other as shown in figure given below.
(i) What is the value of ∠BAC?
(ii) What will be the value of ∠OBC?
OR
What will be the value of ∠OBC?
(iii) Which angle will be equal to ∠OBC?
Answer:
(i) As AB = BC = CA (given)
So, ΔABC is an equilateral triangle
∴ ∠BAC = 60°
(∵ All angles of an equilateral triangle are equal)
(ii) ∠BOC = 2∠BAC
= 2 × 60° (∵ ∠BAC = 60°, as ΔABC is equilateral triangle)
= 120°
(∵ Anlge subtended by an arc at the centre is double the angle subtended by it any point on the remaining part of the circle.)
OR
OB = OC (equal radii)
∴ ∠OBC = ∠OCB = x° (Angle opposite to equal side are equal)
In ΔOBC
∠OBC + ∠OCB + ∠BOC = 180°
⇒ x + x + 120° = 180°
⇒ 2x = 60°
⇒ x = 30°
∴ ∠OBC = 30°
(iii) OB = OC (equal radii)
∴ ∠OBC = ∠OCB (Angle opposite to equal side)
Question 38.
Neeraj has a plot in the shape of a triangle said ABC with AD as the perpendicular bisector of BC such that BD = DC.
(i) Which rule is applied to prove the congruency of ∆ABD and ∆ACD?
(ii) If AB = 10 cm and BD = 6 cm, then find perimeter of ∆ABD?
OR
Find the perimeter of ∆ABC?
(iii) In ∆ADC, find ∠ACD if given ∠BAC = 80° and AD is the angle bisector of ∠ABC.
Answer:
In ∆ABD and ∆ACD
BD = CD (given)
∠ADC = ∠ADC = 90° (AD is altitude)
AD = AD (common)
∴ ∆ABD ≅ ∆ACD (by SAS)
(ii) ∆ABD is a right angled triangle
AB = 10 cm, BD = 6 cm
Using Pythagoras Theorem
(AB)2 = (AD)2 + (BA)2
⇒ (10)2 = (AD)2 + (6)2
⇒ 100 = (AD)2 + 36
⇒ (AD)2 = 100 – 36
⇒ (AD)2 = 64
⇒ AD = 8 cm
Now, perimeter = AB + AD + BD
= (10 + 8 + 6) cm
= 24 cm
OR
As ∆ABD ≅ ∆ACD
∴ AB = AC (Corresponding parts of congruent triangles)
AC = 10 cm
BC = BD = DC
Since BD = DC = 6 cm (given)
∴ BC = (6 + 6) cm = 12 cm
Perimeter of ∆ABC = AB + BC + AC
= (10 + 12 + 10) cm
= 32 cm
(iii) ∠BAD = ∠CAD = 40° (AD is the angle bisector of ∠BAC)
In ∆ADC
∠ACD + ∠ADC + ∠CAD = 180° (angle sum property)
⇒ ∠ACD + 90° + 40° = 180°
⇒ ∠ACD + 130° = 180°
⇒ ∠ACD = 180° – 130°
⇒ ∠ACD = 50°