In the given figure below A D A E = A C B D and ∠ 1 = ∠ 2 . Show that △ B A E ~ △ C A D

Given, A D A E = A C B D and ∠ 1 = ∠ 2 . We have to prove △ B A E ~ △ C A D △ B A E ~ △ C A D B D = A B . . . . . . . . . . . . . ( 1 ) [ Side opposite to equal angles are equal] A D A E = A D B D Given A D A E = A D A B . . . . . . . . . . . . . . . . . ( 2 ) In △ B A E and △ C A D ∠ B A E = ∠ C A D [ Both angle common] a n d A D A E = A C A B [from (2)] ∴ △ B A E ~ △ C A D (By SAS similarity)

If sin A + B = 1 and cos A - B = 3 2 , 0 ° < A + B ≤ 90 ° and A > B , then find the measures of angles A and B .

Given that sin A + B = 1 . . . . 1 cos A - B = 3 2 . . . . 2 0 ° < A + B ≤ 90 ° A > B From equation 1 sin A + B = sin 90 ° ⇒ A + B = 90 ° . . . . 3 From equation 2 cos A - B = cos 30 ° ⇒ A - B = 30 ° . . . . 4 On Adding 3 and 4 , we get 2 A = 120 ° A = 60 ° On putting the value of A in equation 3 , we get B = 90 ° - A B = 90 ° - 60 ° B = 30 ° Hence, A = 60 ° , B = 30 °

Find an acute angle θ when, cos θ - sin θ cos θ + sin θ = 1 - 3 1 + 3 .

Given, cos θ - sin θ cos θ + sin θ = 1 - 3 1 + 3 Taking LHS and dividing numerator and denominator by cos θ , we get, 1 - sin θ cos θ 1 + sin θ cos θ = 1 - tan θ 1 + tan θ = 1 - 3 1 + 3 On comparing both the sides, we get, tan θ = 3 ⇒ θ = 60 ° Hence, the θ is 60 ° .

The median of the following data is 525 . Find the values of x and y , if the total frequency is 100 . Class Interval Frequency 0 - 100 2 100 - 200 5 200 - 300 x 300 - 400 12 400 - 500 17 500 - 600 20 600 - 700 y 700 - 800 9 800 - 900 7 900 - 1000 4

Cumulative frequency for the given data; Class interval Frequency f Cumulative frequency CF 0 - 100 2 2 100 - 200 5 7 200 - 300 x x + 7 300 - 400 12 x + 19 400 - 500 17 x + 36 500 - 600 20 x + 56 600 - 700 y x + y + 56 700 - 800 9 x + y + 65 800 - 900 7 x + y + 72 900 - 1000 4 x + y + 76 total Σ f = 76 + x + y = 100 Given that the median is 525 . So median class is 500 - 600 Now median = l + N 2 - C F f × h where l = 500 , h = 100 , N = 100 , C F = 36 + x , f = 20 so, 525 = 500 + 100 2 - ( 36 + x ) 20 × 100 ⇒ 525 - 500 = ( 50 - ( 36 + x ) ) × 5 ⇒ 25 = ( 14 - x ) × 5 ⇒ x = 9 Also, Σ f = 76 + x + y = 100 ⇒ 9 + y = 100 - 76 ⇒ y = 15 Hence, x = 9 and y = 15

In a flight of 600 km , an aircraft was slowed due to bad weather. Its average speed for the trip was reduced by 200 km / hr and time of flight increased by 30 minutes. Find the original duration of flight.

Let the original duration is x hours & original speed is y km / h . Then, x × y = 600 .....(1) The aircraft was slowed due to bad weather. Its average speed is reduced by 200 km / h and time of flight increased by 30 minutes or 1 2 h . Then, x + 1 2 ( y - 200 ) = 600 .....(2) From equation (1) and (2), we have x + 1 2 ( y - 200 ) = x y ⇒ x y - 200 x + y 2 - 100 = x y ⇒ 200 x - y 2 + 100 = 0 From equation (1), put y = 600 x then, we have ⇒ 200 x - 600 2 x + 100 = 0 ⇒ 200 x - 300 x + 100 = 0 ⇒ 2 x - 3 x + 1 = 0 ⇒ 2 x 2 - 3 + x = 0 ⇒ 2 x 2 + x - 3 = 0 From middle term factorization, we have ⇒ 2 x 2 - 2 x + 3 x - 3 = 0 ⇒ ( x - 1 ) ( 2 x + 3 ) = 0 ⇒ x = 1 , - 3 2 Here, we consider x = 1 hour . Hence, the original duration of the flight is 1 hour .

E M B I B E

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 15 - Sample Paper 1>Exercise>Q 19

EASY

10th Uttar Pradesh Board

IMPORTANT

Earn 100

Sum of powers of prime factors of $196$ is

100% studentsanswered this correctly

Important Questions on Sample Paper 1

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10th Uttar Pradesh Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 15 - Sample Paper 1>Exercise>Q 20

Statement A (Assertion): If the co-ordinates of the mid-points of the sides $A B$ and $A C$ of $△ A B C$ are $D (3, 5)$ and $E (- 3, - 3)$ respectively, then $B C = 20$ units

Statement R (Reason): The line joining the mid-points of two sides of a triangle is parallel to the third side and equal to half of it.

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IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 15 - Sample Paper 1>Exercise>Q 21

If

49 x + 51 y = 499, 51 x + 49 y = 501

, then find the value of

x and y

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10th Uttar Pradesh Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 15 - Sample Paper 1>Exercise>Q 22

In the given figure below $\frac{A D}{A E} = \frac{A C}{B D}$ and $∠ 1 = ∠ 2$ . Show that $△ B A E ~ △ C A D$

Question Image

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10th Uttar Pradesh Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 15 - Sample Paper 1>Exercise>Q 23

In the given figure, $O$ is the centre of circle. Find $∠ A Q B$ , given that $P A and P B$ are tangents to the circle and $∠ A P B = 75 °$ .

Question Image

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IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 15 - Sample Paper 1>Exercise>Q 24

The length of the minute hand of a clock is $6 c m$ . Find the area swept by it when it moves from $7 : 05 p . m .$ to $7 : 40 p . m .$

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10th Uttar Pradesh Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 15 - Sample Paper 1>Exercise>Q 25

In the given figure, arcs have been drawn of radius $7 cm$ each with vertices $A, B, C$ and $D$ of quadrilateral $A B C D$ as centres. Find the area of the shaded region (in ${cm}^{2}$ ).